Mixing
resource id #4 Why am I getting this?
4
I have a pretty basic forum template I am working on for testing purposes
When I create a topic, and press submit, the proccess updates the database but doesn't output on the screen. Why is this? and Why am I getting a Resource id #4 when I echo the $result from this code below:
<?php
$host="server"; // Host name
$username="usernamehere"; // Mysql username
$password=""; // Mysql password
$db_name="forum"; // Database name
$tbl_name="question"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result;
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>
<?php
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
php mysql
edited Nov 2 '12 at 5:41
Sami
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
add a comment |
4
I have a pretty basic forum template I am working on for testing purposes
When I create a topic, and press submit, the proccess updates the database but doesn't output on the screen. Why is this? and Why am I getting a Resource id #4 when I echo the $result from this code below:
<?php
$host="server"; // Host name
$username="usernamehere"; // Mysql username
$password=""; // Mysql password
$db_name="forum"; // Database name
$tbl_name="question"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result;
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>
<?php
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
php mysql
edited Nov 2 '12 at 5:41
Sami
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
1
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07
add a comment |
4
4
4
I have a pretty basic forum template I am working on for testing purposes
When I create a topic, and press submit, the proccess updates the database but doesn't output on the screen. Why is this? and Why am I getting a Resource id #4 when I echo the $result from this code below:
<?php
$host="server"; // Host name
$username="usernamehere"; // Mysql username
$password=""; // Mysql password
$db_name="forum"; // Database name
$tbl_name="question"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result;
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>
<?php
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
php mysql
edited Nov 2 '12 at 5:41
Sami
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
I have a pretty basic forum template I am working on for testing purposes
When I create a topic, and press submit, the proccess updates the database but doesn't output on the screen. Why is this? and Why am I getting a Resource id #4 when I echo the $result from this code below:
<?php
$host="server"; // Host name
$username="usernamehere"; // Mysql username
$password=""; // Mysql password
$db_name="forum"; // Database name
$tbl_name="question"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result;
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>
<?php
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
php mysql
php mysql
edited Nov 2 '12 at 5:41
Sami
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
edited Nov 2 '12 at 5:41
Sami
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
edited Nov 2 '12 at 5:41
Sami
5,20245084
edited Nov 2 '12 at 5:41
Sami
5,20245084
edited Nov 2 '12 at 5:41
Sami
5,20245084
5,20245084
asked Nov 2 '12 at 5:22
noob123noob123
23114
asked Nov 2 '12 at 5:22
noob123noob123
23114
asked Nov 2 '12 at 5:22
noob123noob123
23114
23114
because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
1
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07
add a comment |
because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
1
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07
because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
1
1
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07
add a comment |
5 Answers
5
active
oldest
votes
9
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
Update 2(From OP comments)
You are printing values using field name,In that case you will have to change to
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
|
show 1 more comment
4
Problem is in your code:
$result=mysql_query($sql);
echo $result;
$result is resource type, since mysql_query($sql) returns resource
Stop echoing $result.
edited Nov 2 '12 at 5:38
jogojapan
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
add a comment |
2
If you check the link - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
Hence you are seeing the Resource#4
. What is it you want to achieve?
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
add a comment |
0
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
add a comment |
-2
You don't have to use mysql_fetch_array(). If you want, try something like this:
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a$numvariable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.
– b.enoit.be
Apr 5 '15 at 23:23
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
9
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
Update 2(From OP comments)
You are printing values using field name,In that case you will have to change to
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
|
show 1 more comment
9
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
Update 2(From OP comments)
You are printing values using field name,In that case you will have to change to
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
|
show 1 more comment
9
9
9
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
Update 2(From OP comments)
You are printing values using field name,In that case you will have to change to
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
You are getting resource id #4 because $result is an resource,you must extract the values contained in it by this way,
$result=mysql_query($sql);
$values = mysql_fetch_array($result);
var_dump($values);
More about resource variable
Update 2(From OP comments)
You are printing values using field name,In that case you will have to change to
while($rows=mysql_fetch_array($result,MYSQL_ASSOC))
Or you can directly use mysql_fetch_assoc(),which in your case will be
while($rows=mysql_fetch_assoc($result)){
echo $rows['id'];
}
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
edited Nov 2 '12 at 6:40
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
answered Nov 2 '12 at 5:25
SibuSibu
4,06211836
4,06211836
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
|
show 1 more comment
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
ok, my data shows in the var_dump but it won't show in the table in the above code. Why is this if the information is being passed?
– noob123
Nov 2 '12 at 5:45
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
have you enabled short tags on your server, replace <? with <?php and try
– Sibu
Nov 2 '12 at 5:49
I still get the same result..
– noob123
Nov 2 '12 at 6:31
I still get the same result..
– noob123
Nov 2 '12 at 6:31
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
@noob123 my mistake, i should have seen that earlier, check my updated answer
– Sibu
Nov 2 '12 at 6:40
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
I used the mysql_fetch_assoc in the code above but it doesnt work..I don't know what I'm doing wrong..
– noob123
Nov 2 '12 at 14:50
|
show 1 more comment
4
Problem is in your code:
$result=mysql_query($sql);
echo $result;
$result is resource type, since mysql_query($sql) returns resource
Stop echoing $result.
edited Nov 2 '12 at 5:38
jogojapan
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
add a comment |
4
Problem is in your code:
$result=mysql_query($sql);
echo $result;
$result is resource type, since mysql_query($sql) returns resource
Stop echoing $result.
edited Nov 2 '12 at 5:38
jogojapan
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
add a comment |
4
4
4
Problem is in your code:
$result=mysql_query($sql);
echo $result;
$result is resource type, since mysql_query($sql) returns resource
Stop echoing $result.
edited Nov 2 '12 at 5:38
jogojapan
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
Problem is in your code:
$result=mysql_query($sql);
echo $result;
$result is resource type, since mysql_query($sql) returns resource
Stop echoing $result.
edited Nov 2 '12 at 5:38
jogojapan
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
edited Nov 2 '12 at 5:38
jogojapan
54k977112
edited Nov 2 '12 at 5:38
jogojapan
54k977112
edited Nov 2 '12 at 5:38
jogojapan
54k977112
54k977112
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
answered Nov 2 '12 at 5:30
ujjwalwahiujjwalwahi
127313
127313
add a comment |
add a comment |
2
If you check the link - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
Hence you are seeing the Resource#4
. What is it you want to achieve?
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
add a comment |
2
If you check the link - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
Hence you are seeing the Resource#4
. What is it you want to achieve?
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
add a comment |
2
2
2
If you check the link - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
Hence you are seeing the Resource#4
. What is it you want to achieve?
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
If you check the link - http://php.net/manual/en/function.mysql-query.php
For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error
Hence you are seeing the Resource#4
. What is it you want to achieve?
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
answered Nov 2 '12 at 5:28
JoddyJoddy
1,80511420
1,80511420
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
add a comment |
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
I'm simply trying to output the data in the database into the table in the above code
– noob123
Nov 2 '12 at 5:40
add a comment |
0
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
add a comment |
0
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
add a comment |
0
0
0
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
<?php
$host="server";
$username="usernamehere";
$password="";
$db_name="forum";
$tbl_name="question";
mysql_connect("$host", "$username", "")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
echo $result; //remove it
?>
<html>
<body>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>
<?php
// Start looping table row
while($rows=mysql_fetch_array($result)){
echo"<tr>
<td bgcolor=#FFFFFF>$rows['id']</td>
<td bgcolor=#FFFFFF><a href='view_topic.php?id=$rows['id']'>$rows['topic']</a><BR></td>
<td align=center bgcolor=#FFFFFF>$rows['view']</td>
<td align=center bgcolor=#FFFFFF>$rows['reply']</td>
<td align=center bgcolor=#FFFFFF>$rows['datetime'];</td>
</tr>";
// Exit looping and close connection
}
mysql_close();
?>
<tr>
<td colspan="5" align="right" bgcolor="#E6E6E6"><a href="create_topic.php"><strong>Create New Topic</strong> </a></td>
</tr>
</table>
</body>
</html>
just check above code i think your problem should be done
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
edited Aug 18 '17 at 11:39
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
answered Aug 18 '17 at 11:11
Mustaqeem KhanMustaqeem Khan
11
11
add a comment |
add a comment |
-2
You don't have to use mysql_fetch_array(). If you want, try something like this:
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a$numvariable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.
– b.enoit.be
Apr 5 '15 at 23:23
add a comment |
-2
You don't have to use mysql_fetch_array(). If you want, try something like this:
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a$numvariable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.
– b.enoit.be
Apr 5 '15 at 23:23
add a comment |
-2
-2
-2
You don't have to use mysql_fetch_array(). If you want, try something like this:
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
You don't have to use mysql_fetch_array(). If you want, try something like this:
$sql="SELECT * FROM $tbl_name ORDER BY id DESC"; //that's your query
$result=mysql_query($sql);
$rows=mysql_num_rows($result);
$iteration=0;
echo "<table>";
while($iteration < $rows){
$cell_in_your_html_table = mysql_result($result , $iteration , 'column_name_from_database');
echo "<tr><td>".$cell_in_your_html_table."</td></tr>";
$iteration++;
}
echo "</table>"
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
edited Apr 6 '15 at 16:22
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
answered Apr 5 '15 at 23:02
oxy_moronoxy_moron
11
11
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a$numvariable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.
– b.enoit.be
Apr 5 '15 at 23:23
add a comment |
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a$numvariable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.
– b.enoit.be
Apr 5 '15 at 23:23
1
1
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a
$num variable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.– b.enoit.be
Apr 5 '15 at 23:23
Welcome to Stack Overflow. Please use code block for your code snippet in answers. Please double check your answers (yours does clearly have a
$num variable being null which make your code useless). Please use relevant and readable name for your variable (tada ? Is that the result of an magical trick ?). Please make sure you are actually replying the OP question.– b.enoit.be
Apr 5 '15 at 23:23
add a comment |
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because that's what $result is, a resource.
– user557846
Nov 2 '12 at 5:26
1
Just remove echo $result
– Svetoslav
Nov 2 '12 at 5:27
shouldn't the while loop take care of the resource #4 error? and if I just remove the echo $result..I'm still left with my data not being displayed in the table..
– noob123
Nov 2 '12 at 6:07