Mixing
mysqli_query does not return result
-1
I code php to get returned from mysql.
My query works in mysql workbench, however it doesn't work in php.
sql below is returned well in mysql workbench.
"select * from players_info where name = XXXX and sex = X and season = xxxx and tName = XXXX"
My php is below
function player_info($conn, $name, $season, $sex, $tName){
$output = '';
$sql = "SELECT *
FROM players_info
WHERE name ='".$name."' and sex ='".$sex."'
and season = '".$season."' and tName = '".$tName."'";
$result = mysqli_query($conn, $sql);
if($result -> num_rows > 0){
while($row = mysqli_fetch_assoc($result)){
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
}
} else {
echo "no result!";
}
$conn -> close();
return $output;
}
HTML is below.
<div id = "player_page3">
<?php
echo player_info($conn, $_GET["name"], $_GET["season"], $_GET["sex"], $_GET["tName"]);
?>
</div>
I think my code returns 0 result even though sql is correct.
What is the problem on my code?
UPDATE :
I added code shown below after $sql.
I think it returns nothing.
result is "Error description : no result!"
'no result!' came from the else statement in my code.
if (!mysqli_query($conn, $sql)){
echo("Error description : ". mysqli_error($conn));
}
php mysql function get
asked Jan 1 at 15:27
sk Jinsk Jin
205
|
show 3 more comments
-1
I code php to get returned from mysql.
My query works in mysql workbench, however it doesn't work in php.
sql below is returned well in mysql workbench.
"select * from players_info where name = XXXX and sex = X and season = xxxx and tName = XXXX"
My php is below
function player_info($conn, $name, $season, $sex, $tName){
$output = '';
$sql = "SELECT *
FROM players_info
WHERE name ='".$name."' and sex ='".$sex."'
and season = '".$season."' and tName = '".$tName."'";
$result = mysqli_query($conn, $sql);
if($result -> num_rows > 0){
while($row = mysqli_fetch_assoc($result)){
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
}
} else {
echo "no result!";
}
$conn -> close();
return $output;
}
HTML is below.
<div id = "player_page3">
<?php
echo player_info($conn, $_GET["name"], $_GET["season"], $_GET["sex"], $_GET["tName"]);
?>
</div>
I think my code returns 0 result even though sql is correct.
What is the problem on my code?
UPDATE :
I added code shown below after $sql.
I think it returns nothing.
result is "Error description : no result!"
'no result!' came from the else statement in my code.
if (!mysqli_query($conn, $sql)){
echo("Error description : ". mysqli_error($conn));
}
php mysql function get
asked Jan 1 at 15:27
sk Jinsk Jin
205
You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Usemysqli_error()to get a detailed error message from the database.
– John Conde
Jan 1 at 15:30
5
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
Where do you create the$conn? and how?
– dWinder
Jan 1 at 15:35
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38
|
show 3 more comments
-1
-1
-1
I code php to get returned from mysql.
My query works in mysql workbench, however it doesn't work in php.
sql below is returned well in mysql workbench.
"select * from players_info where name = XXXX and sex = X and season = xxxx and tName = XXXX"
My php is below
function player_info($conn, $name, $season, $sex, $tName){
$output = '';
$sql = "SELECT *
FROM players_info
WHERE name ='".$name."' and sex ='".$sex."'
and season = '".$season."' and tName = '".$tName."'";
$result = mysqli_query($conn, $sql);
if($result -> num_rows > 0){
while($row = mysqli_fetch_assoc($result)){
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
}
} else {
echo "no result!";
}
$conn -> close();
return $output;
}
HTML is below.
<div id = "player_page3">
<?php
echo player_info($conn, $_GET["name"], $_GET["season"], $_GET["sex"], $_GET["tName"]);
?>
</div>
I think my code returns 0 result even though sql is correct.
What is the problem on my code?
UPDATE :
I added code shown below after $sql.
I think it returns nothing.
result is "Error description : no result!"
'no result!' came from the else statement in my code.
if (!mysqli_query($conn, $sql)){
echo("Error description : ". mysqli_error($conn));
}
php mysql function get
asked Jan 1 at 15:27
sk Jinsk Jin
205
I code php to get returned from mysql.
My query works in mysql workbench, however it doesn't work in php.
sql below is returned well in mysql workbench.
"select * from players_info where name = XXXX and sex = X and season = xxxx and tName = XXXX"
My php is below
function player_info($conn, $name, $season, $sex, $tName){
$output = '';
$sql = "SELECT *
FROM players_info
WHERE name ='".$name."' and sex ='".$sex."'
and season = '".$season."' and tName = '".$tName."'";
$result = mysqli_query($conn, $sql);
if($result -> num_rows > 0){
while($row = mysqli_fetch_assoc($result)){
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
}
} else {
echo "no result!";
}
$conn -> close();
return $output;
}
HTML is below.
<div id = "player_page3">
<?php
echo player_info($conn, $_GET["name"], $_GET["season"], $_GET["sex"], $_GET["tName"]);
?>
</div>
I think my code returns 0 result even though sql is correct.
What is the problem on my code?
UPDATE :
I added code shown below after $sql.
I think it returns nothing.
result is "Error description : no result!"
'no result!' came from the else statement in my code.
if (!mysqli_query($conn, $sql)){
echo("Error description : ". mysqli_error($conn));
}
php mysql function get
php mysql function get
asked Jan 1 at 15:27
sk Jinsk Jin
205
asked Jan 1 at 15:27
sk Jinsk Jin
205
edited Jan 1 at 16:29
sk Jin
asked Jan 1 at 15:27
sk Jinsk Jin
205
asked Jan 1 at 15:27
sk Jinsk Jin
205
asked Jan 1 at 15:27
sk Jinsk Jin
205
205
You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Usemysqli_error()to get a detailed error message from the database.
– John Conde
Jan 1 at 15:30
5
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
Where do you create the$conn? and how?
– dWinder
Jan 1 at 15:35
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38
|
show 3 more comments
You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Usemysqli_error()to get a detailed error message from the database.
– John Conde
Jan 1 at 15:30
5
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
Where do you create the$conn? and how?
– dWinder
Jan 1 at 15:35
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38
You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Use
mysqli_error() to get a detailed error message from the database.– John Conde
Jan 1 at 15:30
You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Use
mysqli_error() to get a detailed error message from the database.– John Conde
Jan 1 at 15:30
5
5
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
Where do you create the
$conn? and how?– dWinder
Jan 1 at 15:35
Where do you create the
$conn? and how?– dWinder
Jan 1 at 15:35
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38
|
show 3 more comments
1 Answer
1
active
oldest
votes
0
First of all - as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects.
Here are some points to note:
Your $output variable is not collecting the result because it is overwriting itself over the loop. You need to change that to .=
$output .= $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
.= concatenates new text to the end.
UPDATE
If you want a single row you need to remove the while loop
if($result -> num_rows > 0){
$row = mysqli_fetch_assoc($result)
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
} else {
echo "no result!";
}
answered Jan 1 at 15:40
nazimnazim
8831023
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
First of all - as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects.
Here are some points to note:
Your $output variable is not collecting the result because it is overwriting itself over the loop. You need to change that to .=
$output .= $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
.= concatenates new text to the end.
UPDATE
If you want a single row you need to remove the while loop
if($result -> num_rows > 0){
$row = mysqli_fetch_assoc($result)
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
} else {
echo "no result!";
}
answered Jan 1 at 15:40
nazimnazim
8831023
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
add a comment |
0
First of all - as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects.
Here are some points to note:
Your $output variable is not collecting the result because it is overwriting itself over the loop. You need to change that to .=
$output .= $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
.= concatenates new text to the end.
UPDATE
If you want a single row you need to remove the while loop
if($result -> num_rows > 0){
$row = mysqli_fetch_assoc($result)
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
} else {
echo "no result!";
}
answered Jan 1 at 15:40
nazimnazim
8831023
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
add a comment |
0
0
0
First of all - as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects.
Here are some points to note:
Your $output variable is not collecting the result because it is overwriting itself over the loop. You need to change that to .=
$output .= $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
.= concatenates new text to the end.
UPDATE
If you want a single row you need to remove the while loop
if($result -> num_rows > 0){
$row = mysqli_fetch_assoc($result)
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
} else {
echo "no result!";
}
answered Jan 1 at 15:40
nazimnazim
8831023
First of all - as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects.
Here are some points to note:
Your $output variable is not collecting the result because it is overwriting itself over the loop. You need to change that to .=
$output .= $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
.= concatenates new text to the end.
UPDATE
If you want a single row you need to remove the while loop
if($result -> num_rows > 0){
$row = mysqli_fetch_assoc($result)
$output = $row['name']."n"
."No : ".$row['No']."n"
."Pos : ".$row['pos']."n"
."Height(m) / Weight(kg) : ".$row['height']. " / ".$row['weight']."n"
."Born : ".$row['month']." ".$row['day'].", ".$row['year']."n"
."Shoots / Catches : ".$row['shoots']."n"
."Club : ".$row['team'];
} else {
echo "no result!";
}
answered Jan 1 at 15:40
nazimnazim
8831023
edited Jan 1 at 16:04
answered Jan 1 at 15:40
nazimnazim
8831023
answered Jan 1 at 15:40
nazimnazim
8831023
answered Jan 1 at 15:40
nazimnazim
8831023
8831023
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
add a comment |
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Actually, the result is one row. however I don't know how to return just one row. So I used like that. I mean that is not a big problem so far. How can I change if I hope to result just one row?
– sk Jin
Jan 1 at 15:55
Check the update
– nazim
Jan 1 at 15:59
Check the update
– nazim
Jan 1 at 15:59
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
and BTW as alerted in the comments you need to learn to prepare your sql statements. Use PDO - Your current script IS VULNERABLE to sql injects
– nazim
Jan 1 at 16:00
add a comment |
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You don't know what's wrong because you don't check for errors in your code. Never assume the code is always going to work flawlessly. Use
mysqli_error()to get a detailed error message from the database.– John Conde
Jan 1 at 15:30
5
Your script is at risk of SQL Injection Attack Have a look at what happened to Little Bobby Tables Even if you are escaping inputs, its not safe! Use prepared parameterized statements.
– John Conde
Jan 1 at 15:30
why dont you use the mysqli_error like this . $result = mysqli_query($conn, $sql) or die(mysqli_error($conn)); you wil know the error.
– Anupam Rekha
Jan 1 at 15:33
Where do you create the
$conn? and how?– dWinder
Jan 1 at 15:35
No idea where and how you're using the GET arrays and the return.
– Funk Forty Niner
Jan 1 at 15:38