Mixing
Displaying Not Found Data From MySQL
I wrote simple code to fetch data from MySQL using PHP.
This is the code:
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
{
echo "<p>";
echo "$row[id_kendaraan]";
echo "<p>";
echo "$row[mangkal]";
}
?>
The script is working, if the data is returned by the db call, I can see the results displayed. But, if the query result has no data, the script just displays blank. I want to show a message that says - 'Data not found'. How can I do that?
I have more than one record for a query but the script displays just one data. Please help me to show all records.
php mysqli
edited Jan 2 at 14:56
Funk Forty Niner
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
add a comment |
I wrote simple code to fetch data from MySQL using PHP.
This is the code:
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
{
echo "<p>";
echo "$row[id_kendaraan]";
echo "<p>";
echo "$row[mangkal]";
}
?>
The script is working, if the data is returned by the db call, I can see the results displayed. But, if the query result has no data, the script just displays blank. I want to show a message that says - 'Data not found'. How can I do that?
I have more than one record for a query but the script displays just one data. Please help me to show all records.
php mysqli
edited Jan 2 at 14:56
Funk Forty Niner
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
1
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
Check the answer below :)
– Soheyl
Jan 2 at 12:07
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46
add a comment |
I wrote simple code to fetch data from MySQL using PHP.
This is the code:
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
{
echo "<p>";
echo "$row[id_kendaraan]";
echo "<p>";
echo "$row[mangkal]";
}
?>
The script is working, if the data is returned by the db call, I can see the results displayed. But, if the query result has no data, the script just displays blank. I want to show a message that says - 'Data not found'. How can I do that?
I have more than one record for a query but the script displays just one data. Please help me to show all records.
php mysqli
edited Jan 2 at 14:56
Funk Forty Niner
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
I wrote simple code to fetch data from MySQL using PHP.
This is the code:
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
$row = $result->fetch_array(MYSQLI_BOTH);
{
echo "<p>";
echo "$row[id_kendaraan]";
echo "<p>";
echo "$row[mangkal]";
}
?>
The script is working, if the data is returned by the db call, I can see the results displayed. But, if the query result has no data, the script just displays blank. I want to show a message that says - 'Data not found'. How can I do that?
I have more than one record for a query but the script displays just one data. Please help me to show all records.
php mysqli
php mysqli
edited Jan 2 at 14:56
Funk Forty Niner
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
edited Jan 2 at 14:56
Funk Forty Niner
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
edited Jan 2 at 14:56
Funk Forty Niner
1
edited Jan 2 at 14:56
Funk Forty Niner
1
edited Jan 2 at 14:56
Funk Forty Niner
1
1
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
asked Jan 2 at 11:51
Ahmad ZAhmad Z
124
124
1
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
Check the answer below :)
– Soheyl
Jan 2 at 12:07
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46
add a comment |
1
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
Check the answer below :)
– Soheyl
Jan 2 at 12:07
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46
1
1
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
Check the answer below :)
– Soheyl
Jan 2 at 12:07
Check the answer below :)
– Soheyl
Jan 2 at 12:07
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46
add a comment |
2 Answers
2
active
oldest
votes
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
add a comment |
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
answered Jan 2 at 12:04
dexterdexter
1,28311020
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
add a comment |
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
add a comment |
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
You need to run a loop to iterate through each result. Something similar to this -
<?php
$mangkal = $_POST['mangkal'];
$lat = $_POST['lat'];
$long = $_POST['long'];
$mysqli = new mysqli("localhost", "root", "", "mad");
$query = "SELECT * FROM kendaraan WHERE mangkal LIKE '%$mangkal%' ORDER BY id DESC";
$result = $mysqli->query($query);
while($row = $result->fetch_array(MYSQLI_BOTH)) {
echo "<p>";
echo "$row[id_kendaraan]";
echo "</p><p>";
echo "$row[mangkal]";
echo "</p>";
}
?>
In the code above, a while loop is used to iterate through results one by one. Each time $row will be updated with new data and will be printed.
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
answered Jan 2 at 12:08
Ravi Kumar GuptaRavi Kumar Gupta
7381029
7381029
add a comment |
add a comment |
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
answered Jan 2 at 12:04
dexterdexter
1,28311020
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
add a comment |
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
answered Jan 2 at 12:04
dexterdexter
1,28311020
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
add a comment |
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
answered Jan 2 at 12:04
dexterdexter
1,28311020
Use a while loop like following:
if($result->num_rows > 0)
{
while($row = $result->fetch_array(MYSQLI_BOTH))
{
echo "<p>".$row[id_kendaraan]."</p><br><p>".$row[mangkal]."</p>";
}
} else {
echo "No Record Found.";
}
answered Jan 2 at 12:04
dexterdexter
1,28311020
edited Jan 2 at 12:26
answered Jan 2 at 12:04
dexterdexter
1,28311020
answered Jan 2 at 12:04
dexterdexter
1,28311020
answered Jan 2 at 12:04
dexterdexter
1,28311020
1,28311020
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
add a comment |
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
and how to displaying message that the data not exist or not matches with the query?
– Ahmad Z
Jan 2 at 12:14
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
I have updated the answer. Please check it.
– dexter
Jan 2 at 12:26
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
Oke, thanks you @dexter
– Ahmad Z
Jan 2 at 12:29
add a comment |
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1
To show the message when no row is found you have to check the row count, and to show all the rows, use fetchall and then use a while or foreach loop.
– Soheyl
Jan 2 at 11:56
@Soheyl sorry, i new to php and mysql. How to do that?
– Ahmad Z
Jan 2 at 11:59
Check the answer below :)
– Soheyl
Jan 2 at 12:07
If an answer solved it, just accept it by ticking the checkmark next to it.
– Funk Forty Niner
Jan 2 at 14:55
@FunkFortyNiner sorry, I newbie with stacjoverflow
– Ahmad Z
Jan 4 at 13:46