Reputation: 4851
I have the following PHP code:
<?php
include 'DBConnect.php';
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$query = "SELECT * FROM telejoke.jokes LIMIT 2";
$data = mysql_query($query) or die('Error, insert query failed' . mysql_error());
$info = mysql_fetch_array( $data );
mysql_close($conn);
echo json_encode($info);
?>
For instance, if I put LIMIT 2, I want only the first 2 rows from the table to be gathered, encoded into a JSON Array and echoed. Regardless of this LIMIT number, echo json_encode($info);
it prints out the whole table.
Trying echo json_encode($data);
results in null output.
Help is appreciated. Thanks.
Upvotes: 0
Views: 289
Reputation: 2956
trying doing this:
SELECT * FROM telejoke.jokes LIMIT 0,2
or
SELECT * FROM jokes LIMIT 0,2
Upvotes: 0
Reputation: 21531
Try this...
$info = array();
while ($row = mysql_fetch_array($data)) {
$info[] = $row;
}
echo json_encode($info);
Upvotes: 3