在php和mysql中自动完成文本框

Question

i`m trying to make a auto complete text box in php. here is my JavaScript code

<script src="assets/js/jquery-1.10.2.js"></script>
<script src="assets/js/jquery-1.3.2.min.js"></script>
<script src="assets/js/jquery-ui-1.10.4.custom.min.js"></script>
<script type="text/javascript">
jQuery(document).ready(function(){
$('#search').autocomplete({
source : 'search-job.php',
minLength:2
}); 
});
</script>

here is my html code

<input type="text" id="search" class="form-control input-lg col-xs-4" name="search">

here is my search-job.php

<?php
include 'connectdb.php';
if(isset($_REQUEST['term']))
exit ();
$rs=  mysql_query("select * from jobs where `title` like '%"  . ucfirst($_REQUEST['term']). "%' ORDER BY id ASC LIMIT 0,10") or      die(mysql_error());
$data=  array();
while($row=  mysql_fetch_assoc($rs,MYSQL_ASSOC)){
$data[]=array(
'label'=>$row['title'],
'value'=>$row['title']
);
}
echo json_encode($data);
flush();

i`m getting correct values in search-job.php. but it is not catching by html but when i run code nothing happened? why is that?

this is how it`s looks on network tab

Answer 1

The array format you are passing is wrong.

$data[]=array(
'label'=>$row['title'],
'value'=>$row['title']
);

should be

$data[$row['title']]= >$row['title'];
OR simply
$data[]= >$row['title'];

and

replace

source : 'search-job.php',

with

source: 'search-job.php?term='+document.getElementById('search').value,