Auto-Complete PHP, SQL & Javascript

Question

I am trying to run this Auto-Complete code, but it doesn't seem to be working. Can anyone help?

Thanks in advance.

Here is my Javascript:

<script type="text/javascript">
 $(document).ready(function(){
                $("#search_term").autocomplete({
                    source:'search_lookup.php',
                    minLength:2
                });
            });
</script>

Here is my php (I have removed my Database username and password details):

$get_term= $_GET["term"];

$sql_statement = mysql_query("SELECT pi.name, pi.middle_name, ad1.address_name  FROM person_info pi, address_1 ad1, WHERE pi.name LIKE '%$get_term%' OR  pi.middle_name LIKE '%$get_term%' OR ad1.address_name LIKE '%$get_term%' ORDER BY pi.name");
$json=array();

while ($people= mysql_fetch_array($sql_statement)) {
    $json[]=array(
            'label'=>$people['pi.name'].''.$people['pi.middle_name'].''.$people['ad1.address_name']     'value'=>$people['pi.name'].''.$people['pi.middle_name'].''.$people['ad1.address_name']
            );
}
echo json_encode($json);

?>

Finally, here is my html:

  <input id="search_term" name="search_term" type="text" placeholder="enter here..."/>
   <input id="submit" name="submit" type="submit"/>

Your PHP code has syntax error. Rest looks OK.

Salman Arshad
– Salman Arshad

2014-04-06 11:01:40 +00:00
Commented Apr 6, 2014 at 11:01 — Salman Arshad
– Salman Arshad, Commented Apr 6, 2014 at 11:01

Dov D. · Accepted Answer · 2014-04-06 03:49:54Z

1

I think source needs to be a function which does the ajax request and returns the results.

answered Apr 6, 2014 at 3:49

Dov D.

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Comments

W van Rij · Accepted Answer · 2014-04-06 09:20:05Z

0

The PHP code is wrong. If you have ran that yourself, you would see it gives a parse error.

$json[]=array('label'=>$people['pi.name'].''.$people['pi.middle_name'].''.$people['ad1.address_name'], 'value'=>$people['pi.name'].''.$people['pi.middle_name'].''.$people['ad1.address_name']);

That code should fix it, for returning the correct json data. You missed the "," between the label and value.

answered Apr 6, 2014 at 9:20

W van Rij

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1 Comment

LifeLong Over a year ago

Thanks, this managed to get the job done! Next time I will pay more attention to the syntax.

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