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Asked: 2026-05-20T05:06:02+00:00

am new to jquery! I’m using jquery ui autocomplete in my application where the

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am new to jquery! I’m using jquery ui autocomplete in my application where the auto complete values comes from database. Here are codes that am using but nothing happens

search.php

<?php

    $conn = mysql_connect("localhost", "root", "");
    mysql_select_db("webforum", $conn);

    $q = strtolower($_GET["term"]);

    $query = mysql_query("select name from groups where name like %$q%");
    while ($row = mysql_fetch_array($query)) {
        echo json_encode($row);
    }
    ?>

Here is test.php

<!DOCTYPE html>
<html lang="en">
<head>
    <meta  charset="utf-8">
    <title> jQuery UI Autocomplete - Multiple, remote </title>
    <link rel="stylesheet" href="theme/jquery.ui.all.css">
        <script type="text/javascript" src="jquery/jquery-1.4.4.min.js"></script>
    <script type="text/javascript" src="jquery/ui/jquery.ui.core.js"></script>
    <script type="text/javascript" src="jquery/ui/jquery.ui.widget.js"></script>
    <script type="text/javascript" src="jquery/ui/jquery.ui.position.js"></script>
    <script type="text/javascript" src="jquery/ui/jquery.ui.autocomplete.js"></script>
        <style type="text/css">
    .ui-autocomplete-loading { background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat; }
    </style>
        <script type="text/javascript">
    $(function() {
        function split( val ) {
            return val.split( /,\s*/ );
        }
        function extractLast( term ) {
            return split( term ).pop();
        }

        $( "#birds" )
            // don't navigate away from the field on tab when selecting an item
            .bind( "keydown", function( event ) {
                if ( event.keyCode === $.ui.keyCode.TAB &&
                        $( this ).data( "autocomplete" ).menu.active ) {
                    event.preventDefault();
                }
            })
            .autocomplete({
                source: function( request, response ) {
                    $.getJSON( "http://localhost/webforum/search.php", {
                        term: extractLast( request.term )
                    }, response );
                },
                search: function() {
                    // custom minLength
                    var term = extractLast( this.value );
                    if ( term.length < 2 ) {
                        return false;
                    }
                },
                focus: function() {
                    // prevent value inserted on focus
                    return false;
                },
                select: function( event, ui ) {
                    var terms = split( this.value );
                    // remove the current input
                    terms.pop();
                    // add the selected item
                    terms.push( ui.item.value );
                    // add placeholder to get the comma-and-space at the end
                    terms.push( "" );
                    this.value = terms.join( ", " );
                    return false;
                }
            });
    });
    </script>
</head>
<body>

<div class="demo">

<div class="ui-widget">
    <label for="birds">Birds: </label>
    <input id="birds" size="50" />
</div>

</div>
</body>
</html>

Can anyone help me with this?
Thanx in Advance!

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1 Answer

  1. Editorial Team

    How does your json looks like?
    in order to work wit jquery-ui autocomplete you need to have at least label and value properties:

    {
   'label' : 'your_label',
   'value' : 'your_value'
}

    in your js code you are asking for value property which doesn’t seem to bes set on your json produced by php.

    here is a similar question: Having problems with jQuery UI Autocomplete

    so php has to build the results in the right way:

        $conn = mysql_connect("localhost", "root", "");
    mysql_select_db("webforum", $conn);

    $q = strtolower($_GET["term"]);

   $return = array();

    $query = mysql_query("select name from groups where name like %$q%");
    while ($row = mysql_fetch_array($query))
    {
        //since we have just 1 value from the db just use it as both value and label
        array_push($return,array('label'=>$row['name'],'value'=>$row['name']));
    }

   echo(json_encode($return));

    P.S. it is not that safe make queries with $_GET[] parameters.

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