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Editorial Team
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Asked: 2026-06-08T02:57:46+00:00

I have been messing with this for too long trying to get it to

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I have been messing with this for too long trying to get it to work. Can anyone please see if you have any pointers. 

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <link type="text/css" rel="stylesheet" href="autocomplete.css" />      
        <script src="jquery-1.7.2.min.js" type="text/javascript"></script>
        <script src="jquery-ui-1.8.21.custom.min.js" type="text/javascript"></script>
        <title></title>
    </head>
    <body>
        <style>
            .ui-autocomplete-loading { background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat; }
        </style>
        <script>
            $(function() {                
                $( "#materials" ).autocomplete({
                    source: "autocomplete.php",
                    minLength: 2
                });
            });
        </script>
        <div class="demo">

            <div class="ui-widget">
                <label for="materials">Materials: </label>
                <input id="materials" />
            </div>
        </div><!-- End demo -->            
    </body>
</html>

and the php file is 

require_once "db_con.php"; // Database connection, I know this works.
$q = strtolower($_GET["q"]);
if (!$q)
    return;

$sql = "SELECT * FROM materials WHERE name LIKE '%$q%'"; 
$rsd = mysqli_query($dbc, $sql) or die(mysqli_error($dbc));
while ($rs = mysqli_fetch_array($rsd)) {
    $cname = $rs['name']; // I know this all returns correctly
    echo json_encode($cname); // First time I have ever used json, error might be here.
}

I am trying to have a webpage with an autocomplete powered by Jquery that is supplied data from mysql using PHP. Simples. Only its not working…

Anyone have any ideas what I am missing ?

Regards

—- EDIT —-

In order to check this was working I completed the following: 

<!DOCTYPE html>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <link type="text/css" rel="stylesheet" href="autocomplete.css" />      
        <script src="jquery-1.7.2.min.js" type="text/javascript"></script>
        <script src="jquery-ui-1.8.21.custom.min.js" type="text/javascript"></script>
        <title></title>
    </head>
    <body>
        <style>
            .ui-autocomplete-loading { background: white url('images/ui-anim_basic_16x16.gif') right center no-repeat; }
        </style>
        <script>
            $(function() {                
                $( "#materials" ).autocomplete({
                    source: <?php
include_once 'db_con.php';
$sql = "SELECT name FROM materials";
$rsd = mysqli_query($dbc, $sql) or die(mysqli_error($dbc));
echo '[';
while ($rs = mysqli_fetch_array($rsd)) {
    echo "'" . $rs['name'] . "', "; //add results to array
}
echo ']';
?>,
                        minLength: 2
                    });
                });
        </script>
        <div class="demo">

            <div class="ui-widget">
                <label for="materials">Materials: </label>
                <input id="materials" />
            </div>


        </div><!-- End demo -->        
    </body>
</html>

Which works perfectly. So good infact I think im going to keep this code not quite how its supposed to work but…

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1 Answer

  1. Editorial Team

    Your PHP is all wrong:

    while ($rs = mysqli_fetch_array($rsd)) {
    $cname = $rs['name']; // I know this all returns correctly
    echo json_encode($cname); // First time I have ever used json, error might be here.
}

    Should be:

    $cname = array();
while ($rs = mysqli_fetch_array($rsd)) {
    $cname[]['label'] = $rs['name']; // I know this all returns correctly
    break;
}
echo json_encode($cname); // First time I have ever used json, error might be here.

    label is the default label field within an array row that is used by jqueryautocomplete (I believe). Also the return must be an array of arrays each array row representing a match.

    You can make it more complicated by adding a value field for what to make the textbox to actually equal to by doing:

    $cname = array();
while ($rs = mysqli_fetch_array($rsd)) {
    $cname[]['label'] = $rs['name']; // I know this all returns correctly
    $cname[]['value'] = $rs['id'];
    break;
}
echo json_encode($cname); // First time I have ever used json, error might be here.

    Of course I don’t think your actually wanting the break; I put this in because:

    while ($rs = mysqli_fetch_array($rsd)) {
    $cname = $rs['name']; // I know this all returns correctly
    echo json_encode($cname); // First time I have ever used json, error might be here.
}

    Denotes to me that you are actually returning a single row from your results. If you are not and are actually returning all results then take the break; out.

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