drop down search

**kaushik_acharya** · Jul 1, 2007, 08:48

I tried to make a drop down search as its here in the left side
http://www.indianartcollectors.com/v...s.php?arid=818

my trial is here:
http://www.jalsaghar.com/test.php

but it's not working... can anyone suggest what's wrong?

**php_daemon** · Jul 1, 2007, 09:11

Any relevant php code?

**kaushik_acharya** · Jul 1, 2007, 09:54

whatever i have done can be seen in the Source Code of browser ..... nothing much in php .......

**php_daemon** · Jul 1, 2007, 10:05

Then what is wrong with it?

**kaushik_acharya** · Jul 1, 2007, 10:13

on submitting I m trying to call function displayArtist()
which I have written inside HEAD

<SCRIPT>
function displayArtist()
{
<?php
echo("1234");
?>
}
</SCRIPT>

but it seems to be not calling the function......

here's the complete code:

PHP Code:


<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>

<head>
    <title>
         Jalsaghar: Artist and their Works
    </title>

    <SCRIPT>
            function displayArtist()
            {
                <?php
                    echo("1234");
                ?>
            }
    </SCRIPT>
</head>

<body>

<?php

    $sql = $db_object->query("Select distinct artistid from image");

    echo("<form name=\"frmSearch\" action=\"\" method=\"POST\" enctype=\"multipart/form-data\" onsubmit=\"javascript:displayArtist();\">");

        echo("<select name=\"vchArtist\">");
        while($row = $sql->fetchRow())
        {
            $ArtistId = $row["artistid"];
            echo("<option value=\"$ArtistId\">$ArtistId</option>");
        }
        echo("</select>");

        echo("<input type=\"submit\" name=\"submit\" value=\"Submit\">");

    echo("</form>");

?>

</body>

</html>

**php_daemon** · Jul 1, 2007, 10:19

Look at the source, you've got:

Code:

function displayArtist()
			{
				1234			}

All you've got is a syntax error there. You want to run a php script server-side, look up Ajax.

**kaushik_acharya** · Jul 5, 2007, 09:38

Originally Posted by php_daemon

All you've got is a syntax error there. You want to run a php script server-side, look up Ajax.

can u give me some link where I can learn how to do that .....

though printing 1234 was not my intention for drop down search, I put it to know whether it was calling function displayArtist

**php_daemon** · Jul 5, 2007, 11:09

Try this article: http://www.sitepoint.com/article/bui...-ajax-web-apps

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