Trying to call PHP file using jQuery Ajax

Question

I am developing using jQuery 2.0.2 on Debian 7.0.0.

I am trying to call a PHP script from a JavaScript file using jQuery.ajax. My code is as follows.

    jQuery.ajax({
    type: "POST",
    url: "DisplayParameters.php",
    data: { LUT:lut, 
        Brightness: brightness,
        Gamma: gamma,
        Background: background}
    }).done(function( result ) {
    $("#msg").html( " Ajax called" );
    });

At the beginning of DisplayParameters.php, I have the following.

?>
<script language="JavaScript">
alert('DisplayParameters');
</script>
<?php

but the alert box is not called. Firebug proceeds without any errors and shows the DisplayParameters.php text (with a +/- control to collapse and expand it). But the alert box is not called. I cannot get any feedback from DisplayParameters.php to trace what is happening.

What would be the best way for me to get feedback (like alert boxes or console.log) from DisplayParameters.php?

Please post the full code of DisplayParameters.php

Nir Alfasi
– Nir Alfasi

2013-06-23 03:28:59 +00:00
Commented Jun 23, 2013 at 3:28 — Nir Alfasi
– Nir Alfasi, Commented Jun 23, 2013 at 3:28
What are you trying to achieve?

Jay Bhatt
– Jay Bhatt

2013-06-23 03:37:07 +00:00
Commented Jun 23, 2013 at 3:37 — Jay Bhatt
– Jay Bhatt, Commented Jun 23, 2013 at 3:37

Ast Derek · Accepted Answer · 2013-06-23 03:31:23Z

2

Your ajax call returns the contents of DisplayParameters.php as text, console.log(result) would show:

<script language="JavaScript">
alert('DisplayParameters');
</script>

If you want to add this extra code to the page (which I don't recommend unless you are sure you won't break the current page), you can try:

jQuery.ajax({
    type: "POST",
    url: "DisplayParameters.php",
    data: { LUT:lut, 
        Brightness: brightness,
        Gamma: gamma,
        Background: background}
}).done(function( result ){
    $("#msg").html( " Ajax called" );
    $("body").append(result);
});

Another approach would be to serialize (JSON) the information you need, set the dataType: "json" option, and act based on this information.

If you will send HTML, ensure your PHP code doesn't shows any error or warning messages.

answered Jun 23, 2013 at 3:31

Ast Derek

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1 Comment

OtagoHarbour Over a year ago

Thanks. I did not explain my objective very well. I was initially hoping to get some feedback from DisplayParameters.php to see how to get feedback. The I was hoping to get the strings that were formed from the input arguments. I found that I could this by using echo "<script language=\"JavaScript\">"; with echo "console.log('str='+" . $str . ");"; statements in the PHP file. The results would show up in the DisplayParameters.php text (with +/- controls in Firebug. Thanks, Peter.

kevinandrada · Accepted Answer · 2013-06-23 03:32:58Z

1

The alert() should happen inside .done() on the ajax call. Like the following:

.done(function(result) {
    alert(result);
    $("#msg").html( " Ajax called" );
});

And instead of alerting on DisplayParameters.php, use echo so that that will be the returned value for result. Like the following:

<script language="JavaScript">
    //...some JS codes here
    <?php echo 'DisplayParameters'; ?>
</script>

answered Jun 23, 2013 at 3:32

kevinandrada

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