ajaxForm not working

Question

Here is my ajaxForm code

 var qx = $('#XText').attr('value'); $.ajax({ type: "post", url: "qsubmit.php", data: "q="+qx, success: function() { } });

And the insert code

include('db-config.php'); $q = $_POST['q']; $insert_ann = sprintf("INSERT INTO med_tab (med_title) VALUES ('$q')"); mysql_select_db($database_med_pharm, $med_pharm); $Result1 = mysql_query($insert_ann, $med_pharm) or die(mysql_error());

For some reason this is not working not sure why, any and all assistance would be great.

I want to pass in 2 values in data: "q="+qx, in the ajax js, how do I get that done.

Thanks Jean

@pekka Data is not being inserted @Slaks Could you tell me point me to the correct the error and where the error is exactly, thanks in advance. — X10nD
– X10nD, Commented Jan 2, 2011 at 17:12
$q = $_POST['q']; $insert_ann = sprintf("INSERT INTO med_tab (med_title) VALUES ('$q')"); is the vulnerable code, you're inserting based on POST values which could easily be forged. — Cyclone
– Cyclone, Commented Jan 2, 2011 at 17:14

Darin Dimitrov · Accepted Answer · 2011-01-03 05:17:11Z

If you are talking about the jquery form plugin your code should simply look like this:

$(function() { $('#idofyourform').ajaxForm(function(result) { alert('form successfully submitted'); }); });

If not, then make sure you properly encode the request:

$.ajax({ type: "post", url: "qsubmit.php", data: { q1: 'value 1', q2: 'value 2' }, success: function(result) { alert('form successfully submitted'); } });

or if you want to send the contents of the form:

$.ajax({ type: "post", url: "qsubmit.php", data: $('#idoftheform').serialize(), success: function(result) { alert('form successfully submitted'); } });

Finally, make sure you have installed FireBug to better analyze what's happening under the covers.

For some reason there is no alert from $.ajax{} also no insert, guess the ajaxform is not being triggered for some reason
@Jean, FireBug will indicate if there are some errors. Are there any?
It was not an issue with the coding error, but a braces closed after the $.ajax.
@Darin.. Thanks for helping out.. Does not chrome have a firebug.

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