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
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.