Hi guys i was wondering if you guys could let me know how i can post a form field using jquery and then be declared in a variable after post.
The code for the html:
<table cellpadding="0" cellspacing="0">
<tr>
<td></td>
<td>Name:</td>
<td>Email:</td>
</tr>
<tr>
<td><label for=fri accesskey=f>Friend 1</label></td>
<td class="tablePadding"><input name="friname[]" type="text" id="friname" class="friname" size="25" value="" /></td>
<td><input name="friemail[]" type="text" id="friemail" class="friemail" size="25" value="" /></td>
</tr>
<tr>
<td><label for=fri accesskey=f>Friend 2</label></td>
<td class="tablePadding"><input name="friname[]" type="text" id="friname" class="friname" size="25" value="" /></td>
<td><input name="friemail[]" type="text" id="friemail" class="friemail" size="25" value="" /></td>
</tr>
<tr>
<td><label for=fri accesskey=f>Friend 3</label></td>
<td class="tablePadding"><input name="friname[]" type="text" id="friname" class="friname" size="25" value="" /></td>
<td><input name="friemail[]" type="text" id="friemail" class="friemail" size="25" value="" /></td>
</tr>
</table>
It gets posted to contact.php
here is the php contact.php file
$friname = $_POST['friname'];
foreach ($friname as $item) {
print $item."<br>";
}
Here is the jquery for the post
$('#contactform').submit(function(){
var action = $(this).attr('action');
$("#message").slideUp(750,function() {
$('#message').hide();
$.post(action, {
name: $('#name').val(),
email: $('#email').val(),
friname: $('#contactform input[name="friname[]"]').val(),
friemail: $('.friemail').val(),
comments: $('#comments').val(),
verify: $('#verify').val()
},
function(data){
document.getElementById('message').innerHTML = data;
$('#message').slideDown('slow');
$('#contactform img.loader').fadeOut('fast',function(){$(this).remove()});
if(data.match('success') != null) $('#contactform').slideUp('slow');
}
);
});
return false;
});
Thank you if you guys can help.
Thanks,
Marais