passing input value using jquery on mysql query - php

Question

I have following html form:
<form method="post" action="">
<input type="hidden" name="userid" id="user" value="<?php echo $user ?>"/>
<textarea name="comment" class="subcomment_form" id="ctextarea<?php echo $suggestid ?>"></textarea>
<input type="submit" value="Post" id="<?php echo $suggestid ?>" class="form_btn" style="margin-top: 5px"/>
</form>
following JS code:
$('.form_btn').live("click",function()
{
var ID = $(this).attr("id");
var user= $("input[name=userid]").val();
var comment= $("#ctextarea"+ID).val();
var dataString = 'comment='+ comment + '&suggestid=' + ID + 'user' + user;
if(comment=='')
{
alert("Please Enter Comment Text");
}
else
{
$.ajax({
type: "POST",
url: "action/subcomment.php",
data: dataString,
cache: false,
success: function(html){
$("#commentload"+ID).append("html");
$("#ctextarea"+ID).val('');
$("#ctextarea"+ID).focus();
}
});
}
return false;
});
and following php code for subcomment.php file:
if($_POST['comment'])
{
$comment=$_POST['comment'];
$suggestid=$_POST['suggestid'];
$user=$_POST['user'];
$sql = "INSERT INTO user_suggestions_comments (uvd_id, user_id, usc_comment) VALUES ('".$_POST['suggestid']."', '".$_POST['user']."','".$_POST['comment']."')";
mysql_query( $sql);
}
the problem I have is that value from <input type="hidden" name="userid" id="user" value="<?php echo $user ?>"/> is not passed on php file and textarea content is passed. What I need to change inside that JS code to make it work?

Answer 1

It looks like your dataString is not being built to what your PHP code is expecting.
Try this change.
var dataString = 'comment='+ comment + '&suggestid=' + ID + 'user' + user;
becomes
var dataString = 'comment='+ comment + '&suggestid=' + ID + '&user=' + user;
You are looking for three values from your $_POST and only passing 2
$comment=$_POST['comment'];
$suggestid=$_POST['suggestid'];
$user=$_POST['user'];