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data option in jQuery.ajax() and mysql_query?
I modify a php comment system. I want add it after every article witch are query from database.
this is the php part <?php
...
while($result = mysql_fetch_array($resultset))
{
$article_title = $result['article_title'];
...
?>
<form id="postform" class="postform">
<input type="hidden" name="title" id="title" value="<?=$article_title;?>" />
<input type="text" name="content" id="content" />
<input type="button" value="Submit" class="Submit" />
</form>
...
<?php
}
?>
this is the ajax part.
$(function($) {
$(document).ready(function(){
$(".Submit").click(function(){
var id = $(this).attr("id");
var name = $(this).attr("name");
var dataString = 'id='+ id ;
var parent = $(this);
var anyBlank = 0;
if(anyBlank == "0")
{
var title = $("#title").val();
var content = $("#content").val();
$.ajax({
type: "POST",
url: "ajax_post.php",
data: "title="+title+"&content="+content,
success: function(date_added){
if(date_added != 0)
{
structure = '<div class="comment_date_added">'+date_added+'</div><div id="comment_text"><div id="comment_content">'+content+'</div>';
$("#post_comment").prepend(structure);
}
});
});
ajax_post.php
echo $title; echo $content; How to modify the ajax data part that each article's comment can send each data to the ajax_post.php? thanks a lot.
1 Answer
The "data" property of a $.ajax request is actually a json object. jQuery deals with serializing and getting the data to the server automatically.
You want to do: data: { title: title, content: content } On the PHP end: $title = $_POST['title'] $content = $_POST['content'] Posted: MacOS 1 of 1 people found this answer helpful. Did you? Yes No |
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