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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