Don't print AJAX results

I am working on a script to send data to a mysql table and everything is working correctly, but the successful part of the call is not loading my results into a result column in my page. My code is below.

Any suggestions on what I can do to fix this? I guess the problem is with my "success:" option in my AJAX call.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Facebook like ajax post - jQuery - ryancoughlin.com</title>
<link rel="stylesheet" href="css/screen.css" type="text/css" media="screen, projection" />
<link rel="stylesheet" href="css/print.css" type="text/css" media="print" />
<!--[if IE]><link rel="stylesheet" href="css/ie.css" type="text/css" media="screen, projection"><![endif]-->
<script src="js/jquery.js" type="text/javascript" charset="utf-8"></script>
<script type="text/javascript">
/* <![CDATA[ */
$(document).ready(function(){

    $('p.validate').hide();

    $.getJSON("readJSON.php",function(data){
        $.each(data.posts, function(i,post){
            content = '<div id="post-'+ post.id +'">\n';
            content += '<h3>' + post.author + '</h3>\n';
            content += '<p class="small quiet">' + post.date_added + '</p>\n';
            content += '<p>' + post.post_content + '</p>';
            content += '<hr/>';
            $("#contents").append(content).fadeIn("slow");      
        });
    });  
    $(".reload").click(function() { 
        $("#posts").empty();
    });

    $("#add_post").submit(function() {
        $('p.validate').empty();
        // we want to store the values from the form input box, then send via ajax below
        var author = $('#author').attr('value');
        var post   = $('#post').attr('value'); 

        if(($('#author').val() == "") && ($('#post').val() == "")){
            $("p.validate").fadeIn().append("Both fields are required.");
            $('#author,#post').fadeIn().addClass("error");
        }else{
            $.ajax({
                type: "POST",
                url: "ajax.php",
                data: "author="+ author + "&post=" + post,
                success: function(result){
                    $('#contents').after( "<div>" +result +"</div>" );
                }
            });

        }
        return false;
    });

});
/* ]]> */
</script>
<style type="text/css">
h3{margin:10px 0;}
p{margin:5px 0;}
#posts{display:none;}
</style>
</head>
<body>
        <div class="container">
                <div class="span-24">
                        <div id="post-container" class="span-9 colborder">
                                <h2>Posts loaded via Ajax:</h2>
                                <div id="contents"></div>   
                        </div>
                        <div id="form" class="span-11">
                            <h2>New Post:</h2>

                            <form name="add_post" id="add_post" action="">
                                <label>Author:</label><br />
                                <input type="text" name="author" id="author" size="15" class="text" /><br />
                                <label>Post:</label><br />
                                <textarea name="post" id="post" rows="5" cols="5" class="text"></textarea><br />
                                <input type="submit" value="Post" id="submit" /><br />
                            </form><br />
                            <p class="validate error"></p>

                        </div>  
                </div>
                <div class="span-24">
                    <a href="#" class="reload">Reload</a>
                </div>
        </div>
</body>
</html>

      

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2 answers


Questions to ask yourself ...

  • Does jQuery even do your success callback?
  • If this is so well-formed answer data markup?

To get started, I would add a "debugger"; (if you have firefox and firebug). This will allow you to enter the script console and have a better understanding of what is going on.

The debugger statement will cause the script execution to pause and go to the firebug console. Try the following

  success: function(result){
             debugger;
             $('#contents').after( "<div>" +result +"</div>" );
           }

      



If your script hits this, I suspect your answer markup is not well-formed and jQuery is having trouble parsing into divs, but you can check the whole thing when you are on that breakpoint in firebug.

Another simple thing to check and reject when debugging:

  • Your webserver is serving a valid (status 200) response (check your console or network tab in firebug to see this, or use similar scripts if running, i.e.)

Let me know how you are doing.

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You might have an error. Try adding debug statement to your ajax call using error parameter



$.ajax({
                        type: "POST",
                        url: "ajax.php",
                        data: "author="+ author + "&post=" + post,

                        error: function(XMLHttpRequest, textStatus, errorThrown) 
                        { alert(errorThrown); },

                        success: function(result){
                        $('#contents').after( "<div>" +result +"</div>" );
                        }
                });

      

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