Parse XML code via PHP to jQuery - Getting XML Error

Question

iam trying to update a ajax inbox. I'm having problems with parsing xml to jquery. I'm allways getting the error, that this isnt an xml file. But its (think?) correct and have no code-mistakes in it.

XML Output:

<?xml version='1.0' ?>
<inboxdata>
 <convid>2</convid>
 <sender_id>1</sender_id>
 <reciever_id>2</reciever_id>
 <msg_count>4</msg_count>
 <last_txt_msg>Hello</last_txt_msg>
 <sender_read>1</sender_read>
 <sender_delete>0</sender_delete>
 <time>23.09.2013 - 19:38</time>
 <userpic>img/ava/1.png</userpic>
</inboxdata>
<inboxdata>
 <convid>4</convid>
 <sender_id>1</sender_id>
 <reciever_id>3</reciever_id>
 <msg_count>0</msg_count>
 <last_txt_msg>No msg</last_txt_msg>
 <sender_read>1</sender_read>
 <sender_delete>0</sender_delete>
 <time>23.09.2013 - 20:25</time>
 <userpic>img/ava/1.png</userpic>
</inboxdata>

PHP:

function show_inbox($mysqli)
{
    if(!isset($_SESSION)){
    sec_session_start();
    }
    $userid = $_SESSION['user_id'];
    $get_conversation = $mysqli->query("SELECT * FROM conversation WHERE sender_id =" . $userid);
    $xml_node = "";
    while ($row = mysqli_fetch_array($get_conversation)) {
            $xml_node     .= "<inboxdata>\n";
            $xml_node     .= "<convid>".$row['convid']."</convid>\n";
            $xml_node     .= "<sender_id>".$row['sender_id']."</sender_id>\n";
            $xml_node     .= "<reciever_id>".$row['reciever_id']."</reciever_id>\n";
            $xml_node     .= "<msg_count>".$row['msg_count']."</msg_count>\n";
            $xml_node     .= "<last_txt_msg>".$row['last_txt_msg']."</last_txt_msg>\n";
            $xml_node     .= "<sender_read>".$row['sender_read']."</sender_read>\n";
            $xml_node     .= "<sender_delete>".$row['sender_delete']."</sender_delete>\n";
            $xml_node     .= "<time>".$row['time']."</time>\n";
            $xml_node     .= "<userpic>img/ava/1.png</userpic>\n";
            $xml_node     .= "</inboxdata>\n";
    }
    $returnXML = "<?xml version='1.0' ?>\n".$xml_node;
    echo $returnXML;
}

java:

function show_inbox()
    {
        result = $.ajax({
        type: 'GET',
        async: false,   // WICHTIG! 
        url: 'functions_chat.php',
        data: ({
            a: "show_inbox"
        }),
       success: function(response){
           alert($(response).find('inboxdata').length);
          $(response).find('inboxdata').each(function(){

         var convid   = $(this).find('convid').text();
         $("ul#chatinbox").prepend("<li>"+convid+"</li");

         });
       }
    }).responseText;
    }

If the Alert from alert($(response).find('inboxdata').length); shows up, it allways counts 0. It should count 2?

If I use $.parseXML(response) I'll get the XML Error: Invalid XML... But why its invalid? It seems valid to me!

Wheres the problem in here? Thanks!

You have 2 rootnodes, there can be only 1 [/end Highlander] (create a containing element (<list>?) which holds both <inboxdata> nodes. — Wrikken
– Wrikken, Commented Sep 23, 2013 at 20:41
I think you mean "pass" and "passing", not "parse" and "parsing" - a subtle point, but worth getting right in a context like this where both could apply: To "parse" XML would be to read it, to "pass" it implies sending it somewhere, so they're kind of opposites. — IMSoP
– IMSoP, Commented Sep 23, 2013 at 20:44

ComFreek · Accepted Answer · 2013-09-23 20:41:34Z

1

Your XML is invalid because it does not contain a root element (I use the name 'root' in the sample below, but you can use any name):

<?xml version='1.0' ?>
<root>
  ... your actual content here
</root>

You can check your XML using the W3 Online Validator: http://validator.w3.org/check

answered Sep 23, 2013 at 20:41

ComFreek

29.5k18 gold badges107 silver badges157 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Ponch0 Over a year ago

thats freakin hilarious! LOL! I thought <inboxdata></inboxdata> should be enough. Well, time to learn XML exactly! Thanks alot. Works!

IMSoP Over a year ago

@user2447291 The key is that there must be exactly one outermost element, so if you have a loop, there has to be something outside of the loop too. :)

Collectives™ on Stack Overflow

Parse XML code via PHP to jQuery - Getting XML Error

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related