CODEWITHSUNDEEP
phpjavascriptjqueryajaxjson
domino
domino

Reputation: 7345

jQuery AJAX JSON issue

On the first page I have this function:

 <script>
function update() {
  $("#notice_div").html('Loading..'); 


  $.ajax({
    type: 'GET',
    dataType: 'json',
    data: latestid,
    url: '2includejson.php?lastid='+ latestid + '',
    timeout: 4000,
    success: function(data) {


      $("#cont_div").html(data);
      $("#cont_div").clone().prependTo($("#newdiv")); 
      $("#notice_div").html(''); 
      $("#cont_div").html('');
      window.setTimeout(update, 4000);

    },
    error: function (XMLHttpRequest, textStatus, errorThrown) {
      $("#notice_div").html('Timeout contacting server..');
      window.setTimeout(update, 60000);
    }
});
}
$(document).ready(function() {
    update();

}); 
</script>

And some php.

The included file:

    <? 
     header("Content-Type: application/json", true);

    $la = $_GET['lastid'];  
    include ("../../setup.php");

    $jsonArray[] = array();

      $count = 1; // first message is the newest on load
      $get = $DB->query("SELECT * FROM board WHERE id>'$la' ORDER BY id DESC LIMIT 5", __FILE__, __LINE__);
    while ($msg = $DB->fetch_array($get))
    {   

  if($count == 1){ 
  $latestid = $msg['id']; // newest message - this I need to pass to the other page
  }
  $count++; 

    $jsonArray = "$msg[msg]";
    }
    echo json_encode($jsonArray);
    ?>

I'm just trying to learn how to use ajax and jquery.

As you see, I pass latestid as js variable through the URL url: '2includejson.php?lastid='+ latestid + '',

I need to renew/post pack a newer value from the included page but I have no idea how to do so. Before using json I could just overwrite it with javascript, but now I don't know... The newer value will then be posted again as latestid.

Upvotes: 0

Views: 408

Answers (4)

Nick
Nick

Reputation: 1869

You might have to use the $.parseJSON;

Suppose the server returns a json of the form {"id":"2"}, then in the success function do

success:function(data){
var s=$.parseJSON(data);
alert(s.id)-->will alert 2

And I'm not sure if your server script returns a properly formated json. Use firebug figure this out

Upvotes: 0

Prasanth
Prasanth

Reputation: 3031

Try this

success: function(data) {


      $("#cont_div").html(data);
      $("#cont_div").clone().prependTo($("#newdiv")); 
      $("#notice_div").html(''); 
      $("#cont_div").html('');
  latestid = data.latestid ; // update the value
      window.setTimeout(update, 4000);

    }

Upvotes: 0

Sedat Başar
Sedat Başar

Reputation: 3798

success: function(data) {

the data in here is a json object. so u cant do $("#cont_div").html(data); u have to read the json and then write it to a div like this;

$.each(data,function(key,value){
    $("#cont_div").append(key + " "+ value);
})

Upvotes: 1

pimvdb
pimvdb

Reputation: 154838

You can declare the array without []:

$jsonArray = array();

Also you should then append the data to the array instead of making a string:

$jsonArray[] = $msg['msg'];

And in the end:

$jsonArray['latestid'] = $latestid;

Then in JavaScript, you should declare latestid:

var latestid;

And inside the ajax function, you should just pass the data as an object, and not twice like you're doing now. And just replace latestid there, which has been returned in JSON format:

...
data: {lastid: latestid},
url: '2includejson.php',
timeout: 4000,
success: function(data) {
    latestid = data.latestid; // update latestid
    ...
}
...

Upvotes: 3

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