Adding data to div via ajax

Question

Afternoon

My php and Ajax is now almost complete, but I'm stuck on one thing I'm getting the data sent back through to the ajax but its only showing [object Object] in the div I've specified and I'm wanting the number of results sent back to be in there.

<?  $call="SELECT * FROM notifications WHERE notification_targetuser='$user1_id' ORDER BY notification_id DESC LIMIT 1";
        $chant=mysqli_query($mysqli,$call) or die(mysqli_error($mysqli));
        $num=mysqli_num_rows($chant);


            while($notification_id=mysqli_fetch_array($chant))
            {
            ?>
            <script type="text/javascript">
setInterval(function(){


  var notification_id="<?php echo $notification_id['notification_id'] ;?>"

$.ajax({
type: "GET",
url: "viewajax.php?notification_id="+notification_id,   
dataType:"json",
cache: false,
success: function(response){
$("#mes").prepend(response);

}
});
},20000);

</script>
<? }?>

Vuewajax.php

<?php

 include"database.php";

if(isset($_GET['notification_id'])){

$id=$_GET['notification_id'];
$user1_id=$_SESSION['id'];
$json = array();
$com=mysqli_query($mysqli,"select notification_id from notifications where notification_id > '$id'");
echo mysqli_error($mysqli);

$num = mysqli_num_rows($com);
if($num>0){

    echo '<span id="mes">'.$num.'</span>';
}
$resultArr = mysqli_fetch_array($com);
$json['notification_id'] = $resultArr['notification_id'];

mysqli_free_result($com);


echo json_encode($json);
}
 ?>

My returned response in firebug

<span id="mes">1</span>{"notification_id":"3306"}

Steve · Accepted Answer · 2014-07-13 19:29:53Z

1

You are returning a json object, so you need to access the correct propery, in this case notification_id. Its also good practice to set the correct content-type header:

//php set json header
header('Content-Type: application/json');
echo json_encode($json);

//js access property of returned object
$("#mes").prepend(response.notification_id);

EDIT as per comment - if you are sending json, only send json, not mixed in html:

if($num>0){

    $json['num'] = $num;
}else{
    $json['num'] = 0;
}
$resultArr = mysqli_fetch_array($com);
$json['notification_id'] = $resultArr['notification_id'];

mysqli_free_result($com);
header('Content-Type: application/json');
echo json_encode($json);


//js
 $("#mes").prepend('<span id="mes">'+ response.num + '</span>');

Further EDIT You can check the value of num, and only prepend if its not 0. As 0 evaluates to false, the following will work

if(response.num){
     $("#mes").prepend('<span id="mes">'+ response.num + '</span>');
}

edited Jul 13, 2014 at 19:29

answered Jul 13, 2014 at 18:35

Steve

20.5k5 gold badges47 silver badges71 bronze badges

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

3 Comments

Dave Over a year ago

I'm just needing the actual number of records out of the php that are sent back to be inserted into its div <span id="mes">1</span> Notification_id is only there to find the next available id.

Dave Over a year ago

Fantastic, it works.And as simple as my stupid rookie error of mixing them both. Thanks so much.

Dave Over a year ago

How would I go about showing nothing at all if there is no data? as it throws back undefined in the div if there is no result if I add echo"";

David Jacquel · Accepted Answer · 2014-07-13 19:31:03Z

1

This answer can seems to be off-topic but the code you show us is not secure. He's prone to easy sql injection

eg :

$id  = $_GET['notification_id'];
$com = mysqli_query($mysqli,"select notification_id from notifications where notification_id > '$id'");

Injecting "'; DELETE FROM users WHERE 1 or username = ''" in your notification_id parameter will delete all your users !

Use prepared statements, it's easy and far safer an XKCD ;-)

answered Jul 13, 2014 at 19:31

David Jacquel

53k7 gold badges133 silver badges155 bronze badges

2 Comments

Dave Over a year ago

You're right. But I don't secure my scripts until I have got them working :P

David Jacquel Over a year ago

As an experimented (?) developer, I know that what is not done first shot is difficult to get done once your in a hurry ! It's true for security, documentation and testing. Believe me.

Collectives™ on Stack Overflow

Adding data to div via ajax

2 Answers 2

3 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related