CODEWITHSUNDEEP
phpajaxjsonmysqli
Dave
Dave

Reputation: 97

Adding data to div via ajax

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"}

Upvotes: 1

Views: 86

Answers (2)

David Jacquel
David Jacquel

Reputation: 52809

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 ;-)

Upvotes: 1

Steve
Steve

Reputation: 20469

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>');
}

Upvotes: 1

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