How to stop return undefined for json_encode

Question

I am sent many dynamic post ids from a page and a php server side page(server.php) make a query with those id to find out newly added data in mysql.

If it not found any newly added data in mysql, it's return a undefined value. So as per my script, It's append a undefined one after one at a time interval.

How to stop this undefined value?

my javascript:

var CID = []; // Get all dynamic ids of posts (works well)
$('div[data-post-id]').each(function(i){
CID[i] = $(this).data('post-id');
});

function addrep(type, msg){
CID.forEach(function(id){
    $("#newreply"+id).append("<div class='"+ type +""+ msg.id +"'><ul><div class='cdomment_text'>"+ msg.detail +"</ul></div>");
});
}

function waitForRep(){
    $.ajax({
        type: "GET",
        url: "server.php",
        cache: false,
        contentType: "application/json; charset=utf-8",
        data: {
        // this way array containing all ID's can be sent:
        CID : CID
    },
        timeout:15000, 
        success: function(data){ 
            addrep("postreply", data);
            setTimeout(
                waitForRep, 
                15000 
            );
        },
        error: function(XMLHttpRequest, textStatus, errorThrown){
            setTimeout(
                waitForRep, 
                15000); 
        }
    });
}

$(document).ready(function(){
    waitForRep();
});

server.php

while (true) {
    if($_REQUEST['CID']){  //cid got all dynamic post id as: 1,2,3,4 etc.
      foreach($_REQUEST['CID'] as $key => $value){

        $datetime = date('Y-m-d H:i:s', strtotime('-15 second'));
        $res = mysqli_query($dbh,"SELECT * FROM reply WHERE qazi_id=".$_REQUEST['tutid']."  AND date >= '$datetime' ORDER BY id DESC LIMIT 1") or die(mysqli_error($dbh));
    $data = array();
        while($rows =  mysqli_fetch_assoc($res)){

          $data[]=$rows;

          $data['id'] = $rows['id']; 
          $data['qazi_id'] = $rows['qazi_id'];
          $data['username'] = $rows['username'];
          $data['description'] = $rows['description'];
          $data['date'] = $rows['date'];
          //etc. all
             $id = $rows['id'];
             $qazi_id = $rows['qazi_id'];
             $username = $rows['username'];
             $description = $rows['description'];
             //etc. all
          } //foreach close
      } //foreach close

          if ($description=="") {$detail .= '';}
            else {$detail .=''.$description.'';}
          $data['detail'] = $detail;
          // do somethig

           if (!empty($data)) {
              echo json_encode($data);
              flush();
              exit(0);
           }

    } //request close
    sleep(5);
} //while close

Why not just check if the server returns undefined and ignore it if it does? — Nick Bailey
– Nick Bailey, Commented Mar 12, 2015 at 13:25
I thought about this but I have no idea about this method. Have you any detail solved. Thank u. — user4453049
– user4453049, Commented Mar 12, 2015 at 13:28
PHP does not parse JSON requests sent to it (unless support has been added recently). Remove the contentType option from your ajax request. If you are wanting to accept JSON as a response, set the dataType option to "json" — Patrick Evans
– Patrick Evans, Commented Mar 12, 2015 at 13:31

MatejG · Accepted Answer · 2015-03-12 13:26:36Z

1

Edit your addrep function to not insert if it's undefined:

function addrep(type, msg){
CID.forEach(function(id){
    if(msg.detail != "undefined") {
       $("#newreply"+id).append("+ msg.detail +");
    }
});
}

Or just break it if undefined:

function addrep(type, msg){
    CID.forEach(function(id){
        if(msg.detail == "undefined")
           break;
        $("#newreply"+id).append("+ msg.detail +");
    });
}

answered Mar 12, 2015 at 13:26

MatejG

1,4231 gold badge17 silver badges26 bronze badges

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

3 Comments

user4453049 Over a year ago

Thank u. 1st one work as same. and 2nd one stoped working full script.

user4453049 Over a year ago

I want to say problem is not solved, It's work as same before. Thank u.

MatejG Over a year ago

Try to alert(msg.detail). Replace "undefined" with alerted string.

Collectives™ on Stack Overflow

How to stop return undefined for json_encode

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related