cannot display array values in ajax success function

Question

      $.ajax({
                type: "GET",
                url: 'http://localhost/abc/all-data.php',
                data: {
                    data1: "1"},
                    success: function(response)
                           {
                        alert(response);
                        }          
            });
            return false;

I want to display each element of array one by one in success function of the ajax currently i get all elements to gether

this is my php code

$i=0;
while($row = mysqli_fetch_assoc( $qry )){

$temp[$i]['c_n'] = $row['c_name'];
$temp[$i]['j_t'] = $row['Job_Title'];
$temp[$i]['des'] = $row['description'];
$temp[$i]['req'] = $row['requirments'];
$temp[$i]['dat'] = $row['posted'];

$i++;
}
$data = array('temp'=> $temp);
echo JSON_encode($temp);

I do appreciate your helps

DevJava · Accepted Answer · 2017-11-22 15:30:41Z

2

you probably use something like this in your success function :

   response.temp.forEach(function(element){
         console.log(element.c_n) ;
         console.log(element.j_t) ;
         console.log(element.des) ;
         console.log(element.req) ; 
         console.log(element.dat) ;
    });

answered Nov 22, 2017 at 15:30

DevJava

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Comments

Vyko · Accepted Answer · 2017-11-22 15:19:17Z

1

In your success function, you need to json parse your response

var data = JSON.parse(response);

You can access to your data: data['temp']

If you want your response parsed to json automaticallym you can setup your ajax settings like this:

$.ajaxSetup ({
   contentType: "application/json",
   dataType: 'json'
});

Then you don't need to call JSON.parse anymore.

answered Nov 22, 2017 at 15:19

Vyko

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6 Comments

Ahmad Over a year ago

when i use JSON.parse and alert(data['temp']); it display undefined

Rory McCrossan Over a year ago

@Ahmad please actually show us the code you've used. Simply saying something doesn't work, or shows undefined is useless as it's possible you're not using the method correctly

Vyko Over a year ago

What does console.log(data) print ?

Ahmad Over a year ago

success: function(response) { var data = JSON.parse(response); alert(data['temp']); //console.log(data); }

Vyko Over a year ago

@Ahmad Okay and what about console.log(response) ?

|

Gopal Panadi · Accepted Answer · 2017-11-22 16:09:26Z

0

Your code:

$i=0;  while($row = mysqli_fetch_assoc($qry)){
  $temp[$i]['c_n'] = $row['c_name'];
  $temp[$i]['j_t'] = $row['Job_Title'];
  $temp[$i]['des'] = $row['description'];
  $temp[$i]['req'] = $row['requirments'];
  $temp[$i]['dat'] = $row['posted'];
  $i++;
}      $data = array('temp'=> $temp);   echo JSON_encode($temp);

Please change the last line as return JSON_encode($data);

Hope this helps you :)

answered Nov 22, 2017 at 16:09

Gopal Panadi

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