cannot display array values in ajax success function - php

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

Answer 1

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) ;
});

Answer 2

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.

Answer 3

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 :)