Using ajax to call php and return multiple variables?

Question

I am trying to use javascript to call a php script which then will return multiple variables back to my javascript so I can manipulate them.

This is my JS.

                $.ajax({ 
                    url: 'test.php',
                    data: { id : lastFileId },
                    success: function(output) {
                        alert(output);
                    }
                });

my PHP

<?php
    $fileId = ($_GET['id']);
    $num1 = 1;
    $num2 = 2;

?>

From here, how can I return variables $num1 and $num2 so i can use them in my javascript. Is it possible?

also this is a very basic idea of what I have planned to do if I can achieve this.

return json (use json_encode in PHP) and add the parameter datatype: json to your ajax request. — Pakspul
– Pakspul, Commented Apr 2, 2014 at 19:27

geoff · Accepted Answer · 2017-08-07 18:44:57Z

24

You can return as many variables as you want with json_encode().

Try in your PHP:

<?php
echo json_encode(array($num1, $num2));
?>

You can add to that array , $num3, $num4, ... and so on.

In your JS, you can access each number as follows.

First, you will need this line of code to parse the encoded JSON string, in your success function.

var result = $.parseJSON(output);

That sets result as a JSON object. Now you can access all fields within result:

result[0] -- $num1 in PHP
result[1] -- $num2 in PHP

edited Aug 7, 2017 at 18:44

answered Apr 2, 2014 at 19:31

geoff

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3 Comments

Joe Over a year ago

When i do console.log(output[0]), it gives me ' [ '

Joe Over a year ago

It splits up each char as an element, this is not correct

geoff Over a year ago

My mistake, I forgot a line of code. You'll need var result = $.parseJSON(output);. See corrected post.

Santosh Yadav · Accepted Answer · 2014-04-02 20:21:13Z

5

You can go for Json in PHP and javascript if you want array in response for a ajax request

PHP Code

<?php
    $fileId = isset($_GET['id'])?$_GET['id']:0;
    echo json_encode(array("field"=>$fileId,"num1"=>1,"num2"=>2));
?>

Js Code

jQuery.ajax({
    type: "GET",
    url: 'test.php',
    dataType: "json",
    success: function(response) {
        console.log(response);
        alert(response.num1);
    }
});

edited Apr 2, 2014 at 20:21

answered Apr 2, 2014 at 19:59

Santosh Yadav

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2 Comments

Joe Over a year ago

When I do resp.num1 i get "undefined" in my console

Murhaf Sousli Over a year ago

I'm getting the same "undefined"

tmthydvnprt · Accepted Answer · 2016-06-02 16:00:20Z

1

convert json to object

jQuery.ajax({
    type: "GET",
    url: 'test.php',
    dataType: "json",
    success: function(response) {
       item=JSON.parse(response);
        console.log(item);
        alert(item.num1);
    }
});

edited Jun 2, 2016 at 16:00

tmthydvnprt

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answered Jun 2, 2016 at 14:27

anas eddiani

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Comments

Safeer Ahmed · Accepted Answer · 2019-03-28 11:48:14Z

0

Use a simply string and explode(split) it further in ajax response. Here is PHP code

<?php
   $fileId = ($_GET['id']);
   echo $num1."|".$num2;
?>

Now split(explode) the response with JavaScript

$.ajax({ 
        url: 'test.php',
        data: { id : lastFileId },
        success: function(output) {
             var my_arr = output.split("|");
             console.log(my_arr[0] + "" + my_arr[1]);
        }
});

answered Mar 28, 2019 at 11:48

Safeer Ahmed

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Comments

Dhanushka Udayanga · Accepted Answer · 2019-08-09 09:00:15Z

0

you can simply return it like that return ['num1'=>$num1,'num2' => $num2];

and also, you can access it as followed, respone.num1

answered Aug 9, 2019 at 9:00

Dhanushka Udayanga

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