multiple return from php using ajax

Question

I want to get more then 1 value from my php file.

Here is my ajax code

$.ajax({
    type: "POST",
    url: "insert.php",
    data: $(this).serialize(),
    success: function(response){
        alert(response);
        alert(response.dummy_code);
        alert(response.user_id);
    },
});

and on my insert.php i make this output

echo json_encode(array("dummy_code" => "$dummy_code", "user_id" => "$insertet_user_id"));

As example $dummy_code have to value 5s8374m9f9f3m34mc334 and $user_id the value 14.

With the command

alert(response);

I get this ouput

{"dummy_code":"5s8374m9f9f3m34mc334","user_id":"14"}

And the ouput of the two other commands

undefined

Why response.dummy_code & respone.id dont work?

Just for testing purpose: alert(eval(response.dummy_code)); — Hackerman
– Hackerman, Commented Jun 1, 2015 at 19:21
Read the documentation, in particular, the dataType option. — Jonnix
– Jonnix, Commented Jun 1, 2015 at 19:21

Ramy Deeb · Accepted Answer · 2015-06-01 19:21:44Z

2

Are you setting the correct content type in the response header? if not tru adding this line before the echo:

header('Content-Type: application/json');
echo json_encode(array("dummy_code" => "$dummy_code", "user_id" => "$insertet_user_id"));

answered Jun 1, 2015 at 19:21

Ramy Deeb

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1 Comment

Ramy Deeb Over a year ago

@Name if this helped you, you could mark it as answer. It would help me ;)

Jay Blanchard · Accepted Answer · 2015-06-01 19:27:08Z

2

Because you're getting a string back you have to parse the string as JSON

function(response){
    var myJSON = JSON.parse(response);
    console.log(response);
    console.log(myJSON.dummy_code);
    console.log(myJSON.user_id);
},

If you don't want to do that you can just add a dataType to the jQuery AJAX request -

$.ajax({
    type: "POST",
    url: "insert.php",
    data: $(this).serialize(),
    dataType: 'JSON',

Quit using alert() for troubleshooting., use console.log() instead.

edited Jun 1, 2015 at 19:27

answered Jun 1, 2015 at 19:21

Jay Blanchard

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2 Comments

Name Over a year ago

I tried it now multiple times. But the ouput in the console is every time the same, only {"dummy_code":"5s8374m9f9f3m34mc334","user_id":"14"}. I think there must be 2 lines more (the value of dummy_code & the value of user_id)?

Jay Blanchard Over a year ago

This syntax seems to be working for me just fine. If the last 2 lines were problematic you would get undefined.

Musa · Accepted Answer · 2015-06-01 19:22:45Z

1

your response is not parsed into an object its just a string, if you're returning JSON, tell jQuery and It'll parse it for you.
The are two ways to do it, 1: set the response type to application/json

header('Content-Type: application/json');

or set the dataType in $.ajax to json.

dataType: 'json',

answered Jun 1, 2015 at 19:22

Musa

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1 Comment

Name Over a year ago

Yep that was the problem. THanks!

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