PHP: how to extract ajax array response

Question

I am getting ajax response in array format from php url. How to extract array response values in jQuery? FYI:

PHP array is:

$response = array('msg' => 'Hello', 'html' => '<b>Good bye</b>');

I am getting $response array in my ajax response. i.e.

var promo = "promo=45fdf4684sfd";
$.ajax({
    type: "POST",
    url: baseJsUrl + "/users/calc_discount",
    data: promo,
    success: function (msg) { // I am getting $response here as ajax response.
        //alert(msg);

        // Here I want to check whether response is in array format or not. if it is in array format, I want to extract msg here and want to use response array values.
    }
});

Let me know answer pls. Thanks.

Solution 1

You should echo that $response with json_encode().

You should probably set dataType: 'json' too inside the object literal you send to $.ajax().

Then you can access it natively with JavaScript using the dot operator inside your success callback...

function(msg) {
    alert(msg.html);
}

BTW, this line...

$response = array(['msg'] => 'Hello', 'html' => '<b>Good bye</b>');

... isn't valid PHP. Remove the brackets from the first key.

Solution 2

My favorite solution for this is to encode array with PHP's function json_encode() so jquery will be happy to parse it.

Comments

[Emil Vikström]

PHPJS's json_encode and _decode only works on JS objects, not on PHP equivalents.

[Satya Prakash]

Yes, Emil. You are right. I have not tested every function there but according to the way they are doing that json_encode/decode should work the same way as PHP counterpart is working. I have used few functions from there and that was working in correct way. Thanks for pointing the error.

[Hexodus]

Does your example really work? I have to parse the response first to got a result like this: var obj = jQuery.parseJSON(msg); alert(obj.html);

[user3167101]

@Hexodus Depends on what jQuery function you're using to read it. If you tell it the response type is json, it will automatically parse it (which you should be doing).