How to catch validation failure with laravel when using form requests

Question

I'm using a formrequest file with laravel 5.2 to check input. This form i'm calling using jquery's $.post function. In my console, it's return 422 Unprocessable Entity which I suspect to be coming from the response since i'm not formatting it to json. One way of doing is this

I'd would like to know how I can invoke from my form request, the means to change the output messages to json.

Thanks!

UPDATE 1

JQuery looks like this:

$.ajaxSetup({
    headers: {
        'X-CSRF-TOKEN': $('input[name="csrf-token"]').val()
    }
});

$("#changePassword").on('click', function(e){
    e.preventDefault();

    var data = {};
    data.name = $("input[name='name']").val();
    data.surname = $("input[name='surname']").val();
    data._token = $('input[name="_token"]').val();

    $.post('url',data).done(function(data){
        $(".message").empty().html("Done!");
    }).fail(
        function(response, status){

        }
    );
});

Could you post the code for you $.post section?

Rwd
– Rwd

2016-04-27 07:29:38 +00:00
Commented Apr 27, 2016 at 7:29 — Rwd
– Rwd, Commented Apr 27, 2016 at 7:29

Nehal Hasnayeen · Accepted Answer · 2016-04-27 07:54:48Z

2

Override the response method in your request class like below

    /**
     * Get the proper failed validation response for the request.
     *
     * @param  array  $errors
     * @return \Symfony\Component\HttpFoundation\Response
     */
    public function response(array $errors)
    {
        return Response::json([
            'error' => [
                'message' => $errors,
            ]
        ]);
    }

answered Apr 27, 2016 at 7:54

Nehal Hasnayeen

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1 Comment

Chiko Over a year ago

Right approach I ended up implementing it as: return response()->json([]);

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