Laravel:How to send Ajax to controller without form

Question

I'm using a jquery plugin in my site. I want to validate a form first, Then send another data to controller,But doesn't through form. First validate is return a success feedback already. But when a second one is send, Doesn't have any feedback return. Here is my code

route.php

Route::controller('designs', 'DesignsController');

DesignsController.php

public function postTest() {
    $validator = Validator::make(Input::all(), Design::$rules);
    if ($validator->passes()) {

        $response = array(
            'status' => 'success'
        );
        return Response::json( $response );

    }
    $response = array(
        'status' => 'fail'
    );
    return Response::json( $response );
}
public function postSaveimage() {

    //Save Image complete return $success=1

    if($success)
     $response = array(
        'status' => 'save image success'
    );
    else
     $response = array(
        'status' => 'Fail'
    );
    return Response::json( $response );

}

jquery

$('#design_form').on( 'submit' ,function() {

             //Validate First form data
            $.post(
                $( this ).prop( 'action' ),
                {
                    "_token": $( this ).find( 'input[name=_token]' ).val(),
                    "category_id": $( '#form_category_id' ).val(),
                    "title": $( '#form_title' ).val(),
                    "user_id": $( '#form_user_id' ).val(),
                },
                function( data ) {
                    //if validate fail
                    if(data.status=='fail')
                    {
                        alert('data.status');
                    }
                    //if validate pass
                    else
                    {
                       //Sent second data
                        $.post("designs/saveimage", 
                        { 
                            "_token": $( this ).find( 'input[name=_token]' ).val(),
                            'base64_image': yourDesigner.getProductDataURL() 
                        }, function(data) {
                            if(data) {
                                alert(JSON.stringify(data));
                            }
                            else {

                                alert('fail!');
                            }
                        });
                    }
                },
                'json'
            );
            return false;
        });

Hi, what does var_export( $_POST ); say in your controller code. — ArtisticPhoenix
– ArtisticPhoenix, Commented Nov 7, 2014 at 3:56
Try to use Firebug to see the Ajax response.Then you will know what's the actual return you get from you second POST — Xiao Liu
– Xiao Liu, Commented Nov 7, 2014 at 4:05
It said 500 Internal Server Error "Illuminate\Session\TokenMismatchException" Guess i've to create new token right? — Expl0de
– Expl0de, Commented Nov 7, 2014 at 4:23

jewelnguyen8 · Accepted Answer · 2014-11-07 04:45:58Z

1

$('#design_form').on( 'submit' ,function() {

        var _this = this;
         //Validate First form data
        $.post(
            $( this ).prop( 'action' ),
            {
                "_token": $( this ).find( 'input[name=_token]' ).val(),
                "category_id": $( '#form_category_id' ).val(),
                "title": $( '#form_title' ).val(),
                "user_id": $( '#form_user_id' ).val(),
            },
            function( data ) {
                //if validate fail
                if(data.status=='fail')
                {
                    alert('data.status');
                }
                //if validate pass
                else
                {
                   //Sent second data
                    $.post("designs/saveimage", 
                    { 
                        "_token": $( _this ).find( 'input[name=_token]' ).val(),
                        'base64_image': yourDesigner.getProductDataURL() 
                    }, function(data) {
                        if(data) {
                            alert(JSON.stringify(data));
                        }
                        else {

                            alert('fail!');
                        }
                    });
                }
            },
            'json'
        );
        return false;
    });

Add new line:

var _this = this;

And in the second send:

"_token": $( _this ).find( 'input[name=_token]' ).val()

answered Nov 7, 2014 at 4:45

jewelnguyen8

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1 Comment

Expl0de Over a year ago

THANK YOU!! Work like charms! And thanks for pointing out my mistake.I'm understand now.

noobHere · Accepted Answer · 2014-11-07 05:16:42Z

0

on your second ajax post you have to indicate the host

if it is in a js file you have to specify the host

  $.post("http://www.host.com/public/designs/saveimage")

and if in view

  $.post("{{URL::to('designs/saveimage')}}");

you should check where does the post go and if the route is correct using firebug in firefox or the console in google chrome. .

answered Nov 7, 2014 at 5:16

noobHere

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1 Comment

Expl0de Over a year ago

Firebug said that TokenMismatchException. I think it's about how i pass a data with a token

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