Ajax LARAVEL 419 POST error

Question

I would really appreciate some help on this. I tried tons of solutions as posted in this forum, but I cannot get it to work.

My ajax call is something like

$(document).ready(function() {
    $("#company").click(function() {
        $.ajax({
            type: "POST",
            dataType:'html',
            url : "/company",
            success : function (data) {
                $("#result").html(data);
            }
        });
    });
});

I am calling the view through my route

Route::post('/company', 'Ajaxcontroller@loadContent');

And controller

public function loadContent()
    {
        return view('listing.company')->render();
    }

My company.blade.php is

    @foreach ($companies as $company)
            <div class="posting-description">
            <h5 class="header"><a href="#"></a>{{$company->name}}
            </h5>
            <h5 class="header"> {{$company->streetaddress}} {{$company->postalcode}}</h5>  
            <p class="header">
             <span class="red-text"> <?= $service; ?> </span> is available on <span class="green-text"><?php echo $date; ?></span>
           </p>
    @endforeach

I am getting this error

POST http://127.0.0.1:8234/company 419 (unknown status)

Did you know that this @foreach is a templating engine and not PHP ? and how it is supposed to request the database for companies if you're not calling the database, and also CSS is an interpreted language so saying It compiles the CSS files is incorrect! And finally, yeah you should do the logic in your controller and then pass the result of the view to be rendered to your Ajax! — teeyo
– teeyo, Commented Sep 28, 2017 at 15:25
Yes, I am aware of that. It does not compile something like <?= $service; ?> too. How should I call the database, can you shed some light on that? — Cowgirl
– Cowgirl, Commented Sep 28, 2017 at 15:29
You are using Laravel, an MVC framework, you need to create a route to an action in your controller in this action you will call the database using eloquent and pass the result to the view which should render back HTML. — teeyo
– teeyo, Commented Sep 28, 2017 at 15:31
I am aware of that too. I tried doing that, but in my case, I pass the eloquent to my view, but that specific view needs to get access to the variable in the URL from GET method, it cannot do that. Additionally, I don't want to include the navbar and other things in that view. — Cowgirl
– Cowgirl, Commented Sep 28, 2017 at 15:36
You create your portion of the view without extending the navbar and other componenet and you put only the loop code, I don't get it how you can't access the variable ? you pass the variable to the view, wich means it will be available in the view ! — teeyo
– teeyo, Commented Sep 28, 2017 at 15:38

Abhinav Keshri · Accepted Answer · 2020-06-15 06:32:43Z

143

Laravel 419 post error is usually related with api.php and token authorization

Laravel automatically generates a CSRF "token" for each active user session managed by the application. This token is used to verify that the authenticated user is the one actually making the requests to the application.

Add this to your ajax call

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

or you can exclude some URIs in VerifyCSRF token middleware

 protected $except = [
        '/route_you_want_to_ignore',
        '/route_group/*
    ];

edited Jun 15, 2020 at 6:32

Abhinav Keshri

6451 gold badge7 silver badges23 bronze badges

answered Sep 29, 2017 at 16:42

Dhiraj

2,7672 gold badges13 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

B3none Over a year ago

Worth noting that some people change the meta name attribute. The one in the project I was working on was renamed to _token.

Adnan Rasheed · Accepted Answer · 2018-11-07 12:14:08Z

14

419 error happens when you don`t post csrf_token. in your post method you must add this token along other variables.

answered Nov 7, 2018 at 12:14

Adnan Rasheed

8768 silver badges12 bronze badges

Comments

EXayer · Accepted Answer · 2019-01-31 10:25:30Z

3

Had the same problem, regenerating application key helped - php artisan key:generate

answered Jan 31, 2019 at 10:25

EXayer

1401 silver badge9 bronze badges

Comments

Gorka · Accepted Answer · 2021-07-19 12:20:39Z

2

Step 1: Put the csrf meta tag in head

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <meta http-equiv="X-UA-Compatible" content="ie=edge">
    <meta name="csrf-token" content="{{ csrf_token() }}">
    <title>Document</title>
</head>
<body>

Step 2: Use this ajax format

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script>
    $(document).ready(function(){
      $("#frm").submit(function(e){
        e.preventDefault();
        $.ajaxSetup({
            headers: {
                'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
            }
            });
        $.ajax({
                url:"{{ url('form_submit') }}",
                data:$('frm').serialize(),
                type:'post',
                success: function(result){
                    console.log(result);
                }
        });
      });
    });
</script>

edited Jul 19, 2021 at 12:20

Gorka

2,0891 gold badge19 silver badges34 bronze badges

answered Jul 18, 2021 at 18:06

Agni Sankar Chakraborty

191 bronze badge

1 Comment

Richard Hpa Over a year ago

Can you give some more explanation to your answer rather than just supplying some code. Especially since this is quite a old question and already has an question with quite a few upvotes

Lulceltech · Accepted Answer · 2017-09-28 15:27:18Z

1

You don't have any data that you're submitting! Try adding this line to your ajax:

data: $('form').serialize(),

Make sure you change the name to match!

Also your data should be submitted inside of a form submit function.

Your code should look something like this:

<script>
	$(function () {
		$('form').on('submit', function (e) {
			e.preventDefault();
			$.ajax({
				type: 'post',
				url: 'company.php',
				data: $('form').serialize(),
				success: function () {
					alert('form was submitted');
				}
			});
		});
	});
</script>

edited Sep 28, 2017 at 15:27

answered Sep 28, 2017 at 15:22

Lulceltech

1,72313 silver badges23 bronze badges

2 Comments

teeyo Over a year ago

Still, he does not have any logic in the controller from what I understand :)

Lulceltech Over a year ago

I mean he does have more issues beyond just the AJAX, but he's asking about the AJAX portion of it so i'm just giving him that answer :P

Francisco Isidori · Accepted Answer · 2018-11-06 14:02:48Z

1

I had the same issue, and it ended up being a problem with the php max post size. Increasing it solved the problem.

answered Nov 6, 2018 at 14:02

Francisco Isidori

1611 silver badge6 bronze badges

Comments

eli · Accepted Answer · 2018-12-12 15:25:57Z

1

I received this error when I had a config file with <?php on the second line instead of the first.

answered Dec 12, 2018 at 15:25

eli

1872 silver badges15 bronze badges

Comments

eldorjon · Accepted Answer · 2019-11-21 09:25:28Z

1

You may also get that error when CSRF "token" for the active user session is out of date, even if the token was specified in ajax request.

answered Nov 21, 2019 at 9:25

eldorjon

3423 silver badges15 bronze badges

Comments

Александр Волошиновский · Accepted Answer · 2017-09-28 15:29:07Z

0

In laravel you can use view render. ex. $returnHTML = view('myview')->render(); myview.blade.php contains your blade code

answered Sep 28, 2017 at 15:29

Александр Волошиновский

1,1518 silver badges23 bronze badges

7 Comments

Cowgirl Over a year ago

Even if I do render, it throws 500 error because that view does not have the access to the current URL get for the search form.

Александр Волошиновский Over a year ago

remove this one dataType:'html',

Cowgirl Over a year ago

I get unknown 419 status

Александр Волошиновский Over a year ago

istead of this url : "company.php",, must be a route name, you use the Laravel MVC.

Cowgirl Over a year ago

Yes, I am including the route name in the URL.

|

teeyo · Accepted Answer · 2017-09-28 15:50:54Z

0

In your action you need first to load companies like so :

$companies = App\Company::all();
return view('listing.company')->with('companies' => $companies)->render();

This will make the companies variable available in the view, and it should render the HTML correctly.

Try to use postman chrome extension to debug your view.

answered Sep 28, 2017 at 15:50

teeyo

3,7573 gold badges25 silver badges38 bronze badges

2 Comments

Cowgirl Over a year ago

I am doing that too, I am just using compact function to pass that companies variable to the listing.company view

teeyo Over a year ago

Well the ->render() should render the HTML, you should get a string with HTML inside, no blade tags! I'm sorry I have no Laravel installaion available so I can debug with you.

Witold · Accepted Answer · 2022-06-03 16:24:47Z

0

for me this happens now and then when running a unit test

php artisan config:clear

helped me

answered Jun 3, 2022 at 16:24

Witold

8935 silver badges6 bronze badges

Collectives™ on Stack Overflow

Ajax LARAVEL 419 POST error

11 Answers 11

1 Comment

Comments

Comments

1 Comment

2 Comments

Comments

Comments

Comments

7 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

11 Answers 11

1 Comment

Comments

Comments

1 Comment

2 Comments

Comments

Comments

Comments

7 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related