在laravel中使用ajax提交表单时找不到404

Question

I am new with laravel, I am having an issue when posting my data to controller in laravel while using ajax. My view contains a select box which is populated from database when a user select a value from select box i am trying to call ajax which will return details of salary of the user.The issue is it gives me 404 not found error.

The Code for the controller file is shown below.

    this function is defined inside a controller

    public function getPostSalary()
        {
            echo "ajax called";
            return 'true';
        }

The Routes file is

Route::post('company/salary-user', 'CompanyController@getPostSalary');

this is my ajax code from where the controller is being calleD

<script>
    $(document).ready(function() {
        $('#employee').change(function () {
            var values = $('#employee').val();
            if (values == '') {
                $.pnotify({
                    title: 'Message',
                    text: 'Please Select a User.',
                    type: 'error',
                    delay: 3000
                });
                return false;
            }
            var csrf = $('#csrf').val();
            $.ajax({
                url: '{!! URL::to("company/salary-user")!!}',
                type: "POST",
                data: { user_id: values, _token: csrf },
                cache: false,
                beforeSend: function () {

                },
                success: function (data) {
                    console.log(data);
                    $('#loader').hide();
                }
            });
        }); 
    });
</script>

Can someone help me to figure out what is the issue and what should be done to call my function.

Answer 1

Single quotes is for string so your url is not generating as you expecting. Use like this

  url : "{!! URL::to('company/salary-user')!!}",

Answer 2

replace this line:

{!! URL::to("company/salary-user")!!}

by :

{{ url('/company/salary-user') }}

or

{{ route('/company/salary-user') }}

Answer 3

You can use direct url like

 $.ajax({
            url:'company/salary-user',
            type: "POST",
            data: { user_id: values, _token: csrf },
            cache: false,
            beforeSend: function () {

            },
            success: function (data) {
                console.log(data);
                $('#loader').hide();
            }
        });