在 Laravel 5 中使用 Ajax 并返回 json 数组

Question

I am new to "AJAX" and I have been trying to send a request "ONSELECT" using "AJAX" and receive a "JSON" response in "laravel 5".

Here is my View

<select>
<option data-id="a" value="a">a</option>
<option data-id="b" value="b">b</option>
<option data-id="c" value="c">c</option>
</select>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script>

<script type="text/javascript">
$('select').change(function(){
var data = $(this).children('option:selected').data('id');

$.ajax({
    type    :"POST",
    url     :"http://localhost/laravel/public/form-data",
    dataType:"html",
    data    :{ data1:data },

    success :function(response)
    alert("thank u");
    }),
});
</script>

Here is my Controller to receive ajax request

public function formdata(){
    $data = Input::get('data1');

    //somecodes

    return Response::json(array(
                    'success' => true,
                    'data'   => $data
                )); 
}

Here is my Route

 Route::post('form-data',array('as'=>'form-data','uses'=>'FormController@formdata'));

I also have tried to change the URL of ajax with just only form-data and {{Url::route('form-data')}} .

Answer 1

Laravel 5 uses csrf token validation for security reasons....try this...

1. In routes.php

    route post('form-data', 'FormController@postform');

2. In master layout file

   <meta name="csrf-token" content="{{ csrf_token() }}" />

3.  var CSRF_TOKEN = $('meta[name="csrf-token"]').attr('content'); 

    $.ajax({ url: '/form-data/', type: 'POST', data: {_token: CSRF_TOKEN}, dataType: 'JSON', success: function (data) { console.log(data); } });

Answer 2

Add error callback to your ajax request to find if an error is thrown,

$.ajax({
  type    :"POST",
  url     :"http://localhost/laravel/public/form-data",
  dataType:"json",
  data    :{ data1:data },
  success :function(response) {
    alert("thank u");
  },
  error: function(e) {
    console.log(e.responseText);
  }
});

its better to use console.log() to see detailed information even if the response is a json string. Try the code and let us know if something is logged to the browser console

Answer 3

Your jQuery code has syntax error in success callback, that's why its not making any post request to Laravel, please try below javascript it will work

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.0.3/jquery.min.js"></script> <select> <option data-id="a" value="a">a</option> <option data-id="b" value="b">b</option> <option data-id="c" value="c">c</option> </select> <script type="text/javascript"> $(function () { $('select').on('change', function (e) { var data = $(this).children('option:selected').data('id'); $.ajax({ type :"POST", dataType:"json", url :"http://localhost/laravel/public/form-data", data :{ data1:data }, success :function(response) { alert("thank u"); } }); }); }) </script>

In Laravel , you can just return array or object and it will automatically convert it to json response

return ['success' => true, 'data' => $data];

Answer 4

You made error in code ,please write it properly.

$.ajax({
    type    :"POST",
    url     :"http://localhost/laravel/public/form-data",
    dataType:"json",
    data    :{ data1:data },
    success :function(response){
    alert("thank u");
    }
});

Update

I just saw your Returning datatype is json , so use

 dataType:"json",

or

 dataType:"jsonp",

Answer 5

The problem was type should be "GET" instead of "POST" and

route get('form-data', 'FormController@postform');

Thank U every one for your help

Answer 6

如果您发送所有表单数据则更好，然后在 ajax 中使用data: $(this).serialize() ，并在表单内部使用{{ csrf_field() }}