Question

我在laravel方面没有那么先进。我想使用jquery ajax插入数据而不刷新我也试图在google中找到解决方案。我知道我的帖子看起来像零努力。但我有几天试着这个。这是我的代码。提前致谢。

recruitment.blade.php

<form id="frm" method="POST" action="{{ url('post_job') }}">
     {{ csrf_field() }}

     <input name="title" type="text" required /> <br>
     <input name="nature" type="text" required /> <br>
     <textarea rows="4" cols="50" name="description" required></textarea>

     <input type="submit" name="" value="Update post">
</form>

web.php（路线）

Route::POST('/post_job', 'RecruitmentController@store');

控制器

public function store(Request $request)
{
 $recruitment = new recruitment() ;
 $recruitment->title = $request->get('title');
 $recruitment->nature = $request->get('nature');
 $recruitment->description = $request->get('description');
 $recruitment->save();

 Session::flash('msg','Post done successfully');
 return redirect('/recruitment');
 }

的Ajax

<script>

function saveToDatabase()
{

    select = $(this).serialize();
    $('#frm').live("change", function ()
    {
        // POST to php script
        $.ajax
        ({
            type: 'POST',
            url: '/post_job',
            data:{selected:select}
        }).then(function(data){alert(data)});
    });
}

$(document).ready(function()
{
    saveToDatabase();
});

Answer 1

控制器：有关Json的回复，请参阅doc

public function store(Request $request)
{
 $recruitment = new Recruitment;
 $recruitment->title = $request->get('title');
 $recruitment->nature = $request->get('nature');
 $recruitment->description = $request->get('description');
 $recruitment->save();

 // Session::flash('msg','Post done successfully');
 // return redirect('/recruitment');
   return response()->json( [ 'msg' => 'Post done successfully' ] );
 }

头标记

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

的Ajax 有关ajax请求，请参阅X-CSRF-TOKEN

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

function saveToDatabase() {
    $( '#frm' ).submit( function( e ) {
        e.preventDefault();
        var select = $( this ).serialize();
        // POST to php script
        $.ajax( {
            type: 'POST',
            url: '/post_job',
            data: select
        }).then( function( data ) {
            alert( data )
        } );
        return false;
    } );
}

$(document).ready(function()
{
    saveToDatabase();
});
</script>

在laravel 5.4中使用ajax插入数据

1 个答案: