Question

控制器：UserController.php

<?php
    namespace App\Http\Controllers;
    use Illuminate\Http\Request;
    use DB;
    use App\Http\Requests;
    use App\Http\Controllers\Controller;

    class UserController extends Controller {
        public function insertform() {
            return view('contact-us');
        }
        public function enq()
        {
            $name = $this->input->post('name');
            $phone = $this->input->post('phone');
            $email = $this->input->post('email');
            $msg = $this->input->post('msg');
            $data=array('name'=>$name,"phone"=>$phone,"email"=>$email,"msg"=>$msg);
            DB::table('enquiry')->insert($data);
            echo "Record inserted successfully";
        }
    }

查看：Contact-us.blade.php

<script>
    $(document).ready(function(){
        $("#submit").click(function(e){
            e.preventDefault();
            name = $("#name").val(); 
            phone = $("#phone").val();
            email = $("#email").val();
            msg = $("#msg").val();
            $.ajax({
                type:"POST",
                data:{"name":name,"phone":phone,"email":email,"msg":msg},
                url:"/enq",
                success:function(data){
                    $("#success").html(data);
                }
            });
        });
    });
</script>
<div id="success"></div>
<form method="post">
    <input type="text" placeholder="Your Name" name="name" id="name">
    <input type="text" placeholder="Phone Number" name="phone" id="phone">
    <input type="text" placeholder="Email Adress" name="email" id="email">
    <textarea placeholder="Massege" name="msg" id="msg"></textarea>
    <input type="submit" name="submit" id="submit" value="submit now" class="btn-blue">
</form>

web.php

<?php
Route::get('contact-us', function () {
    return view('contact-us');
});
Route::post('enq','UserController@enq');

我是laravel的新手，在这里，我要在数据库中插入一个简单的表单值。现在，当我单击“提交”按钮时，什么都没有显示。我对此一无所知。那么，我该怎么做？请帮助我。

谢谢

Answer 1

更改您的enq函数

public function enq(Request $request)
       {
           $name  = $request->name;
           $phone = $request->phone;
           $email = $request->email;
           $msg   = $request->msg;
           $data=array('name'=>$name,"phone"=>$phone,"email"=>$email,"msg"=>$msg);
           DB::table('enquiry')->insert($data);
           echo "Record inserted successfully";
       }

并配置CSRF令牌

添加 <meta name="_token" content="{!! csrf_token() !!}"/>在您的头部

然后

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

在您的ajax调用之前添加此代码

Answer 2

在ajax请求中添加其他数据，令牌。

    $.ajax({
        type:"POST",
        data:{"name":name,"phone":phone,"email":email,"msg":msg,"_token":"{{csrf_token()}}"},
        url:"{{URL::to('enq')}}",
        success:function(data){
            $("#success").html(data);
        }
    });

要检查错误，可以按键盘上的F12键并查看控制台。错误显示为红色，如果错误代码为500，则错误与php

如何使用laravel通过jquery ajax将表单值插入数据库？

2 个答案: