Question

我是Laravel 5.3的新手并且正在学习。现在我已经创建了表单，我想通过控制器和模型将表单值插入数据库。现在我从表单到控制器获取所有数据，我想将它传递给模型。但是当我将数据传递给模型时我得到一个错误< / p>

"BadMethodCallException in Controller.php line 82:
Method [savemodel] does not exist"

这是我的查看页面代码：

{{Form::open(array('action' => 'Testing@store', 'method' => 'post'))}}

<div class="form-group">
    {!! Form::label('title', 'Title:', ['class' => 'control-label']) !!}
    {!! Form::text('title', null, ['class' => 'form-control']) !!}
</div>

<div class="form-group">
    {!! Form::label('description', 'Description:', ['class' => 'control-label']) !!}
    {!! Form::textarea('description', null, ['class' => 'form-control']) !!}
</div>

{!! Form::submit('Submitform', ['class' => 'btn btn-primary']) !!}

{!! Form::close() !!}

这是我的Controller（Testing.php）：

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;

use App\TestingModel;
class testing extends Controller
{
    /**
     * Display a listing of the resource.
     *
     * @return \Illuminate\Http\Response
     */
    public function index()
    {
        //
    }

    /**
     * Show the form for creating a new resource.
     *
     * @return \Illuminate\Http\Response
     */
    public function create()
    {
        return view('pages.create');
    }

    /**
     * Store a newly created resource in storage.
     *
     * @param  \Illuminate\Http\Request  $request
     * @return \Illuminate\Http\Response
     */
    public function store(Request $request)
    {
        echo 'hi';
      $storeform=new TestingModel();
      $storeform->savemodel($request);


    }




}

这是我的模型（TestingModel）：

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class TestingModel extends Model
{
   public function savemodel()
   {
       echo 'hello';exit;
   }


}

我不知道我做得对不对。如何从模型中插入数据？请帮帮我

Answer 1

这是我的模态。

`

<?php`

`namespace App;`

`use Illuminate\Database\Eloquent\Model;`

`use App\UserUploads;`
`use App\FileUplods;`

`class FileUplods extends Model`  
`{`   
    `protected $table = 'upload_files';`  
     `protected $fileable = ['user_id', 'origanal_file_name' , 'sortby' , 'new_file_name_after_upload'];`

`}`


and here is my controller

`<?php`

`namespace App\Http\Controllers;`

`use Illuminate\Http\Request;`

`use App\Http\Requests;`

`use App\FileUplods;`

`class ImageController extends Controller`
`{`

    `public function ajaxfileupload(Request $request)`
        `{`   
            `$fileupload = new FileUplods();`
            `$fileupload->user_id = $user_id;`
            `$fileupload->origanal_file_name = $request->origanal_file_name;`
            `$fileupload->sortby = $request->sortby;`
            `$fileupload->new_file_name_after_upload = $request->new_file_name_after_upload;`
            `$fileupload->save();`
        `}`
`}

这是我的控制器尝试这样做会非常有效。

Answer 2

首先在控制器中为模型添加命名空间

use App\TestingModel

然后你的控制器功能就像这样

public function store(Request $request)
{
    //echo 'hi';
  $storeform=new TestingModel;
  $storeform->title = $request->input('title');
  $storeform->description = $request->input('description');
  $storeform->save();
 }

并像这样编辑你的模型

$table = 'table_name';
$fillable = [ 'title', 'description' ];

如果你想从模型中运行函数，那么就像这样编写代码

$storeform = TestingModel::savemodel();

Testing.php第39行中的FatalThrowableError：Class＆＃39; App \ TestingModel＆＃39;找不到laravel 5.3

2 个答案: