Question

我是Laravel框架中的新手，我正在尝试构建一个管理数据库的工具。我已经创建了我的表，我设法从视图中填充它，我没有得到的是将表显示回视图。我构建了两个函数，一个只显示表单'Project_create@showpform'，另一个应该使用插入的数据'Project_create@showtable'更新相同的视图。我收到错误："Undefined variable: project_data (View:/var/www/tcc/src/resources/views/form_project.blade.php)"。

在这里你可以看到我的路线：

Route::get('/', 'DataController@overview');

Route::get('/myproject', 'Project_create@showpform');
Route::post('/myproject', 'Project_create@showtable');

Route::get('/upload','DataController@showform');
Route::post('/upload', 'DataController@read_xsl');

Route::get('/scrape', 'DataController@scrape');

这是我的Project_create控制器：

<?php

namespace App\Http\Controllers;
use Illuminate\Http\Request;
use Illuminate\Support\Facades\Input;

class project_create extends Controller
{
    public function showpform()
    {
        return view('form_project');
    }

    public function showtable(Request $request)
    {
        $name = $request->input('name');
        $description = $request->input('description');

        \DB::table('projects')->insert(
            [
                'project_id' => 1,
                'name' => $rawUrl,
                'description' => $hashedUrl,
                'created_at' =>  \Carbon\Carbon::now(),
            ]);

        $project_data = \DB::table('projects')->select('*')->get();
        return View::make('form_project', compact('project_data'));
    }
}

以下是我的观点：

<html lang="en">
    <head>
        <title>File Upload</title>
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.js"></script>
        <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">
        <link rel="stylesheet" href="https://www.w3schools.com/w3css/4/w3.css">
    </head>
<body>

    <div class="container">
        <h3 class="jumbotron">Create here your project</h3>

        <form class="w3-container w3-light-grey" action={{action('DataController@showform')}} enctype="multipart/form-data">
        {{csrf_field()}}
            <p>
            <label>Project Name</label>
            <input class="w3-input w3-border w3-round" name="name" type="text"></p>
            <p>
            <label>Project Description</label>
            <input class="w3-input w3-border w3-round" name="description" type="text"></p>

        <button type="submit" class="btn btn-primary" style="margin-top:10px">Create Project</button>
        </form>

    </div>

    <table>
        <thead>
            <tr>
                <th> id</th>
                <th> name</th>
                <th> description</th>
                <th> created_at </th>
            </tr>
        </thead>
        <tbody>
             @foreach($project_data as $data)
              <tr>
                  <td> {{$data->id}} </td>
                  <td> {{$data->name}} </td>
                  <td> {{$data->description}} </td>
                  <td> {{$data->created_at}} </td>
              </tr>
             @endforeach
       </tbody>
   </table>


</body>
</html>

感谢您的帮助。

Answer 1

您的视图有一行

@foreach($project_data as $data)

但是当你打电话时

public function showpform()

你没有通过

$project data

喜欢

public function showtable(Request $request)

...

return View::make('form_project', compact('project_data'));

在Blade文件中添加isset功能。 e.g。

@if(isset($project_data)) @foreach($project_data as $data) <tr> <td> {{$data->id}} </td> <td> {{$data->name}} </td> <td> {{$data->description}} </td> <td> {{$data->created_at}} </td> </tr> @endforeach @endif

Laravel从视图中的表单填充数据库表并将其显示在同一视图中

1 个答案: