Laravel不返回页面

Question

我通过将变量传递给ajax并在控制器中使用它来将数据保存到数据库中。

保存工作正常，它将数据保存到数据库中，但问题是显示保存的页面。数据库有2个字段，名称和html。名称显然是网站的名称，而html是纯粹的html代码。因此，当用户保存页面名称“随机”我想用本地主机/随机

错误消息我得到的是显示HTML页面，他说：

试图获得的非对象（View属性：C ：\ XAMPP \ htdocs中\ fyproject \资源\意见\布局\ website.blade.php）

这是我迄今所做的：

阿贾克斯：

var web_name; 
function updateDatabase(newCode, name_website) 
{ 
    code2 = document.getElementById("content-link2").innerHTML; 
    web_name = ($('#website_name').val()); 
    // make an ajax request to a PHP file 
    // on our site that will update the database 
    // pass in our lat/lng as parameters 
    $.post('http://localhost/template', { 
      _token: $('meta[name=csrf-token]').attr('content'), 
      newCode: (code2), 
      name_website: (web_name), 
     }) 
     .done(function() { 
     }) 
     .fail(function() { 
      alert("error"); 
     }); 
} 

控制器：

 public function postDB(Request $request) { 
     $newName = $request->input('name_website'); 
     $newLat = $request->input('newCode'); 
     $websites = new Website(); 
     $websites->name = $newName; 
     $websites->html = $newLat; 
     $websites->save(); 
     $name = $newName; 
     return redirect($name); 
    } 
    public function website($name) { 
    $website = Website::where('name', $name)->first()->html; 

    // Render resources/views/template.blade.php or any view you want 
    // and pass the data. E.g. $website, so you can access $website->html in your view. 
    return view('layouts/website', compact('name')); 
} 
} 

路线：

Route::get('home', 'BuilderController@homepage'); 
Route::get('template', 'BuilderController@templates'); 
Route::post('template', 'BuilderController@postDB'); 
Route::get('logout', 'BuilderController@getLogout'); 
Route::get('/{name}', 'BuilderController@website'); 

website.blade.php：

@extends('layouts.master') @section('title', 'Website Builder') @section('content') 
<meta name="csrf-token" content="{{ csrf_token() }}" /> 


{{html_entity_decode($name->html)}} 
<script src="https://unpkg.com/axios/dist/axios.min.js"></script> 
<script> 
    axios.get('/localhost/name').then(
     html => document.querySelector('html').innerHTML = html  
    ); 
</script> 
</html> 
@endsection @show 

这是怎么回事？

Answer 1

问题是，当你需要传递$网站时，你传递的变量名称为你的视图。此外，您的Eloquent查询中有一个错误。

试图改变以下，

$name->html // Change this 
$website->html // to this 
return view('layouts/website', compact('name')); // Change this 
return view('layouts/website', compact('website')); // To this 
$website = Website::where('name', $name)->first()->html; // Change this 
$website = Website::where('name', $name)->first(); // To this 

Answer 2

首先而不是在postDB重定向，你必须重定向通过JavaScript

控制器

public function postDB(Request $request) { 

    $website = new Website(); 
    $website->name = $request->input('name_website'); 
    $website->html = $request->input('newCode'); 
    $website->save(); 

    return response()->json(['url' => url($website->name)]); 
} 

的Javascript

$.post('http://localhost/template', { 
    _token: $('meta[name=csrf-token]').attr('content'), 
    newCode: (code2), 
    name_website: (web_name), 
}, function(response){ 
    // Redirecting to the new page here. 
    window.location = response.url; 
}) 
.done(function() { 

}) 
.fail(function() { 
    alert("error"); 
}); 

控制器

public function website($name) { 
    $website = Website::where('name', $name)->first(); 
    return view('layouts/website', compact('website')); 
} 

现在，您可以通过访问$网站变量像$网站 - > HTML