编辑现有帖子时 ::find() 中的 Laravel 未定义变量

Question

我创建了一个功能来编辑我的 laravel 网站上现有的 post。但是，当我尝试提交时该网站出现错误。它声明了一个 Undefined 变量：slug 如果您在下面看到我的示例，它就没有意义，因为我的 Post 日志的响应显示了目标文件中的变量。

 public function adminEditPostsPost(Request $request)
     {


        $user = Auth::user()->first();
        // Log::info($request->all());

        $update = false;

       // $token = $request->_token;
        $token = $request->session()->token();

          $postId = $request->input('pxp-submit-comm-id');
          $title = $request->input('pxp-submit-comm-title');
          $overview = $request->input('pxp-submit-comm-overview');
          $post_status = $request->input('pxp-submit-post-status');
          //$images_array = $request->input('pxp-submit-comm-images-array');
          $photo = $request->input('pxp-submit-comm-photo-featured');

          // Log::info($slug);

        $id = $postId;
        $prop_id = 0;

        if ($request->is('admin/*')) {

               $posts = Posts::find($postId);               
               $posts->$title = $title;
               Log::info($posts);
               $posts->$slug = $request->input('pxp-submit-comm-slug');
               $posts->$content = $overview;
               $posts->$featured_url = $photo;
               $posts->$post_status = $post_status;
                

                $posts->save();
         
            }


        $photos = Photos::where('enabled', '<=', '4')->get()->sortByDesc("created_at");
        $posts = Posts::orderBy('created_at', 'DESC')->where('post_status', '<=', '1')->get();

        return view('admin.posts')->with('photos', $photos)->with('posts', $posts);


     }

这是 post 变量的日志：

{
  "id": 8,
  "slug": "two-homes-now-under-construction-for-late-spring-move-in-dates",
  "title": "Two Homes NOW Under Construction for Late Spring Move-in Dates",
  "featured_url": "146",
  "content": "<p></p>",
  "excerpt": null,
  "category": null,
  "tags": null,
  "created_at": "2021-02-16T15:49:44.000000Z",
  "updated_at": "2021-02-16T15:49:44.000000Z",
  "post_status": "1",
  "Two Homes NOW Under Construction for Late Spring Move-in Dates": "Two Homes NOW Under Construction for Late Spring Move-in Dates"
}

在对象的底部，我注意到它有“两栋正在建设中的房屋，以备晚 Spring 入住日期”作为关键，这对我来说没有意义，我想知道是否我的代码中保存了一些本不应该保存的内容，或者由于它不在数据库中，所以一开始是如何保存的。

Answer 1

如果 $title 变量包含 notVariable，则 $posts->$title 等同于 $posts->notVariable。所以你只需要在 $posts key/attribute.

上删除 $

改变

$posts->$title = $title;
Log::info($posts);
$posts->$slug = $request->input('pxp-submit-comm-slug');
$posts->$content = $overview;
$posts->$featured_url = $photo;
$posts->$post_status = $post_status;

到

$posts->title = $title;
Log::info($posts);
$posts->slug = $request->input('pxp-submit-comm-slug');
$posts->content = $overview;
$posts->featured_url = $photo;
$posts->post_status = $post_status;

Answer 2

PHP 允许使用 variable variables as well as using variables as object keys (see this post 以获得更多解释）。您似乎无意中采用了这一原则。

在您设置的 Log::info() 语句之前

$posts->$title = $title

在 PHP 中，这实际上是将键设置为 $titles 值，这解释了对象中的“现在正在建设的两栋房屋，晚些 Spring 入住日期”键.您没有收到 Undefined variable 错误 $title 因为该变量存在。您收到 Undefined variable: slug 错误，因为您试图使用不存在的变量 $slug、$content 等来设置其他键.您需要更新代码以删除箭头后的 $。

$posts->title = $title;
$posts->slug = $request->input('pxp-submit-comm-slug');
$posts->content = $overview;
$posts->featured_url = $photo;
$posts->post_status = $post_status;

编辑现有帖子时 ::find() 中的 Laravel 未定义变量

Lavavel Undefined Variable in ::find() when editing existing posts

php

laravel

post

laravel-7