我尝试了一个不在数据中的段,并收到了类似这样的错误: 试图获取非对象的属性“ id”
videodetail.php
public function detail($slug)
{
$video = Video::where('slug', $slug)->first();
$blogKey = 'blog_' . $video->id;
if(!Session::has($blogKey)) {
$video->increment('view_count');
Session::put($blogKey, 1);
}
$randomvideo = Video::whereHas('categories', function ($q) use ($video) {
return $q->whereIn('name', $video->categories->pluck('name'));
})
->where('id', '!=', $video->id)
->approved()
->published()
->take(8)->get();
return view('video.detail', compact('video', 'randomvideo'));
}
我在类别中收到此错误“在null上调用成员函数video()”
public function videoByCategory($slug)
{
$category = Category::where('slug', $slug)->first();
$videos = $category->videos()->approved()->published()->paginate(21);
return view('category.index', compact('category', 'videos'));
}
答案 0 :(得分:1)
将检查添加到您的类别。
public function videoByCategory($slug)
{
$category = Category::where('slug', $slug)->first();
if ($category) {
$videos = $category->videos()->approved()->published()->paginate(21);
return view('category.index', compact('category', 'videos'));
} else {
// whatever you want to do
}
}