【发布时间】:2020-03-19 12:48:03
【问题描述】:
我是 laravel 的新手。我有一张桌子,但没有这个模型。我正在使用下面来获取记录。
$classes = DB::table('school_classes')->get();
return view('classes', ['allClass' => $classes]);
它在下面返回:
[{"id":1,"register_school_id":1,"class":"I","section":"A","created_at":"2020-03-19 00:00:00","updated_at":"2020-03-19 00:00:00"},
{"id":2,"register_school_id":1,"class":"I","section":"B","created_at":"2020-03-19 00:00:00","updated_at":"2020-03-19 00:00:00"},
{"id":3,"register_school_id":1,"class":"I","section":"C","created_at":"2020-03-19 00:00:00","updated_at":"2020-03-19 00:00:00"}, {"id":4,"register_school_id":1,"class":"I","section":"D","created_at":"2020-03-19 00:00:00","updated_at":"2020-03-19 00:00:00"},
{"id":5,"register_school_id":1,"class":"I","section":"E","created_at":"2020-03-19 00:00:00","updated_at":"2020-03-19 00:00:00"}]
如何分别选择每个项目的值...我尝试了 foreach 但出现错误。 请帮忙。提前致谢。
【问题讨论】:
-
在 Laravel 中,如果你回显对象,它们会自动转换为 json。尝试转储和模具助手
dd($allClass);以查看真实对象。利用模型。它将大大加快您项目的编码速度。
标签: laravel-5 eloquent laravel-query-builder