Laravel displaying/updating 来自 DB 的数据在仅查看状态下显示数据 "active" 和 "deactive" 按 $id 查找

Question

我尝试了下面的代码，但它不起作用。

在此代码中，当我在浏览器中输入 "DELETED"、"id" 号码时 "URL" 它也显示 "DELETED" 状态用户详细信息，我只想显示和访问"ACTIVATE" AND "DEACTIVATE" users only if I entered "DELETED" Users "id" in "URL" 它应该显示错误：

public function staff_status($id)
{
    $user = User::where('status', '=', 'ACTIVATE')->orWhere('status', '=', 'DEACTIVATE')->find($id);

    return view('home', compact('user'));
}

public function status_update(Request $request, $id)
{
    $user = User::where('status', '=', 'ACTIVATE') > orWhere('status', '=', 'DEACTIVATE')->find($id);

    $user['status'] = $request->status;

    $user->save();

    return back()->with('success', 'SUBMITED SUCCESSFULLY .');
}

Answer 1

正如@Chay22 提到的，您应该添加另一个 where 条件

只需用此替换您的代码

public function staff_status($id)
{
    $user = User::where('status', '!=', 'DELETED')
                  ->where(function($query){
                      $query->where('status','=','ACTIVATE');
                      $query->orWhere('status','=','DEACTIVATE');
                  })->find($id);
    if (! $user) {
        return // return your error here
    }

    return view('home',compact('user'));
}

public function status_update(Request $request, $id)
{
    $user = User::where('status', '!=', 'DELETED')
                  ->where(function($query){
                      $query->where('status','=','ACTIVATE');
                      $query->orWhere('status','=','DEACTIVATE');
                  })->find($id);

    if ($user) {
        $user->status = $request->status;
        $user->save();

        return back()->with('success','SUBMITED SUCCESSFULLY .');
    } else {
        return // return your error here
    }
}

Laravel displaying/updating 来自 DB 的数据在仅查看状态下显示数据 "active" 和 "deactive" 按 $id 查找

Laravel displaying/updating data from DB show data in view only status "active" and "deactive" find by $id

laravel

laravel-5.8