Question

我想在laravel上更新相关模型（1到1）时改进这段代码，并且知道这是正确，错误，正确但不推荐等等。谢谢大家。

    $user = User::find($id);
    $user->name = $request->input('name');
    $user->email = $request->input('email');
    $user->update();

    $tenant = Tenant::where('user_id', $id)->first();
    $tenant->suite_number = $request->input('suite_number');
    $tenant->update();

Answer 1

首先，您必须在您的用户模型中定义用户和租户的关系：

// User.php
public function tenant()
{
    return $this->hasOne('App\Tenant');
}

由于这是一对一型号，您可以更新您的相关型号，如下所示。确保关系模型也存在。

$user = User::find($id);
$user->name = $request->input('name');
$user->email = $request->input('email');
$user->tenant->suite_number = $request->input('suite_number');
$user->push();

Answer 2

我相信它可能是

// you can omit the input() if u want to
$user = User::find($id);
$user->update([
    'name' => $request->name;
    'email' => $request->email;    
]);

$tenant = Tenant::where('user_id', $id)->first();
$tenant->update([
    'suite_number' => $request->suite_number;
]);

对于急切加载，它将是

$user = User::with('tenant')->find($id);
$user->update([
    'name'                => $request->name;
    'email'               => $request->email;   
    'tenant.suite_number' => $request->suite_number;
]);

或反向关系

$tenant = Tenant::with('user')->where('user_id', $id)->first();
$tenant->update([
    'suite_number' => $request->suite_number;
    'user.name'    => $request->name;
    'user.email'   => $request->email;  
]);

更新属于关系laravel 5.2

2 个答案: