我有以下路線：Route::apiResource('payments', 'PaymentController', ['except' => ['store', 'destroy']]);Route::get('users/{user}/payments', [ 'as' => 'users.payments', 'uses' => 'PaymentController@index',]);這個控制器：class PaymentController{ public function index(Request $request, User $user) { // Initialize query builder $query = Payment::query(); // If a user was provided, return all payments belonging to that user // if condition is satisfied if ($user) { if (condition) { $query = $customer->payments(); } else { // Some code that causes no results $query->whereNull('id'); // <----- This hits the database } } return PaymentResource::collection($query->paginate(10)); }}如果您點擊它，則應返回所有用戶進行的所有付款。/payments如果命中它，則僅當為 true 時，它才應返回用戶進行的所有付款，否則應返回空列表。/users/id/paymentscondition此代碼有效，但是是實際命中數據庫的解決方法。$query->whereNull('id');有沒有辦法避免命中數據庫并仍然返回空列表？
2 回答
猛跑小豬
TA貢獻1858條經驗 獲得超8個贊
您可以避免命中數據庫，但仍然使用初始化的變量返回空列表到空集合或數組
class PaymentController
{
public function index(Request $request, User $user)
{
// Initialize query builder
$query = Payment::query();
// If a user was provided, return all payments belonging to that user
// if condition is satisfied
$return = collect();
if ($user) {
if (condition) {
$return = $customer->payments()->paginate(10);
}
}
return PaymentResource::collection($return);
}
}
縹緲止盈
TA貢獻2041條經驗 獲得超4個贊
我試圖在函數中有一個單一的返回，但更容易遵循@Chin梁給出的建議（謝謝！
class PaymentController
{
public function index(Request $request, User $user)
{
// Initialize query builder
$query = Payment::query();
// If a user was provided, return all payments belonging to that user
// if condition is satisfied
if ($user) {
if (condition) {
$query = $customer->payments();
} else {
// Some code that causes no results
return PaymentResource::collection(collect());
}
}
// Other code here, for example a where statement
// depending on what is passed in $request
return PaymentResource::collection($query->paginate(10));
}
}
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