four-eyes's questions

我有这些界面

interface CommonAnimalProps {
    common: {
        size: number, 
        weight: number,
        origin: string,
    }
}

interface DogProps extends CommonAnimalProps {
    name: string
}

interface CatProps extends CommonAnimalProps {
    furious: boolean
}

interface BirdProps extends CommonAnimalProps {
    canFly: boolean
}

export enum AnimalType {
    dog = "dog",
    cat = "cat",
    bird = "bird",
}

export interface AnimalTypeToAnimalPropsMap {
    [AnimalType.dog]: (DogProps)[],
    [AnimalType.cat]: (CatProps)[],
    [AnimalType.bird]: (BirdProps)[]
}

export type AllAnimalValues<T> = T[keyof T]
export type AllAnimalTypesArray = AllAnimalValues<AnimalTypeToAnimalPropsMap>
export type AllAnimalTypes = AllAnimalValues<AnimalTypeToAnimalPropsMap>[0]
export type AllAnimalTypesArrayDiscriminating = AllAnimalTypes[]

这AllAnimalTypesArrayDiscriminating将导致数组看起来像这样

const foo: AllAnimalTypesArrayDiscriminating = [
    {
        name: "Bello",
        common: {
            size: 10,
            weight: 25,
            origin: "Knowhwere"

        },
    },
    {
        furious: true,
        common: {
            size: 3,
            weight: 5,
            origin: "Anywhere"

        },
    },
    {
        canFly: false,
        common: {
            size: 39,
            weight: 50,
            origin: "Somewhere"

        },
    },    
]

是否可以使用AllAnimalTypesArrayDiscriminating并从中创建一个新的类型/接口（AllAnimalTypesArrayDiscriminatingModified），其中结果数组中的每个元素bar都有另一个 prop（color: string）添加，CommonAnimalProps以便结果如下所示

const bar: AllAnimalTypesArrayDiscriminatingModified = [
  {
    name: "Bello",
    common: {
      size: 10,
      weight: 25,
      origin: "Knowhwere",
      color: 'red',

    },
  },
  {
    furious: true,
    common: {
      size: 3,
      weight: 5,
      origin: "Anywhere",
      color: 'brown',

    }
  }, {
    canFly: false,
    common: {
      size: 39,
      weight: 50,
      origin: "Somewhere",
      color: 'white',

    },
  }
];

请注意，我无法修改原始内容CommonAnimalProps以将其添加color为可选参数。我也无法编辑DogProps, CatProps, BirdProps。我正在寻找一种使用AllAnimalTypesArrayDiscriminating（或任何其他我未排除的类型）来添加此属性的方法。有办法做到这一点吗？

我正在运行版本 15.4 (15F31d)。昨天我安装了一个更新，再次打开 XCode，出现一条消息，如“模拟器需要更新”。我点击了确定，什么也没发生。现在，我所有的模拟器都消失了，我无法找回它们。我已经重新启动了 XCode 和我的电脑。

在这里，在我安装更新之前，我可以选择不同的模拟器

当我点击“管理运行目标...”时，它看起来是这样的。它们都选择“显示运行目标：始终”

Product - Destination在我已经选择的菜单下Show all run destinations。

我不明白的是，我可以启动模拟器应用程序

但是 XCode 无法识别任何模拟器......

当我连接 iPhone 时，它​​没有显示在这里

但我可以在这里看到

我也尝试从命令行运行它

$ xcrun simctl list devices

这给了我

    iPhone Xs (6723AC4A-4B19-46F5-BAE5-6C3DF2BC253B) (Shutdown)
    iPhone SE (3rd generation) (CA9C9B3E-B542-412C-B845-71A0B7D9D797) (Booted)
    iPhone SE (3rd generation) (DB2FF536-D849-4786-B3BD-8FB7068F212A) (Shutdown)
    iPhone 15 (B039DEC7-D015-4313-A7D9-2CB2324E232B) (Shutdown)
    iPhone 15 Plus (797BBF51-A7A9-4F41-A532-424B775AB14D) (Shutdown)
    iPhone 15 Pro (36E9933C-D4FB-4818-B4EA-F3ADE9BC5AFF) (Shutdown)
    iPhone 15 Pro Max (4CDB180C-79F2-4827-B2AB-256F1812FE98) (Shutdown)
    iPad Air (5th generation) (AE3E77BB-137D-42EC-B2C0-2179320216F6) (Shutdown)
    iPad (10th generation) (8FD9CB57-CEBA-4570-B7F0-FDEFAAF20683) (Shutdown)
    iPad mini (6th generation) (0B6E1473-1864-49B0-BB91-4A2C4F692424) (Shutdown)
    iPad Pro (11-inch) (4th generation) (6B9E5069-A52A-4772-96D9-68FFCF496BEC) (Shutdown)
    iPad Pro (12.9-inch) (6th generation) (1E45FFF9-E844-4EFC-9938-59C80F8721D7) (Shutdown)

然后我跑

$ yarn ios --udid "CA9C9B3E-B542-412C-B845-71A0B7D9D797"

返回

error Failed to build iOS project. "xcodebuild" exited with error code '70'. To debug build logs further, consider building your app with Xcode.app, by opening 'Foo.xcworkspace'.
Command line invocation:
    /Applications/Xcode.app/Contents/Developer/usr/bin/xcodebuild -workspace Foo.xcworkspace -configuration Debug -scheme Foo -destination id=CA9C9B3E-B542-412C-B845-71A0B7D9D797

User defaults from command line:
    IDEPackageSupportUseBuiltinSCM = YES


2024-07-25 15:38:32.989 xcodebuild[2255:51186] Writing error result bundle to /var/folders/_j/5k_qtwqn5k39srgfty_40xmc0000gp/T/ResultBundle_2024-25-07_15-38-0032.xcresult
xcodebuild: error: Unable to find a destination matching the provided destination specifier:
        { id:CA9C9B3E-B542-412C-B845-71A0B7D9D797 }

    Ineligible destinations for the "Foo" scheme:
        { platform:iOS, id:dvtdevice-DVTiPhonePlaceholder-iphoneos:placeholder, name:Any iOS Device, error:iOS 17.5 is not installed. To use with Xcode, first download and install the platform }

error Command failed with exit code 1.

为什么我不能再在模拟器上运行我的应用程序了？我的应用程序iOS Deployment target设置为13...

我找不到可以让我回到上一个文件的快捷方式。我知道这一点，但是，这太过精细了。它总是跳转到上一个光标位置。如果我在文件中多次移动光标，这意味着我必须经过所有这些位置才能转到我来自的文件。我希望能够直接转到我来自/之前所在的文件。

有捷径吗？

我正在尝试在 Spring Data JPA 中的查询中使用动态列名和值

interface CustomFooRepository {
    fun findByValueIgnoreCase(query: String, type: RequestType): List<Foo>
}

class CustomFooRepositoryImpl(
    @PersistenceContext private val entityManager: EntityManager
) : CustomFooRepository {

    override fun findByValueIgnoreCase(query: String, type: RequestType): List<Foo> {
        val column = when (type) {
            RequestType.first_name -> "first_name"
            RequestType.last_name -> "last_name"
            RequestType.email -> "email"
        }

        val sql = """
            SELECT 
                *,
                SIMILARITY(:column, :query) AS score
            FROM foo
            WHERE :column iLIKE %:query%
            ORDER BY score DESC NULLS LAST, :column
            LIMIT 20;
        """

        val constructedQuery = entityManager.createNativeQuery(sql, Foo::class.java)
        constructedQuery.setParameter("column", column)
        constructedQuery.setParameter("query", query)
        return constructedQuery.resultList as List<Foo>
    }
}


interface FooRepository : JpaRepository<Foo, Long>, CustomFooRepository

但这让我

org.springframework.dao.InvalidDataAccessResourceUsageException：没有命名参数'：query％'的参数

为什么？

我想从 Spring 实现滚动 API。这样，当用户在前端滚动时，前端每次发出请求时都会加载新数据。

我尝试限制 Spring 返回的结果，并且正在尝试各种方法。

1. 页面请求

    // Service
    fun getOrders(customerId: Long): List<OrderDto> {
        val customer = customerRepository.findById(customerId).getOrElse { return emptyList() }
        val orders: WindowIterator<Order> = WindowIterator.of<Order> { position ->
            orderRepsitory.findAllByCustomerOrderByCreated(
                 customer, 
                 position as KeysetScrollPosition?,
                 PageRequest.of(0, 20)
            )
        }
            .startingAt(ScrollPosition.keyset())
        val ordersResult: MutableList<Orders> = ArrayList<Orders>()
        orders.forEachRemaining { ordersResult.add(it) }
        return ordersResult.map {
            OrderDto(...)
        }
    }


    // Order Respository
    interface OrderRepository : JpaRepository<Order, Long> {
    fun findAllByCustomerOrderByCreated(
        customer: Customer, 
        position: KeysetScrollPosition?,
        pageable: Pageable,
        ): Window<Order>
    }

这总是返回整个表。

2. 调整 JPA 查询

    // Order Respository
    interface OrderRepository : JpaRepository<Order, Long> {
    fun findFirst5ByCustomerOrderByCreated(
        customer: Customer, 
        position: KeysetScrollPosition?,
        ): Window<Order>
    }


    // call to repository in service
    orderRepsitory.findFirst5ByCustomerOrderByCreated(
        customer, 
        position as KeysetScrollPosition?
    )

结果相同，它返回整个表。

3. 使用限制

    // Order Respository
    interface OrderRepository : JpaRepository<Order, Long> {
    fun findAllByCustomerOrderByCreated(
        customer: Customer, 
        limit: Limit,
        ): Window<Order>
    }

    // call to repository in service
    orderRepsitory.findAllByCustomerOrderByCreated(
        customer, 
        position as KeysetScrollPosition?,
         Limit.of(20)
    )

结果相同，它返回整个表。

如何使用 Spring Scroll API 限制结果？

我有两个classesDTO

class FooDto (
    val a: String,
    val b: Double,
)

class BarDto (
    val c: String,
    val d: String,
)

现在，我想定义第三个AnotherDto，将属性的类型设置为BartDto或FooDto。

class AnotherDto (
    val e: FooDto || BarDto
)

这显然是行不通的。我读到，sealed classes它看起来像这样

sealed class Menu{
   data class PIZZA(val name: String, val size:String, val quantity:Int):Menu()
   data class BURGER(val quantity:Int, val size:String): Menu()
   data class CHICKEN(val quantity:Int, val pieces:String):Menu()
}

但是，这需要我再次定义所有变量（例如name: String和quanitity: Int）。这不是我想要的，我只是想重用我之前定义的FooDto而不BarDto重复。

所以这

sealed class FooBar{
   data class Foo(val a: String, val b: Double):FooBar()
   data class Bar(val c: String, val d: String): FooBar()
}

有点毫无意义，因为我FooDto已经BarDto写过并且它们在其他地方使用。我想要这样的东西

sealed class FooBar{
   FooDto:FooBar()
   BarDto: FooBar()
}

这显然是行不通的。

如何重用我已经编写的DTOsFooDto并BarDto创建一个“联合类型”，我可以将其作为数据类型分配给 Kotlin 中的变量？