从kotlin中的单个函数返回不同对象类型之一

Question

我目前有以下结构：

@Entity
@Table(name = "table_app_settings")
data class AppSetting(
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "app_setting_id")
val id: Long? = null,
@Column(name = "app_setting_name")
val name: String = "",
@Column(name = "app_setting_value")
var value: String = "",
@Column(name = "app_setting_type")
val type: AppSettingType,
)
enum class AppSettingType {
CHAR,
STRING,
BYTE,
SHORT,
INT,
LONG,
DOUBLE,
FLOAT,
BOOLEAN,
}

然后将其与以下内容一起保存到数据库中：

override fun saveAppSetting(setting: AppSetting): DatabaseResult<AppSetting> {
log.info("Saving App Setting ${setting.name} to database.")
return try {
// Attempt to save the entity to the database. If we do not throw an exception, return success.
val savedSetting = appSettingsRepository.save(setting)
DatabaseResult(
code = ResultCode.CREATION_SUCCESS,
entity = savedSetting
)
} catch(exception: DataAccessException) {
log.error("Unable to save App Setting ${setting.name} to database. Reason: ${exception.message}")
DatabaseResult(
code = ResultCode.CREATION_FAILURE
)
}
}

现在，假设我希望将Char类型保存到数据库中，我想我会使用以下内容：

override fun saveAppSetting(name: String, value: Char): DatabaseResult<Char> {
val appSettingResult = saveAppSetting(AppSetting(
name = name,
value = value.toString(),
type = AppSettingType.CHAR,
))
return if(appSettingResult.code != ResultCode.CREATION_FAILURE) {
val entity = getAppSetting<Char>(appSettingResult.entity?.name!!).entity.toString().first()
DatabaseResult(
code = appSettingResult.code,
entity = entity
)
} else {
DatabaseResult(
code = ResultCode.CREATION_FAILURE,
)
}
}

我还认为，为了检索正确的对象类型，我需要执行以下操作：

override fun getAppSetting(name: String): DatabaseResult<Any?> {
log.info("Getting App Setting $name from database.")
val appSetting = appSettingsRepository.findAppSettingByName(name)
return if(appSetting != null) {
log.info("App Setting $name has ID of ${appSetting.id} within the database")
when(appSetting.type) {
AppSettingType.CHAR -> {
DatabaseResult<Char>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.first(),
)
}
AppSettingType.STRING -> {
DatabaseResult<String>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value,
)
}
AppSettingType.BYTE -> {
DatabaseResult<Byte>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toByte(),
)
}
AppSettingType.SHORT -> {
DatabaseResult<Short>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toShort(),
)
}
AppSettingType.INT -> {
DatabaseResult<Int>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toInt(),
)
}
AppSettingType.LONG -> {
DatabaseResult<Long>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toLong(),
)
}
AppSettingType.DOUBLE -> {
DatabaseResult<Double>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toDouble(),
)
}
AppSettingType.FLOAT -> {
DatabaseResult<Float>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toFloat()
)
}
AppSettingType.BOOLEAN -> {
DatabaseResult<Boolean>(
code = ResultCode.FETCH_SUCCESS,
entity = appSetting.value.toBoolean()
)
}
}
} else {
log.error("App Setting $name does not seem to exist within the database.")
DatabaseResult(
code = ResultCode.FETCH_FAILURE
)
}

然而，当我希望使用所述对象时，我仍然需要写以下内容：

val newBarcode = getAppSetting("barcode_value").entity.toString().toInt()

假设我已经"；初始化的"；值为177的CCD_ 2。

如何在不执行.toString.to...()的情况下使函数返回所需内容？

Answer 1

是的，这一切都有可能，这里有一个简化的演示，首先是

import kotlin.reflect.KClass
data class AppSetting(
val id: Long? = null,
val name: String = "",
var value: String = "",
val type: AppSettingType,
)
enum class AppSettingType(val clazz: KClass<out Any>) {
CHAR(Char::class),
STRING(String::class),
INT(Int::class),
}

所以我添加了一个clazz，所以从枚举中我们知道Kotlin类型的

现在是一个模拟你的存储库提取的功能

fun findAppSettingByName(name: String): AppSetting? {
return when(name) {
"Char thing" -> AppSetting(value= "C", type = AppSettingType.CHAR)
"String thing" -> AppSetting(value= "Str", type = AppSettingType.STRING)
"Int thing" -> AppSetting(value= "42", type = AppSettingType.INT)
else -> throw IllegalArgumentException()
}
}

接下来，在函数声明中，我将其与T通用，并为了演示的目的删除了DatabaseResult容器。然后我添加了一个clazz参数，这是将所需类信息携带到函数中的典型Java方式：

fun <T : Any> getAppSetting(name: String, clazz: KClass<T>): T? {
val appSetting: AppSetting? = findAppSettingByName(name)
return appSetting?.let {
require(clazz == appSetting.type.clazz) {
"appSetting.type=${appSetting.type.clazz} mismatched with requested class=${clazz}"
}
when (appSetting.type) {
AppSettingType.CHAR -> appSetting.value.first()
AppSettingType.STRING -> appSetting.value
AppSettingType.INT -> appSetting.value.toInt()
} as T
}
}

as T对于将值强制转换为所需的返回类型很重要——这是未选中的，但when((子句应该创建正确的类型。

现在让我们测试一下：

val c1: Char? = getAppSetting("Char thing", Char::class)
val s1: String? = getAppSetting("String thing", String::class)
val i1: Int? = getAppSetting("Int thing", Int::class)
println("c1=$c1 s1=$s1 i1=$i1")
val c2: Char? = getAppSetting("Char thing")
val s2: String? = getAppSetting("String thing")
val i2: Int? = getAppSetting("Int thing")
println("c2=$c2 s2=$s2 i2=$i2")
}

输出为

c1=C s1=Str i1=42
c2=C s2=Str i2=42

但是c2/s2/i2是如何工作的，最后一部分是这个函数

inline fun <reified T : Any> getAppSetting(name: String) = getAppSetting(name, T::class)

这是具体化的通用参数。。。不需要通过CCD_ 9，因为这可以从接收变量的数据类型中找到。

有很多关于这个高级主题的文章，例如

• https://typealias.com/guides/getting-real-with-reified-type-parameters/
• https://medium.com/kotlin-thursdays/introduction-to-kotlin-generics-reified-generic-parameters-7643f53ba513

现在，我没有完全回答您想要什么，因为您想要接收DatabaseResult<T>包装。可能的是，有一个返回DatabaseResult<T>的函数，您可以从中获得T作为"；clazz"；参数，但我会把它留给其他人来改进：-(但我认为这会让你非常接近。