如何在這個擴展函式中回傳正確的型別

我知道我可能做錯了，但我有點堅持在這個函式中回傳正確的型別。

 val dbFields = dbInfo.equipmentList

        fun DbEquipmentWithMediaList.toApiUpdateDeviceList(): ApiUpdateDeviceList {
            val newUpdate: ApiUpdateDeviceList
           newUpdate = dbFields.forEach { ApiUpdateDeviceList(
                    deviceEquipmentId = it.sSEquipmentDbInfo.deviceEquipmentId,
                    secondaryDeviceId = it.sSEquipmentDbInfo.secondaryDeviceId,
                    deviceInstanceId = it.sSEquipmentDbInfo.deviceInstanceId.toString(),
                    deviceSerialNumber = it.sSEquipmentDbInfo.deviceSerialNumber,
                    devicePartNumber = it.sSEquipmentDbInfo.devicePartNumber,
                    deviceManufacturerName = it.sSEquipmentDbInfo.deviceManufacturer,
                    deviceInstallationDate = DateUtil.dateToStringUTCSiteSurvey(it.sSEquipmentDbInfo.deviceInstallationDate),
                    deviceLocation = it.sSEquipmentDbInfo.locationInfo,
                    deviceTechnicianNotes = it.sSEquipmentDbInfo.deviceTechnicianNotes
            )
            }
        }

我需要在 forEach 回圈之后將該函式作為 ApiUpdateDeviceList 回傳。任何幫助表示贊賞，謝謝。

這是有關 DbEquipmentWithMediaList 類和 ApiUpdateDeviceList 兩個類的更多資訊

對于 DbEquipmentWithMediaList

data class DbEquipmentWithMediaList(
    @Embedded val sSEquipmentDbInfo: SSEquipmentDbInfo,
    @Relation(
        parentColumn = "id",
        entityColumn = "media_id",
        entity = SurveySiteDbMedia::class
    )
    val mediaList: List<SurveySiteDbMedia>
)

data class SSEquipmentDbInfo(
        @PrimaryKey(autoGenerate = true)
    @ColumnInfo(name = "id") val id: Long,
        @ColumnInfo(name = "equipment_id", index = true) val equipmentId: Long,
        @ColumnInfo(name = "actual_device_equipmentId") val deviceEquipmentId: Int?,

        @ColumnInfo(name = "secondary_device_id") val secondaryDeviceId: Int,
        @ColumnInfo(name = "secondary_device_checked_status") val secondaryDeviceCheckedStatus: Int?,
        @ColumnInfo(name = "device_instance_id") val deviceInstanceId: Int?,
        @ColumnInfo(name = "media_names_list") val mediaNames: List<String>? = null,
        @ColumnInfo(name = "device_serial_number") var deviceSerialNumber: String? = null,
        @ColumnInfo(name = "device_part_number") var devicePartNumber: String? = null,
        @ColumnInfo(name = "device_manufacturer") var deviceManufacturer: String? = null,
        @ColumnInfo(name = "device_installation_date") var deviceInstallationDate: Date? = null,
        @ColumnInfo(name = "location") var locationInfo: String? = null,
        @ColumnInfo(name = "device_technician_notes") var deviceTechnicianNotes: String? = null,
        @ColumnInfo(name = "image_start_count") val imageStartCount: Int?,
        @ColumnInfo(name = "video_start_count") val videoStartCount: Int?,
        @ColumnInfo(name = "device_equipment_id") val deviceEquipmentMediaId: Int? = 0,
        @ColumnInfo(name = "device_update_mode") val isDeviceUpdateMode: Boolean
)

對于 ApiUpdateDeviceList

data class ApiUpdateDeviceList(
        @SerializedName("DeviceEquipmentId") val deviceEquipmentId: Int?,
        @SerializedName("DeviceId") val secondaryDeviceId: Int?,
        @SerializedName("DeviceInstance") val deviceInstanceId: String?,
        @SerializedName("SerialNumber") val deviceSerialNumber: String?,
        @SerializedName("PartNumber") val devicePartNumber: String?,
        @SerializedName("ManufacturerName") val deviceManufacturerName: String?,
        @SerializedName("InstallationDate") val deviceInstallationDate: String?,
        @SerializedName("Location") val deviceLocation: String?,
        @SerializedName("TechnicianNotes") val deviceTechnicianNotes: String?
)

uj5u.com熱心網友回復：

所以，我會重命名這個類，讓它的名字中沒有 List，因為它只代表一個專案：

ApiUpdateDeviceList -> ApiUpdateDevice

由于這是一個一對多的關系，你從這個擴展函式回傳的型別應該是List<ApiUpdateDevice>.

我實際上會首先將一對一的關系分解為它自己的擴展函式，使用with范圍函式來簡化代碼：

fun SurveySiteDbMedia.toApiUpdateDevice(): ApiUpdateDevice = with(sSEquipmentDbInfo) {
    ApiUpdateDevice(
        deviceEquipmentId = deviceEquipmentId,
        secondaryDeviceId = secondaryDeviceId,
        deviceInstanceId = deviceInstanceId.toString(),
        deviceSerialNumber = deviceSerialNumber,
        devicePartNumber = devicePartNumber,
        deviceManufacturerName = deviceManufacturer,
        deviceInstallationDate = DateUtil.dateToStringUTCSiteSurvey(deviceInstallationDate),
        deviceLocation = locationInfo,
        deviceTechnicianNotes = deviceTechnicianNotes
    )
}

您可以使用該map函式而不是forEach從另一個串列中生成一個串列。所以我們用它和上面的函式來簡潔地創建你的串列轉換：

fun DbEquipmentWithMediaList.toApiUpdateDeviceList(): List<ApiUpdateDevice> =
    dbFields.map { it.toApiUpdateDevice() }

標籤：安卓算法科特林

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