Android Fragments 示例化ViewModel中的类

wztqucjr  于 2023-06-23  发布在  Android
关注(0)|答案(1)|浏览(198)

此应用程序创建贺卡。流程是在片段#1上选择照片,在片段#2上选择联系人,在片段#3上输入消息,在片段#4上查看、确认并发送。
此应用程序的数据存储在SQLite中(使用Room)。
当前状态是使用bundle/arguments的组合在片段之间共享数据,并保存到SQLite。太乱了
未来状态是将这些片段的所有数据放入现有的ViewModel中,以便所有片段中的所有数据都可用(然后在完成卡后保存到SQLite)。
我想在ViewModel中使用SQLite结构所基于的同一个类--基本上,只要调用ViewModel,就示例化它。这是一个类:

class DBClasses {

    @Entity(tableName="photos")
    data class photos(

        @ColumnInfo(name = "user_id")
        val user_id: String,

        @PrimaryKey(autoGenerate = true)
        val photo_id: Int,

        @ColumnInfo(name = "photo_name")
        val photo_name: String,

        @ColumnInfo(name = "upload_date")
        val upload_date: String,

        @ColumnInfo(name = "status")
        val status: String

    )

    @Entity(tableName="cards")
    data class postcards(

        @ColumnInfo(name = "user_id")
        val user_id: String,

        @PrimaryKey(autoGenerate = true)
        val postcard_id: Int,

        @ColumnInfo(name = "photo_id")
        val photo_id: Int,

        @ColumnInfo(name = "status")
        val status: String,

        @ColumnInfo(name = "message_id")
        val message_id: Int,

        @ColumnInfo(name = "contact_id")
        val contact_id: Int

    )

    @Entity(tableName="contacts")
    data class contacts(

        @ColumnInfo(name = "user_id")
        val user_id: String,

        @PrimaryKey(autoGenerate = true)
        val contact_id: Int,

        @ColumnInfo(name = "status")
        val status: String,

        @ColumnInfo(name = "addressee")
        val addressee: String,

        @ColumnInfo(name = "first_name")
        var first_name: String,

        @ColumnInfo(name = "last_name")
        val last_name: String,

        @ColumnInfo(name = "address")
        var address: String,

        @ColumnInfo(name = "city")
        val city: String,

        @ColumnInfo(name = "state")
        val state: String,

        @ColumnInfo(name = "zip_code")
        val zip_code: String

    )

    @Entity(tableName="messages")
    data class messages(

        @ColumnInfo(name = "user_id")
        val user_id: String,

        @PrimaryKey(autoGenerate = true)
        val message_id: Long,

        @ColumnInfo(name = "status")
        val status: String,

        @ColumnInfo(name = "message")
        val message: String,

        @ColumnInfo(name = "string")
        val string: String,

    )

    @Entity(tableName = "owner")
    data class owner(

        @PrimaryKey(autoGenerate = true)
        val owner_id: Long,

        @ColumnInfo(name = "user_id")
        val user_id: String,

        @ColumnInfo(name = "status")
        val status: String,

        @ColumnInfo(name = "sender")
        val sender: String,

        @ColumnInfo(name = "first_name")
        val first_name: String,

        @ColumnInfo(name = "last_name")
        val last_name: String,

        @ColumnInfo(name = "address")
        val address: String,

        @ColumnInfo(name = "city")
        val city: String,

        @ColumnInfo(name = "state")
        val state: String,

        @ColumnInfo(name = "zip_code")
        val zip_code: String

    )

}

有办法做到这一点吗?否则,我认为我必须在ViewModel中重新创建类,这似乎是不必要的重复。我尝试在ViewModel中示例化现有的类(DBClasses),这允许我这样做,但是当我尝试实际引用片段中的空值时,我会遇到错误。
编辑:为了访问值,我采取了另一种方法,在ViewModel中设置它-请参阅下面的联系人示例:

private val _user_id = MutableLiveData<String>("")
    val user_id: LiveData<String> = _user_id

    private val _contact_id = MutableLiveData<Int>(0)
    val contact_id: LiveData<Int> = _contact_id

    private val _status = MutableLiveData<String>("")
    val status: LiveData<String> = _status

    private val _addressee = MutableLiveData<String>("")
    val addressee: LiveData<String> = _addressee

    private val _first_name = MutableLiveData<String>("")
    val first_name: LiveData<String> = _first_name

    private val _last_name = MutableLiveData<String>("")
    val last_name: LiveData<String> = _last_name

    private val _address = MutableLiveData<String>("")
    val address: LiveData<String> = _address

    private val _city = MutableLiveData<String>("")
    val city: LiveData<String> = _city

    private val _state = MutableLiveData<String>("")
    val state: LiveData<String> = _state

    private val _zip_code = MutableLiveData<String>("")
    val zip_code: LiveData<String> = _zip_code

    fun setUserID(user_id: String) {
        _user_id.value = user_id
    }

    fun setStatus(status: String) {
        _status.value = status
    }

    fun setAddressee(addressee: String) {
        _addressee.value = addressee
    }

    fun setFirstName(first_name: String) {
        _first_name.value = first_name
    }

    fun setLastName(last_name: String) {
        _last_name.value = last_name
    }

    fun setAddress(address: String) {
        _address.value = address
    }

    fun setCity(city: String) {
        _city.value = city
    }

    fun setState(state: String) {
        _state.value = state
    }

    fun setZip(zip_code: String) {
        _zip_code.value = zip_code
    }
dfty9e19

dfty9e191#

通过使用by activityViewModels()初始化视图模型,可以在活动级别中仅使用一个视图模型。这个视图模型将保持现有状态,即使您的片段正在改变。
如果您觉得您的片段中有一个复杂的状态管理,您也可以将它与每个片段视图模型结合起来。

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