java.lang.IllegalArgumentException: 尝试通过 bundle 传递对象时出现非法值。

huangapple go评论100阅读模式
英文:

java.lang.IllegalArgumentException: Ilegal value, when I try to pass object via bundle

问题

以下是翻译好的内容:


我正试图将一个来自Java类"Transfer"的对象传递到Kotlin类"Limit",如果我尝试使用data.name通过"Bundle"获取传递的值,编译器不会报错,但是当我进入这个问题所在的活动时,我该如何通过Bundle()在我的意图中获取一个对象?

public class Transfer {
    
    private void sendBundle(){
        Usermodel doTransfer = fillUserModel();
        Limit.intentTransfer(doTransfer, "France");
    }
    
    
    private UserModel fillUserModel() {
        Usermodel newUserModel = new Usermodel();
        usermodel.setName("Jonas");
        userModel.setAge("30");
        usermodel.setIdNumber("123458");
        userModel.setOccupation("dev");
        userModel.setFormation("CC");
        
        return newUserModel ;
    }

}


class UserModel(

    val name: String? = "",
    val age: String? = "",
    val idNumber: String? = "",
    val occupation: String? = "",
    val formation: String? = "",
)

class Limit {

     private val data: Usermodel by bindBundle(DATA)

     private val country: String by bindBundle(COUNTRY)

     override fun onCreate(savedInstanceState: Bundle?) {

      //here I can get the values ​​using  data.name or data.age 
      //  and android studio does not point out error

}


companion object {
  
    const val DATA = "data";
    const val COUNTRY= "CountryUser";
}


fun intentTransfer (test : UserModel, CountryUser : String)
 : Intent {
        return Intent(context, Limit::class.java).apply {
            putExtras(
                BundleOf(
                    DATA to test,
                    COUNTRY to CountryUser 
                )
            )
        }
    }

OUTPUT
WHEN I ENTER ACTIVITY:

 java.lang.IllegalArgumentException: Ilegal value type android.model.UserModel for key "data"
英文:

I'm trying to pass an object from the class "Transfer" (java) to the class "Limit" (Kotlin), if I try to get the values ​​that come through the "Bundle" using data.name, the compiler does not give an error, but when I enter in the activity of this problem, how can I get an object via Bundle () in my intent?

public class Transfer {
    
    private void sendBundle(){
        Usermodel doTransfer = fillUserModel();
        Limit.intentTransfer(doTransfer, "France");
    }
    
    
    private UserModel fillUserModel() {
        Usermodel newUserModel = new Usermodel();
        usermodel.setName("Jonas");
        userModel.setAge("30");
        usermodel.setIdNumber("123458");
        userModel.setOccupation("dev");
        userModel.setFormation("CC");
        
        return newUserModel ;
    }

}


class UserModel(

    val name: String? = "",
    val age: String? ="",
    val idNumber: String? ="",
    val occupation: String? ="",
    val formation: String? ="",
)

class Limit {

     private val data: Usermodel by bindBundle(DATA)

     private val country: String by bindBundle(COUNTRY)

     override fun onCreate(savedInstanceState: Bundle?) {

      //here I can get the values ​​using  data.name or data.age 
      //  and android studio does not point out error

}


companion object {
  
    const val DATA = "data"
    const val COUNTRY= "CountryUser"
}


fun intentTransfer (test : UserModel, CountryUser : String)
 : Intent {
        return Intent(context, Limit::class.java).apply {
            putExtras(
                BundleOf(
                    DATA to test,
                    COUNTRY to CountryUser 
                )
            )
        }
    }

OUTPUT
WHEN I ENTER ACTIVITY:

 java.lang.IllegalArgumentException: Ilegal value type android.model.UserModel for key "data"

答案1

得分: 6

你需要创建一个实现了 Parcelable 接口的类,该类的对象将通过 Bundle 进行传递。

在你的 gradle 文件中:

apply plugin: 'org.jetbrains.kotlin.android.extensions'

你的数据类应该像这样:

@Parcelize
data class UserModel(

    val name: String? = "",
    val age: String? = "",
    val idNumber: String? = "",
    val occupation: String? = "",
    val formation: String? = ""
) : Parcelable
英文:

You will need to make the class Parcelable whose object you are passing through the Bundle.

In your gradle:

apply plugin: 'org.jetbrains.kotlin.android.extensions'

Your data class should be like:

@Parcelize
data class UserModel(

    val name: String? = "",
    val age: String? ="",
    val idNumber: String? ="",
    val occupation: String? ="",
    val formation: String? ="",
) : Parcelable

</details>



# 答案2
**得分**: 3

你的 `UserModel` 类无法被捆绑包识别为可序列化或可传递。尝试实现其中一个,错误将消失。

<details>
<summary>英文:</summary>

Your `UserModel` class cannot be recognized by bundle as a Serializable or Parcelable. Try to implement one of them and error will disappear.

</details>



# 答案3
**得分**: 2

我花了半天时间来解决这个问题。我的应用因为相同的错误而崩溃。
但是我使用了`kotlinx.serialization`的注解,甚至无法想象我还需要额外提及序列化的事情。
即使AS(Android Studio)也能将这个参数视为可序列化的!
只需在代码中添加`java.io.Serializable`,问题就解决了。

<details>
<summary>英文:</summary>

I spent half day with this issue. My app crashed with same error.   
But I use kotlinx.serialization annotations and could not even imagine that I need additionally mention about Serialization   
Even AS can see this parameter as Serializable!  
Just added `java.io.Serializable` to the code and fixed[![enter image description here][1]][1]


  [1]: https://i.stack.imgur.com/ae3pA.png

</details>



huangapple
  • 本文由 发表于 2020年9月9日 00:21:12
  • 转载请务必保留本文链接:https://go.coder-hub.com/63797893.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定