Question

我正在尝试通过Room Persistence库将数据库添加到我的android应用中，并且出现此错误：

错误：实体和Pojos必须具有可用的公共构造函数。您可以有一个空的构造函数或一个其参数与字段匹配的构造函数（按名称和类型）。尝试了以下构造函数，但它们不匹配： User（int，java.lang.String，java.lang.String，int，int，int，java.lang.String）-> [param：id->匹配字段：unmatched，param：name->匹配字段：unmatched ，param：gender->匹配字段：unmatched，param：age->匹配字段：unmatched，param：weight->匹配字段：unmatched，param：height->匹配字段：unmatched，param：workout->匹配字段：unmatched ]

这是我的代码：

    @Entity
    public class User {

@PrimaryKey
private int userId;
private String userName;
private String userGender;
private int userAge;
private int userWeight;
private int userHeight;
private String workoutPlan;


public User(int id, String name, String gender, int age, int weight, int height, String workout) {

    this.userId = id;
    this.userName = name;
    this.userGender = gender;
    this.userAge = age;
    this.userWeight = weight;
    this.userHeight = height;
    this.workoutPlan = workout;

} ...

有人可以告诉我我做错了什么或错过了什么吗？

Answer 1

请更改参数名称，使其与实体属性匹配。

  public User(int userId, String userName, String userGender, int userAge, int userWeight, int userHeight, String workoutPlan) {
    this.userId = userId;
    this.userName = userName;
    this.userGender = userGender;
    this.userAge = userAge;
    this.userWeight = userWeight;
    this.userHeight = userHeight;
    this.workoutPlan = workoutPlan;
  } ...

为了保持持久性，它在Room中使用JavaBeans约定。了解更多信息： https://developer.android.com/training/data-storage/room/defining-data#java

Answer 2

您也可以尝试以下操作：-

@Ignore
public User(int id, String name, String gender, int age, int 
      weight, int height, String workout) {
   this.userId = id;
   this.userName = name;
   this.userGender = gender;
   this.userAge = age;
   this.userWeight = weight;
   this.userHeight = height;
   this.workoutPlan = workout;
 }

 public User(String name, String gender, int age, int 
      weight, int height, String workout) {
   this.userName = name;
   this.userGender = gender;
   this.userAge = age;
   this.userWeight = weight;
   this.userHeight = height;
   this.workoutPlan = workout;
 }

Answer 3

我只是通过添加一个空的构造函数来解决这个问题。 public User(){}

Answer 4

我也遇到了这个错误。我的构造函数中只有我的一个属性不匹配，我发现它相当混乱。我的变量（在构造函数之外）被命名为mURL，我试图将其与构造函数中的url匹配。

相反，我应该将外部变量设置为mUrl。是的，最后两个字母大写给了我这个错误。设置this.mUrl = url代替this.mURL = url解决了该错误。

Answer 5

科特琳：

@Entity(tableName = "t_article_tabs")
data class WxArticleTabsEntity(
    @ColumnInfo(name = "tabId") @PrimaryKey @SerializedName("id") val id: Int?,
    @ColumnInfo(name = "tabName") @SerializedName("name") val name: String?,
    ...
    @Ignore  @SerializedName("children") val children: List<Any>?,
)

更改为：

@Entity(tableName = "t_article_tabs")
data class WxArticleTabsEntity(
    @ColumnInfo(name = "tabId") @PrimaryKey @SerializedName("id") val id: Int?,
    @ColumnInfo(name = "tabName") @SerializedName("name") val name: String?,
    ...
){
    @Ignore  @SerializedName("children") val children: List<Any>? = null
}

Answer 6

我解决了仅从默认构造函数中删除@Ignore的问题。

如果您从某个地方复制了代码，并在默认构造函数中使用了@Ignore，则需要从默认构造函数中删除@Ignore。

之前：

@Ignore
public Card(){}

之后：

public Card(){}

我刚刚添加是因为它与我同在。

房间持久性：实体和Pojos必须具有可用的公共构造函数

6 个答案: