这个类有一个内部类,那么如何用内部类包裹一个类?
public class UserModel
{
private User user;
public User getUser() {
return user;
}
public class User {
@SerializedName("id")
@Expose
private int id;
@SerializedName("first_name")
@Expose
private String first_name;
@SerializedName("last_name")
@Expose
private String last_name;
@SerializedName("email")
@Expose
private String email;
@SerializedName("mobile_no")
@Expose
private String mobile_no;
@SerializedName("created_at")
@Expose
private String created_at;
@SerializedName("updated_at")
@Expose
private String updated_at;
@SerializedName("user_role_id")
@Expose
private int user_role_id;
public int getId() {
return id;
}
public String getFirst_name() {
return first_name;
}
public String getLast_name() {
return last_name;
}
public String getEmail() {
return email;
}
public String getMobile_no() {
return mobile_no;
}
public String getCreated_at() {
return created_at;
}
public String getUpdated_at() {
return updated_at;
}
public int getUser_role_id() {
return user_role_id;
}
}
}
答案 0 :(得分:0)
您必须使内部类可以是Parcelable或Serializable
public class UserModel implements Parcelable {
private User user;
public User getUser() {
return user;
}
public class User implements Serializable{
@SerializedName("id")
@Expose
private int id;
@SerializedName("first_name")
@Expose
private String first_name;
@SerializedName("last_name")
@Expose
private String last_name;
@SerializedName("email")
@Expose
private String email;
@SerializedName("mobile_no")
@Expose
private String mobile_no;
@SerializedName("created_at")
@Expose
private String created_at;
@SerializedName("updated_at")
@Expose
private String updated_at;
@SerializedName("user_role_id")
@Expose
private int user_role_id;
public int getId() {
return id;
}
public String getFirst_name() {
return first_name;
}
public String getLast_name() {
return last_name;
}
public String getEmail() {
return email;
}
public String getMobile_no() {
return mobile_no;
}
public String getCreated_at() {
return created_at;
}
public String getUpdated_at() {
return updated_at;
}
public int getUser_role_id() {
return user_role_id;
}
}
protected UserModel(Parcel in) {
user = (User) in.readValue(User.class.getClassLoader());
}
@Override
public int describeContents() {
return 0;
}
@Override
public void writeToParcel(Parcel dest, int flags) {
dest.writeValue(user);
}
@SuppressWarnings("unused")
public static final Parcelable.Creator<UserModel> CREATOR = new Parcelable.Creator<UserModel>() {
@Override
public UserModel createFromParcel(Parcel in) {
return new UserModel(in);
}
@Override
public UserModel[] newArray(int size) {
return new UserModel[size];
}
};
}
答案 1 :(得分:-1)
您希望这两个类都实现Parcelable接口。
资料来源:this