Room数据库按用户名删除用户失败问题求助

使用Room持久化库开发用户与库存管理应用，已实现用户数据查询并展示到RecyclerView，但按用户名删除用户时，能检测到目标用户存在，却无法完成删除操作。

删除用户相关代码 #

String usrNam = edtTxtUsrNam2.getText().toString();

// Get all users from database + log
List<Users> users = LoginMenu.mainDatabase.dao().getUsers();
Log.d(LoginMenu.TAG, "Searching for " + usrNam);
boolean userFound = false;

// Check for User from text entry
for (Users usr : users) {
    // User is successfully found
    if (usr.getName().equals(usrNam)) {
        userFound = true;
        break;
    }
}
if (!userFound) {
    Toast.makeText(getBaseContext(), "User not found", 
    Toast.LENGTH_SHORT).show();
    Log.d(LoginMenu.TAG, " User " + usrNam + " Not found");
    edtTxtUsrNam2.getText().clear();
} else {
    // Build Messagebox Confirmation
    builder.setTitle("User " + usrNam + " found!");
    builder.setMessage("Confirm user deletion?");

    // Selecting yes in Alert
    builder.setPositiveButton("Yes", (dialogInterface, i) -> {
        // Delete User from database
        Users activeUser = new Users();
        activeUser.setName(usrNam);
        LoginMenu.mainDatabase.dao().delUser(activeUser);
        
        // Toast message + Log
        Toast.makeText(getApplicationContext(),"User " + usrNam +
                " successfully deleted!",Toast.LENGTH_SHORT).show();
        
        Log.d(LoginMenu.TAG,"User " + usrNam + " Successfully deleted");
        
        edtTxtUsrNam2.getText().clear();
        dialogInterface.dismiss();
    });
    // Selecting no in Alert
    builder.setNegativeButton("No", (dialogInterface, i) -> dialogInterface.dismiss());
    // Create Alert Dialog
    AlertDialog alert = builder.create();
    alert.show();
}

DAO代码 #

@androidx.room.Dao
public interface Dao {

    // Add Users to Database
    @Insert
    void AddUsers(Users usr);

    // Request all users from DB
    @Query("SELECT * FROM Users")
    List<Users> getUsers();

    // Delete user from DB
    @Delete
    void delUser (Users usr);
}

User实体类代码 #

@Entity(tableName = "Users")
public class Users {

    // Set User Entries for Database
    @ColumnInfo (name = "ID")
    @PrimaryKey (autoGenerate = true)
    private int id;

    @ColumnInfo(name = "EmployeeId")
    private String  employeeId;

    @ColumnInfo(name = "Name")
    private String name;

    @ColumnInfo(name = "Password")
    private String password;

    @ColumnInfo(name = "DateOfBirth")
    private String dateOfBirth;

    @ColumnInfo(name = "ContactNumber")
    private String contactNumber;

    @ColumnInfo(name = "Address")
    private String address;

    // 需补充所有字段的getter/setter方法，Room依赖这些方法访问字段
    public int getId() { return id; }
    public void setId(int id) { this.id = id; }
    public String getName() { return name; }
    public void setName(String name) { this.name = name; }
    // 其他字段的getter/setter省略，需自行补充
}

问题原因及解决方案 #

核心原因 #

Room的@Delete注解默认通过实体的主键匹配数据库记录。你新建的activeUser仅设置了name，未设置主键id，Room无法找到对应主键的记录，因此删除操作无效。

方案一：使用完整用户对象删除 #

遍历用户列表时，保存找到的完整Users对象，删除时传入该真实对象：

// 修改遍历逻辑，保存目标用户对象
Users targetUser = null;
for (Users usr : users) {
    if (usr.getName().equals(usrNam)) {
        userFound = true;
        targetUser = usr; // 保存完整的用户对象
        break;
    }
}

// 在确认删除的回调中使用targetUser
builder.setPositiveButton("Yes", (dialogInterface, i) -> {
    if (targetUser != null) {
        LoginMenu.mainDatabase.dao().delUser(targetUser);
        // 后续提示、日志操作不变
    }
});

方案二：自定义@Query删除语句（更高效） #

直接在DAO中添加按用户名删除的方法，无需查询所有用户：

@Dao
public interface Dao {
    // 保留原有方法，新增按用户名删除的方法
    @Query("DELETE FROM Users WHERE Name = :userName")
    void deleteUserByName(String userName);
}

删除回调中直接调用该方法：

builder.setPositiveButton("Yes", (dialogInterface, i) -> {
    LoginMenu.mainDatabase.dao().deleteUserByName(usrNam);
    // 后续提示、日志操作不变
});

此方案无需查询全表，直接执行删除SQL，性能更优。

额外注意 #

确保Users实体类包含所有字段的getter和setter方法，Room需要通过这些方法访问实体字段，否则可能出现数据匹配异常。

内容的提问来源于stack exchange，提问作者Ethan Cunningham