房間資料庫查詢-洗掉用戶

我使用 Room 持久性庫構建了一個充當用戶和庫存資料庫的應用程式。目前，我能夠進行查詢以查看/輸出用戶到回收站視圖，但我需要能夠使用用戶名作為查詢物件來洗掉用戶。目前我有以下。

洗掉用戶類

                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();

道

@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);

用戶類

@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;

它檢測到用戶完全沒有問題，但不會洗掉 User 物件，我對此很陌生，并且希望得到任何幫助，因為我確信它很簡單。

uj5u.com熱心網友回復：

那是因為您正在創建新用戶并且它只有“名稱”值。

    Users activeUser = new Users();
    activeUser.setName(usrNam); // This user is not equal what's in db.
    LoginMenu.mainDatabase.dao().delUser(activeUser);

相反，當您在資料庫中回圈用戶時，如果您找到了用戶，請從那里獲取它，如下所示：

      Users activeUser = new Users();
      for (Users usr : users) {
           // User is successfully found
           if (usr.getName().equals(usrNam)) {
               userFound = true;
               activeUser = usr; // now we got the exact item on db
               break;
           }
      }

現在您的函式將洗掉正確的用戶。

      //Users activeUser = new Users();
      //activeUser.setName(usrNam); //you don't need them anymore.
      LoginMenu.mainDatabase.dao().delUser(activeUser);

uj5u.com熱心網友回復：

我將 Dao 查詢更改為：

// Delete user from DB
    @Query("DELETE FROM Users WHERE name =:userId")
    abstract void deleteByUserId(String userId);

以及 RemoveUsers 類下的查詢本身

// Delete User from database
LoginMenu.mainDatabase.dao().deleteByUserId(usrNam);

查詢現在按預期作業

標籤：爪哇机器人房间

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