数据库SQLite应用程序

时间:2017-05-01 15:45:33

标签: java android sqlite insert-update

当我从应用程序退出并再次运行时,我的应用程序不会插入值,我的数据库出现此问题。我确信数据只存储在第一次运行中,我不能再添加数据了。另外,由于这个原因,我无法更新。

有人可以帮忙吗?

public static final String TABLE_NAME = "User_table";
public static final String COL_1 = "ID";
public static final String COL_2 = "NAME";
public static final String COL_3 = "Password";
public static final String TABLE_NAME2 = "Courses";
public static final String COL_1C = "ID";
public static final String COL_2C = "Course_Name";


public DatabaseHelper(Context context) {
    super(context, DATABASE_NAME, null, 1);
}

@Override
public void onCreate(SQLiteDatabase db) {

    //create first table
    String CREATE_USER_TABLE= "CREATE TABLE " + TABLE_NAME + "(" + COL_1 + " INTEGER PRIMARY KEY AUTOINCREMENT," + COL_2+ " TEXT," + COL_3 + " TEXT" +")";


    db.execSQL(CREATE_USER_TABLE);

    // create second table
    String CREATE_COURSE_TABLE= "CREATE TABLE " + TABLE_NAME2 + "(" + COL_1C + " INTEGER PRIMARY KEY AUTOINCREMENT," + COL_2C+ " TEXT" +")";

    db.execSQL(CREATE_COURSE_TABLE);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    // Drop older table if existed
    //first table
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME);

    db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME2);
    // Create tables again
    onCreate(db);
}

public boolean Inser_to_user(String name,String password){
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues Values = new ContentValues();
    Values.put(COL_2, name);
    Values.put(COL_3, password);
    long Result = db.insert(TABLE_NAME, null, Values);
    if (Result == -1) {
        return false;
    } else {
        return true;
    }
}

public boolean Inser_to_courses(String name) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues Values = new ContentValues();
    Values.put(COL_2C, name);

    long Result = db.insert(TABLE_NAME2, null, Values);
    if (Result == -1) {
        return false;
    } else {
        return true;
    }
}

public Cursor showdata(){
    SQLiteDatabase db = this.getReadableDatabase();
    Cursor Result=db.rawQuery("select * from" +TABLE_NAME,null);
    return Result;
}

public boolean updateuser(String id,String name,String password){
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues Values=new ContentValues();
    Values.put(COL_1,id);
    Values.put(COL_2,name);
    Values.put(COL_3,password);
    db.update(TABLE_NAME,Values,COL_1 +"=?",new String[]{Integer.toString(Integer.parseInt(id))});
    return true;
}

0 个答案:

没有答案