Question

我正在尝试将两个名为“Friend”和“Social”的对象写入数据库。朋友写没有问题，但'活动'不会工作。

DatabaseHandler代码：

    public DatabaseHandler(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}

// Creating Tables
@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_FRIENDS + "("
            + KEY_ID + " INTEGER PRIMARY KEY," + KEY_FIRSTNAME + " TEXT," + KEY_LASTNAME + " TEXT,"
            + KEY_EMAIL + " TEXT" + ")";
    db.execSQL(CREATE_CONTACTS_TABLE);

    String CREATE_ACTIVITIES_TABLE = "CREATE TABLE " + TABLE_ACTIVITIES + "("
            + KEY_ID + " INTEGER PRIMARY KEY," + KEY_FRIEND + " TEXT," + KEY_DESCRIPTION + " TEXT,"
            + KEY_DAY + " TEXT" + KEY_MONTH + " TEXT" + KEY_YEAR + " TEXT"+ KEY_FRIEND_ID + " TEXT" +")";
    db.execSQL(CREATE_ACTIVITIES_TABLE);


}

// Upgrading database
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    // Drop older table if existed
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_FRIENDS);
    db.execSQL("DROP TABLE IF EXISTS " + TABLE_ACTIVITIES);
    // Create tables again
    onCreate(db);
}

/**
 * All CRUD(Create, Read, Update, Delete) Operations
 */

// Adding new contact
void addContact(Friend friend) {
    SQLiteDatabase db = this.getWritableDatabase();

    ContentValues values = new ContentValues();
    values.put(KEY_FIRSTNAME, friend.getFirstName());
    values.put(KEY_LASTNAME, friend.getLastName());
    values.put(KEY_EMAIL, friend.getEmail());

    // Inserting Row
    db.insert(TABLE_FRIENDS, null, values);
    db.close(); // Closing database connection
}

// Adding new social
void addSocial(Social social) {
    SQLiteDatabase db = this.getWritableDatabase();

    ContentValues values = new ContentValues();
    values.put(KEY_FRIEND_ID, social.getFriend_ID());
    values.put(KEY_DESCRIPTION, social.getDescription());
    values.put(KEY_FRIEND, social.getFriend());
    values.put(KEY_DAY, social.getDay());
    values.put(KEY_MONTH, social.getMonth());
    values.put(KEY_YEAR, social.getYear());

    // Inserting Row
    db.insert(TABLE_ACTIVITIES, null, values);
    db.close(); // Closing database connection
}



// Getting All Contacts
public List<Friend> getAllContacts() {
    List<Friend> contactList = new ArrayList<Friend>();
    // Select All Query
    String selectQuery = "SELECT  * FROM " + TABLE_FRIENDS;

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);

    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Friend friend = new Friend();
            friend.setId(Integer.parseInt(cursor.getString(0)));
            friend.setFirstName(cursor.getString(1));
            friend.setLastName(cursor.getString(2));
            friend.setEmail(cursor.getString(3));
            // Adding contact to list

            contactList.add(friend);
        } while (cursor.moveToNext());
    }

    // return contact list
    return contactList;
}




// Getting All Activities
public List<Social> getAllActivities() {
    List<Social> activityList = new ArrayList<Social>();
    // Select All Query
    String selectQuery = "SELECT  * FROM " + TABLE_ACTIVITIES;

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
    Social social = new Social();
    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {

            social.setId(Integer.parseInt(cursor.getString(0)));
            social.setFriend(cursor.getString(1));
            social.setDescription(cursor.getString(2));
            social.setDay(cursor.getString(3));
            social.setMonth(cursor.getString(4));
            social.setYear(cursor.getString(5));
            social.setFriend_ID(cursor.getString(6));


            // Adding contact to list

            activityList.add(social);
        } while (cursor.moveToNext());
    }


    return activityList;
}

这是在尝试向DB添加新社交对象的活动中。当我尝试在数据库中显示所有社交对象时，它不会给我和错误，但会显示空白。

            Social social = new Social();
            List<Friend> friends = db.getAllContacts();
            for(Friend friend: friends)
            {
                if(friend.getFirstName().equalsIgnoreCase(String.valueOf(spinner2.getSelectedItem())))
                {
                  social.setFriend(friend.getFirstName() + friend.getLastName());
                  social.setFriend_ID(String.valueOf(friend.getId()));
                  social.setDescription(rs3.getText().toString());
                  social.setDay(String.valueOf(simpleDatePicker.getDayOfMonth()));
                  social.setMonth(String.valueOf(simpleDatePicker.getMonth()));
                  social.setYear(String.valueOf(simpleDatePicker.getYear()));
                  found = true;

                }
            }

            if(found)
            {
                db.addActivity(social);
            }

Answer 1

create table SQL中的语法错误。缺少逗号。

Answer 2

由于省略了逗号，以下代码不会像您期望的那样创建活动表，

String CREATE_ACTIVITIES_TABLE = "CREATE TABLE " + TABLE_ACTIVITIES + "("
                + KEY_ID + " INTEGER PRIMARY KEY," + KEY_FRIEND + " TEXT," + KEY_DESCRIPTION + " TEXT,"
                + KEY_DAY + " TEXT" + KEY_MONTH + " TEXT" + KEY_YEAR + " TEXT"+ KEY_FRIEND_ID + " TEXT" +")";
        db.execSQL(CREATE_ACTIVITIES_TABLE);

在KEY_DAY列的列类型（TEXT），KEY_MONTH列和KEY_YEAR列之后，省略了逗号。

相反，它会在的行中创建一个表（注意变量，例如KEY_ID将被相应变量的值替换，这可能会影响结果）： -

列： -

KEY_ID（TYPE INTEGER和PRIMARY KEY）
KEY_FRIEND（TYPE as TEXT）
KEY_DESCRIPTION（TYPE as TEXT）
KEY_DAY（类型为TEXT KEY_MONTH TEXT KEY_YEAR TEXTKEY_FRIEND_ID TEXT）

因此插入不会以KEY_MONTH列的形式插入行.... KEY_FRIEND_ID将不存在。

更正的代码可能是： -

        String CREATE_ACTIVITIES_TABLE = 
            "CREATE TABLE " + TABLE_ACTIVITIES + 
            "(" + 
            KEY_ID + " INTEGER PRIMARY KEY," + 
            KEY_FRIEND + " TEXT," +
            KEY_DESCRIPTION + " TEXT," + 
            KEY_DAY + " TEXT," +    //<<<<<<<< added comma 
            KEY_MONTH + " TEXT," +  //<<<<<<<< added comma
            KEY_YEAR + " TEXT," +   //<<<<<<<< added comma 
            KEY_FRIEND_ID + " TEXT" +
            ")";
        db.execSQL(CREATE_ACTIVITIES_TABLE);

更改代码后，您需要执行以下一个（注意数据可能丢失）： -

删除应用程序的数据
卸载应用
增加数据库版本

完成上述一次后，您可以重新运行该应用。

请注意，上述代码尚未经过测试，因此可能会出现一些错误。

不会写入数据库

2 个答案: