Question

如何从sqlite数据库中获取数据并显示在listview中

Answer 1

在登录页面后，我想在listview中把Apple显示成A，Boy显示成B等等，直到F。但是在程序中当我完全登录后，只有登录表成功创建，主菜单还是没有创建。
我想在test database中创建主菜单，然后我想从主菜单表（mainmenu table）中获取数据再显示在listview中。
我使用了下面的代码：
if(username.length()>0&&password.length()>0)
{
SQLiteAdapter db=new SQLiteAdapter(Main.this);
db.openToWrite();
if(db.Login(username,password))
{
System.out.println("goutham");
Intent intent=new Intent(getApplicationContext(),ExampleActivity.class);

startActivity(intent);
}

SQLiteAdapter.java
}

public Cursor queueAll() {
String[] columns = new String[] { KEY_ID, KEY_CONTENT };
Cursor cursor = sqLiteDatabase.query(MYDATABASE_TABLE, columns, null,
null, null, null, null);

return cursor;
}

private static class SQLiteHelper extends SQLiteOpenHelper {

public SQLiteHelper(Context context, String name,
CursorFactory factory, int version) {
super(context, name, factory, version);
}

@Override
public void onCreate(SQLiteDatabase db) {
// TODO Auto-generated method stub
db.execSQL(SCRIPT_CREATE_DATABASE);
db.execSQL(DATABASE_CREATE);
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
// TODO Auto-generated method stub

}

}

public long AddUser(String username, String password) {
ContentValues initialValues = new ContentValues();
initialValues.put(KEY_USERNAME, username);
initialValues.put(KEY_PASSWORD, password);
return sqLiteDatabase.insert(DATABASE_TABLE, null, initialValues);

}

public boolean Login(String username, String password) {
// TODO Auto-generated method stub
Cursor mCursor = sqLiteDatabase.rawQuery("SELECT * FROM "
+ DATABASE_TABLE + " WHERE username=? AND password=?",
new String[] { username, password });
if (mCursor != null) {
if (mCursor.getCount() > 0) {
return true;
}
}
return false;
}

}

ExampleActivity.java
public class ExampleActivity extends Activity {
private SQLiteAdapter mySQLiteAdapter;

/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
ListView listContent = (ListView) findViewById(R.id.contentlist);

/*
* Create/Open a SQLite database and fill with dummy content and close
* it
*/
mySQLiteAdapter = new SQLiteAdapter(this);
mySQLiteAdapter.openToWrite();
// mySQLiteAdapter.deleteAll();

mySQLiteAdapter.insert("A for Apply");
mySQLiteAdapter.insert("B for Boy");
mySQLiteAdapter.insert("C for Cat");
mySQLiteAdapter.insert("D for Dog");
mySQLiteAdapter.insert("E for Egg");
mySQLiteAdapter.insert("F for Fish");

mySQLiteAdapter.close();

/*
* Open the same SQLite database and read all it's content.
*/
mySQLiteAdapter = new SQLiteAdapter(this);
mySQLiteAdapter.openToRead();

Cursor cursor = mySQLiteAdapter.queueAll();
startManagingCursor(cursor);

String[] from = new String[] { SQLiteAdapter.KEY_CONTENT };
int[] to = new int[] { R.id.text };

SimpleCursorAdapter cursorAdapter = new SimpleCursorAdapter(this,
R.layout.row, cursor, from, to);

listContent.setAdapter(cursorAdapter);

mySQLiteAdapter.close();

}
}

运行程序后，登录表在数据库中创建了，但是主菜单表没有创建。运行程序后，显示一个错误：
sqlite returned code=1 no such a table in MY_TABLE

通过互联网整理获得以下解决方法:

=================1楼=====================
Database class：
public String getData1() throws SQLException{
// TODO Auto-generated method stub
String[] columns1 = new String[] { KEY_DATE };
Cursor c1 = ourDatabase.query(DATABASE_MARKSTABLE, columns1, null, null, null,
null, KEY_ENDINGTIME+" DESC", " 30");
String result1 = "";

int isName = c1.getColumnIndex(KEY_DATE);

for (c1.moveToFirst(); !c1.isAfterLast(); c1.moveToNext()) {

result1 = result1 + c1.getString(isName)
+ " " + "\n";
}
c1.close();
return result1;
}

Answer 2

代码：
if(username.length()>0&&password.length()>0)
{
SQLiteAdapter db=new SQLiteAdapter(Main.this);
db.openToWrite();
if(db.Login(username,password))
{
System.out.println("goutham");
Intent intent=new Intent(getApplicationContext(),ExampleActivity.class);

startActivity(intent);
}