问题描述
- 如何从sqlite数据库中获取数据并显示在listview中?
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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
解决方案
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;
}
时间: 2024-12-10 08:14:56