我在Android中创建了一个示例SQLite数据库,它将条目读入ListView。我将添加代码,以便单击时,每个ListView项目都会启动一个新的活动以显示更多信息。我如何从另一个活动中对此数据库运行查询?由于从另一个类访问SQLite数据库
public class Database extends ListActivity {
private final String SAMPLE_DB_NAME = "myFriendsDb";
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
ArrayList<String> results = new ArrayList<String>();
SQLiteDatabase db = null;
try {
db = this.openOrCreateDatabase(SAMPLE_DB_NAME, MODE_PRIVATE, null);
db.execSQL("CREATE TABLE IF NOT EXISTS people" +
" (LastName VARCHAR, FirstName VARCHAR," +
" Country VARCHAR, Age INT(3));");
db.execSQL("INSERT INTO people" +
" Values ('Jones','Bob','UK',30);");
db.execSQL("INSERT INTO people" +
" Values ('Smith','John','UK',40);");
db.execSQL("INSERT INTO people" +
" Values ('Thompson','James','UK',50);");
Cursor c = db.rawQuery("SELECT FirstName, LastName FROM people", null);
if (c != null) {
if (c.moveToFirst()) {
do {
String firstName = c.getString(c.getColumnIndex("FirstName"));
String lastName = c.getString(c.getColumnIndex("LastName"));
results.add("" + firstName + " " + lastName);
}while (c.moveToNext());
}
}
this.setListAdapter(new ArrayAdapter<String>(this, android.R.layout.simple_list_item_1,results));
} catch (SQLiteException se) {
Log.e(getClass().getSimpleName(), "Could not create or Open the database");
} finally {
if (db != null)
db.execSQL("DELETE FROM people");
db.close();
}
}
http://www.vogella.com/articles/AndroidSQLite/article.html#databasetutorial是关于各种SQLite相关事情的不错教程。您应该特别关注用于打开数据库的'SQLiteOpenHelper'和用于访问数据的DataSource类。 – zapl 2012-04-28 22:55:35