我正在开发一个Sms应用程序, 我正试图从每次谈话中得到最后的短信。
这是我的SQL语句:
SELECT MAX(smsTIMESTAMP) AS smsTIMESTAMP,_id, smsID, smsCONID, smsMSG, smsNUM, smsREAD, smsTYPE, smsSHORTMSG, COUNT(*) AS smsNUMMESSAGES FROM sms GROUP BY smsCONID ORDER BY smsTIMESTAMP desc
我在SQLite Expert中运行了查询,得到了正确的答案:
然而,当我在我的应用程序中运行它时,我得到:
这是我的表
这是我的Datamanipulator类:
public class DataManipulator {
private static final String DATABASE_NAME = "smsapplication.db";
private static final int DATABASE_VERSION = 3;
static final String TABLE_NAME = "sms";
private static Context context;
static SQLiteDatabase db;
private static DataManipulator instance;
private SQLiteStatement insertStmt;
private static final String INSERT = "insert or ignore into "
+ TABLE_NAME + " (smsID, smsCONID, smsMSG, smsNUM, smsREAD, smsTIMESTAMP, smsTYPE, smsSHORTMSG) values (?,?,?,?,?,?,?,?)";
public DataManipulator(Context context) {
DataManipulator.context = context;
OpenHelper openHelper = new OpenHelper(DataManipulator.context);
DataManipulator.db = openHelper.getWritableDatabase();
this.insertStmt = DataManipulator.db.compileStatement(INSERT);
}
public static DataManipulator getInstance(Context mContext)
{
if(instance == null)
{
instance = new DataManipulator(mContext);
}
return instance;
}
public long insert(int smsID, int smsCONID, String smsMSG,
String smsNUM, int smsREAD, long smsTIMESTAMP, String smsTYPE, String smsSHORTMSG) {
this.insertStmt.bindLong(1, smsID);
this.insertStmt.bindLong(2, smsCONID);
this.insertStmt.bindString(3, smsMSG);
this.insertStmt.bindString(4, smsNUM);
this.insertStmt.bindLong(5, smsREAD);
this.insertStmt.bindString(6, String.valueOf(smsTIMESTAMP));
this.insertStmt.bindString(7, smsTYPE);
this.insertStmt.bindString(8, smsSHORTMSG);
return this.insertStmt.executeInsert();
}
public void deleteAll() {
db.delete(TABLE_NAME, null, null);
}
public Cursor getLastSmsForAllConversations()
{
Cursor cursor = db.query(TABLE_NAME, new String[] {"_id"," MAX(smsTIMESTAMP) AS smsTIMESTAMP ", "smsCONID", "smsNUM", "smsREAD", "smsTYPE","smsSHORTMSG","COUNT(*) AS smsNUMMESSAGES" },
null, null, "smsCONID", null, "smsTIMESTAMP desc");
return cursor;
}
public Cursor getConversationMessages(int conID)
{
Cursor cursor = db.query(TABLE_NAME, new String[] {"_id", "smsID", "smsCONID", "smsMSG", "smsNUM", "smsREAD", "smsTIMESTAMP", "smsTYPE","smsSHORTMSG" },
"smsCONID="+conID, null, null, null, "smsTIMESTAMP asc");
return cursor;
}
public void printCursor()
{
Cursor cursor = db.rawQuery("SELECT MAX(smsTIMESTAMP) AS smsTIMESTAMP,_id, smsID, smsCONID, smsMSG, smsNUM, smsREAD, smsTYPE, smsSHORTMSG, COUNT(*) AS smsNUMMESSAGES FROM sms GROUP BY smsCONID ORDER BY smsTIMESTAMP desc", null);
// db.query(TABLE_NAME, new String[] {"_id","smsTIMESTAMP AS smsTIMESTAMP ", "smsCONID", "smsNUM", "smsREAD", "smsTYPE","smsSHORTMSG" },"smsCONID=40", null, null, null, "smsTIMESTAMP desc");
if (cursor.moveToFirst()){
do{
String data ="";
data= cursor.getString(cursor.getColumnIndex("smsTIMESTAMP"))+" - " +cursor.getString(cursor.getColumnIndex("smsSHORTMSG"));
Log.i("data", data);
}while(cursor.moveToNext());
}
cursor.close();
}
private static class OpenHelper extends SQLiteOpenHelper {
OpenHelper(Context context) {
super(context, DATABASE_NAME, null, DATABASE_VERSION);
}
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL("CREATE TABLE " + TABLE_NAME + " (_id INTEGER , smsID INTEGER PRIMARY KEY, smsCONID INTEGER, smsMSG TEXT,smsNUM TEXT, smsREAD INTEGER, smsTIMESTAMP INTEGER, smsTYPE TEXT, smsSHORTMSG TEXT)");
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
db.execSQL("DROP TABLE IF EXISTS " + TABLE_NAME);
onCreate(db);
}
@Override
public synchronized void close() {
if (db != null)
db.close();
super.close();
}
}
}
答案 0 :(得分:2)
当您使用GROUP BY
时,结果的每一行都对应于原始表的多行。
如何计算这些结果有三种可能性:
MAX
或COUNT
等聚合函数的列计算组中所有行的值; GROUP BY
子句中的MIN
或MAX
匹配的行中获取值;这是您在SQLite Exprert中看到的内容,但不是使用较旧的SQLite的Androids。要解决您的问题,您必须先使用GROUP BY
获取足够的信息来识别您想要的记录:
SELECT smsCONID, -- OK: used in GROUP BY
MAX(smsTIMESTAMP) AS smsTIMESTAMP, -- OK: aggregate MAX
COUNT(*) AS smsNUMMESSAGES -- OK: aggregate COUNT
FROM sms
GROUP BY smsCONID
ORDER BY smsTIMESTAMP DESC
然后,将结果表与原始sms
表连接,以获取这些记录的其他列:
SELECT *
FROM (SELECT smsCONID,
MAX(smsTIMESTAMP) AS smsTIMESTAMP,
COUNT(*) AS smsNUMMESSAGES
FROM sms
GROUP BY smsCONID) AS grouped
JOIN sms
ON grouped.smsCONID = sms.smsCONID
AND grouped.smsTIMESTAMP = sms.smsTIMESTAMP
ORDER BY smsTIMESTAMP DESC
答案 1 :(得分:0)
您的查询和代码是正确的我没有发现任何问题而不是上面的查询试试这个..
SELECT _id,smsID,smsCONID,smsMSG,smsNUM,smsREAD,smsTYPE,smsSHORTMSG,FROM sms GROUP BY smsCONID ORDER BY smsTIMESTAMP desc LIMIT 1
如果你想要数,那就这样做..
SELECT COUNT(*)AS smsNUMMESSAGES,_id,smsID,smsCONID,smsMSG,smsNUM,smsREAD,smsTYPE,smsSHORTMSG,FROM sms GROUP BY smsCONID ORDER BY smsTIMESTAMP desc LIMIT 1