Android SQLite查詢ContentProvider案例語句語法

我收到一個錯誤，當我運行此查詢，任何線索我的語法錯誤是什麼？Android SQLite查詢ContentProvider案例語句語法

final Cursor c = getContentResolver().query(
Provider.GAME_URI, new String[]{"CASE WHEN " +DBHelper.COLUMN_CRE_USER + " = " + preferences.getInt(LoginScreen.ID, 0) + " THEN (" + DBHelper.COLUMN_CRE_PTS + " AS you AND " + DBHelper.COLUMN_CON_PTS + " AS them) ELSE (" + DBHelper.COLUMN_CON_PTS + " AS you AND " + DBHelper.COLUMN_CRE_PTS + " AS them)"},"turn = ?", 
new String[] { preferences.getInt(LoginScreen.ID, 0) + ""}, null);

04-24 19：57：48.345：E/AndroidRuntime（18775）：android.database.sqlite.SQLiteException：致近「AS」：語法錯誤（碼1）：，而編譯：SELECT CASE WHEN createduser = 112 THEN（AS你createdplayerpts並作爲他們connectedplayerpts）ELSE（connectedplayerpts你和createdplayerpts AS他們）從博弈WHERE（轉=？）

來源

2013-04-25 Aaron Smentkowski

試試這個。

SELECT 
    CASE WHEN createduser = 112 THEN createdplayerpts ELSE connectedplayerpts END AS you, 
    CASE WHEN createduser = 112 THEN connectedplayerpts ELSE createdplayerpts END AS them 
FROM 
    game WHERE (turn = ?)

來源

2013-04-25 03:21:18 appsroxcom

救主！我欠你一杯啤酒。 – 2013-04-25 06:26:19

歡迎！任何時候 ;-） – appsroxcom 2013-04-25 06:30:42

Android SQLite查詢ContentProvider案例語句語法

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