Question

我在运行此查询时收到错误，任何线索我的语法错误是什么？

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）：引起：android.database.sqlite.SQLiteException：near“AS”：语法错误（代码1）:,编译时：SELECT CASE WHEN createduser = 112然后（创建的播放器，你和他们连接的播放器）ELSE（你和他们创建的播放器是连接的播放器）来自游戏WHERE（转=？）

Answer 1

试试这个。

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 = ?)

Android SQLite查询ContentProvider Case Statement语法

1 个答案: