+--------------------+---------------------+----+----+----+----+----+
| DATA_SUBSCRIBER_ID | MONTHLY_CALENDAR_ID | F1 | F2 | F3 | F4 | F5 |
+--------------------+---------------------+----+----+----+----+----+
| 3699060 | 01/07/2015 | 0 | 1 | 0 | 0 | 0 |
| 3699247 | 01/07/2015 | 0 | 0 | 1 | 1 | 0 |
| 3699342 | 01/08/2015 | 0 | 0 | 0 | 0 | 0 |
| 3699380 | 01/08/2015 | 1 | 0 | 0 | 0 | 1 |
| 3699506 | 01/09/2015 | 1 | 1 | 1 | 0 | 0 |
| 3699526 | 01/09/2015 | 0 | 1 | 1 | 1 | 1 |
+--------------------+---------------------+----+----+----+----+----+
我要进行查询，以使新表如下所示：
MONTHLY_CALENDAR_ID, SUM(F1+F2+F3+F4+F5), SUM(F1), SUM(F2), SUM(F3), SUM(F4), SUM(F5), USERS_WITH_0_FLAGS, USERS_WITH_1_FLAGS, USERS_WITH_2_FLAGS, USERS_WITH_3_FLAGS, USERS_WITH_4_FLAGS, USERS_WITH_5_FLAGS
+---+------------+---+---+---+---+---+---+---+---+---+---+---+---+
| | A | B | C | D | E | F | G | H | I | J | K | L | M |
+---+------------+---+---+---+---+---+---+---+---+---+---+---+---+
| 1 | 01/07/2015 | 3 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 2 | 01/08/2015 | 5 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
| 3 | 01/09/2015 | 7 | 1 | 2 | 2 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
+---+------------+---+---+---+---+---+---+---+---+---+---+---+---+
基本上，我可以使用SUM完成所有列。但是我想统计每一行，并记录每个月有多少个标志。有什么想法如何查询吗？
SELECT
MONTHLY_CALENDAR_ID,
SUM(F1 + F2 + F3 + F4 + F5 ),
SUM(F1), SUM(F2), SUM(F3), SUM(F4), SUM(F5),
(SELECT COUNT(*) FROM a WHERE SUM(F1 + F2 + F3 + F4 + F5 )) AS USERS_WITH_0_FLAGS
FROM ANA_SUB a GROUP BY MONTHLY_CALENDAR_ID ORDER BY MONTHLY_CALENDAR_ID;
答案 0 :(得分：0)
您可以通过以下方式执行此操作：首先计算每个用户的标志数，然后使用该数来填充列以指示该用户具有的标志数（例如，如果用户有0个标志，请用1个其他0、1个标志填充users_with_0_flags ，在users_with_1_flags中填充1个else 0，以此类推），然后在分组依据中使用它，例如：
WITH ana_sub AS (SELECT 3699060 data_subscriber_id, to_date('01/07/2015', 'dd/mm/yyyy') monthly_calendar_id, 0 f1, 1 f2, 0 f3, 0 f4, 0 f5 FROM dual UNION ALL
SELECT 3699247 data_subscriber_id, to_date('01/07/2015', 'dd/mm/yyyy') monthly_calendar_id, 0 f1, 0 f2, 1 f3, 1 f4, 0 f5 FROM dual UNION ALL
SELECT 3699342 data_subscriber_id, to_date('01/08/2015', 'dd/mm/yyyy') monthly_calendar_id, 0 f1, 0 f2, 0 f3, 0 f4, 0 f5 FROM dual UNION ALL
SELECT 3699380 data_subscriber_id, to_date('01/08/2015', 'dd/mm/yyyy') monthly_calendar_id, 1 f1, 0 f2, 0 f3, 0 f4, 1 f5 FROM dual UNION ALL
SELECT 3699506 data_subscriber_id, to_date('01/09/2015', 'dd/mm/yyyy') monthly_calendar_id, 1 f1, 1 f2, 1 f3, 0 f4, 0 f5 FROM dual UNION ALL
SELECT 3699526 data_subscriber_id, to_date('01/09/2015', 'dd/mm/yyyy') monthly_calendar_id, 0 f1, 1 f2, 1 f3, 1 f4, 1 f5 FROM dual),
user_res AS (SELECT user_id,
monthly_calendar_id,
total_flags_per_user,
f1,
f2,
f3,
f4,
f5,
CASE WHEN total_flags_per_user = 0 THEN 1 ELSE 0 END users_with_0_flags,
CASE WHEN total_flags_per_user = 1 THEN 1 ELSE 0 END users_with_1_flags,
CASE WHEN total_flags_per_user = 2 THEN 1 ELSE 0 END users_with_2_flags,
CASE WHEN total_flags_per_user = 3 THEN 1 ELSE 0 END users_with_3_flags,
CASE WHEN total_flags_per_user = 4 THEN 1 ELSE 0 END users_with_4_flags,
CASE WHEN total_flags_per_user = 5 THEN 1 ELSE 0 END users_with_5_flags
FROM (SELECT data_subscriber_id user_id,
monthly_calendar_id,
f1,
f2,
f3,
f4,
f5,
f1 + f2 + f3 + f4 + f5 total_flags_per_user
FROM ana_sub))
SELECT row_number() OVER (ORDER BY monthly_calendar_id) rn,
monthly_calendar_id a,
sum(total_flags_per_user) b/*total_flags_per_month*/,
SUM(f1) c/*f1_total*/,
SUM(f2) d/*f2_total*/,
SUM(f3) e/*f3_total*/,
SUM(f4) f/*f4_total*/,
SUM(f5) g/*f5_total*/,
SUM(users_with_0_flags) h/*users_with_0_flags_total*/,
SUM(users_with_1_flags) i/*users_with_1_flags_total*/,
SUM(users_with_2_flags) j/*users_with_2_flags_total*/,
SUM(users_with_3_flags) k/*users_with_3_flags_total*/,
SUM(users_with_4_flags) l/*users_with_4_flags_total*/,
SUM(users_with_5_flags) m/*users_with_5_flags_total*/
FROM user_res
GROUP BY monthly_calendar_id;
给出：
RN A B C D E F G H I J K L M
-- ---------- -- -- -- -- -- -- -- -- -- -- -- --
1 01/07/2015 3 0 1 1 1 0 0 1 1 0 0 0
2 01/08/2015 2 1 0 0 0 1 1 0 1 0 0 0
3 01/09/2015 7 1 2 2 1 1 0 0 0 1 1 0
我在这里假设（data_subscriber_id，monthly_calendar_id）上有一个唯一键或主键