考虑下表“观点”
user_id _date cnt
------------------------
1 2011-02-10 123
1 2011-02-11 99
1 2011-02-12 100
1 2011-02 13 12
2 2011-02-10 32
2 2011-02-11 433
2 2011-02-12 222
2 2011-02 13 334
3 2011-02-10 766
3 2011-02-11 654
3 2011-02-12 43
3 2011-02 13 27
...
100 2011-02-13 235
如您所见,该表包含每位用户(user_id)的每日综合浏览量(cnt)(_date)。我正在寻找一个SELECT查询,它将user_ids作为列输出,因此table-data将采用矩阵形式,如下所示:
_date 1 2 3 ... 100
---------------------------------
2011-02-10 123 32 766
2011-02-11 99 433 654
2011-02-12 100 222 43
2011-02-13 12 334 27 235
这可以用SELECT语句吗?
答案 0 :(得分:4)
如果你正在处理一组有限的用户ID,你可以这样做:
SELECT _date,
SUM(CASE WHEN _user_id = 1 THEN cnt ELSE 0 END) AS user1,
SUM(CASE WHEN _user_id = 2 THEN cnt ELSE 0 END) AS user2,
SUM(CASE WHEN _user_id = 3 THEN cnt ELSE 0 END) AS user3,
...
FROM views
GROUP BY _date
不过,它更像是一个黑客而不是一个好的查询。
答案 1 :(得分:0)
您似乎有一长串要转换的值。如果是这种情况,那么您可以使用prepared statements。您的代码将如下所示(请参阅SQL Fiddle with Demo):
CREATE TABLE Table1
(`user_id` int, `_date` datetime, `cnt` int)
;
INSERT INTO Table1
(`user_id`, `_date`, `cnt`)
VALUES
(1, '2011-02-09 17:00:00', 123),
(1, '2011-02-10 17:00:00', 99),
(1, '2011-02-11 17:00:00', 100),
(1, '2011-02-13 00:00:00', 12),
(2, '2011-02-09 17:00:00', 32),
(2, '2011-02-10 17:00:00', 433),
(2, '2011-02-11 17:00:00', 222),
(2, '2011-02-13 00:00:00', 334),
(3, '2011-02-09 17:00:00', 766),
(3, '2011-02-10 17:00:00', 654),
(3, '2011-02-11 17:00:00', 43),
(3, '2011-02-13 00:00:00', 27),
(100, '2011-02-12 17:00:00', 235)
;
SET @sql = NULL;
SELECT
GROUP_CONCAT(DISTINCT
CONCAT(
'sum(case when user_id = ''',
user_id,
''' then cnt else 0 end) AS ''',
user_id, ''''
)
) INTO @sql
FROM Table1;
SET @sql = CONCAT('SELECT _Date, ', @sql, '
FROM table1
GROUP BY _Date');
PREPARE stmt FROM @sql;
EXECUTE stmt;
DEALLOCATE PREPARE stmt;