3

考虑下表“视图”

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，该表包含每个用户 (user_id) 每天 (_date) 的综合浏览量 (cnt)。我正在寻找一个 SELECT 查询，它将 user_ids 输出为列，因此表数据将采用矩阵形式，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

这可能与 SELECT 语句有关吗？

4

2 回答 2

4

如果您正在处理一组有限的用户 ID，则可以执行以下操作：

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，这更像是一个 hack，而不是一个好的查询。

于 2012-04-08T00:01:28.667 回答
0

看起来您有一长串要转换的值。如果是这种情况，那么您可以使用准备好的语句。您的代码将如下所示（参见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;
于 2012-10-01T21:10:40.053 回答