MySQL离开外连接麻烦

这是一个按小时将交易按价格点分组的查询：

SELECT hour(Stamp) AS hour, PointID AS pricepoint, count(1) AS counter
FROM Transactions
GROUP BY 1,2;

样本输出：

+------+------------+---------+
| hour | pricepoint | counter |
+------+------------+---------+
|    0 |         19 |       5 |
|    0 |         20 |      14 |
|    1 |         19 |       3 |
|    1 |         20 |      12 |
|    2 |         19 |       2 |
|    2 |         20 |       8 |
|    3 |         19 |       2 |
|    3 |         20 |       4 |
|    4 |         19 |       1 |
|    4 |         20 |       1 |
|    5 |         19 |       4 |
|    5 |         20 |       1 |
|    6 |         20 |       2 |
|    8 |         19 |       1 |
|    8 |         20 |       4 |
|    9 |         19 |       2 |
|    9 |         20 |       5 |
|   10 |         19 |       6 |
|   10 |         20 |       1 |
|   11 |         19 |      10 |
|   11 |         20 |       2 |
|   12 |         19 |      10 |
|   12 |         20 |       3 |
|   13 |         19 |      10 |
|   13 |         20 |      10 |
|   14 |         19 |       8 |
|   14 |         20 |       3 |
|   15 |         19 |       6 |
|   15 |         20 |       8 |
|   16 |         19 |      11 |
|   16 |         20 |      10 |
|   17 |         19 |       7 |
|   17 |         20 |      17 |
|   18 |         19 |       7 |
|   18 |         20 |       9 |
|   19 |         19 |      10 |
|   19 |         20 |      12 |
|   20 |         19 |      17 |
|   20 |         20 |      11 |
|   21 |         19 |      12 |
|   21 |         20 |      29 |
|   22 |         19 |       6 |
|   22 |         20 |      21 |
|   23 |         19 |       9 |
|   23 |         20 |      23 |
+------+------------+---------+

如您所见,有些小时没有交易(例如早上7点),有些小时仅有一个价格点的交易(例如6am,只有价格点20,但没有价格点19的交易).

我想在没有事务时显示结果集为“ 0”,而不是像现在这样不存在.

尝试在那里使用左外连接. inHour表包含值0..23

SELECT H.hour, PointID AS Pricepoint, COALESCE(T.counter, 0) AS Count
FROM inHour H
LEFT OUTER JOIN
(
 SELECT hour(Stamp) AS Hour, PointID, count(1) AS counter
 FROM Transactions
 GROUP BY 1,2
 ) T
ON T.Hour = H.hour;

产生以下输出(为简洁起见,将其截断)：

|    5 |         19 |     4 |
|    5 |         20 |     1 |
|    6 |         20 |     2 |
|    7 |       NULL |     0 |
|    8 |         19 |     1 |
|    8 |         20 |     4 |

我实际上想要的是：

|    5 |         19 |     4 |
|    5 |         20 |     1 |
|    6 |         19 |     0 |
|    6 |         20 |     2 |
|    7 |         19 |     0 |
|    7 |         20 |     0 |
|    8 |         19 |     1 |
|    8 |         20 |     4 |

在我期望的输出中,将值“ 0”放在给定时间内没有交易的价格点旁边.

您的建议将受到欢迎！谢谢.

解决方法:

SELECT h.Hour, p.Pricepoint, COUNT(t.*) AS Count
FROM inHour h,
(SELECT DISTINCT PointId AS Pricepoint FROM Transactions) p
LEFT OUTER JOIN Transactions t
ON h.Hour = hour(t.Stamp) AND p.Pricepoint = t.PointID
GROUP BY h.Hour, p.Pricepoint
ORDER BY h.Hour, p.Pricepoint

我目前没有时间尝试该操作,因此请告知它是否无效,我将尝试进行调整.

标签：left-join,mysql,group-by
来源： https://codeday.me/bug/20191208/2090338.html