MySQL 左外連接故障

Question

這是一個按價格點按小時對交易進行分組的查詢：

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

樣品 output：

+------+------------+---------+
| 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 |
+------+------------+---------+

如您所見，有些時間沒有交易（例如早上 7 點），有些時間只有一個價格點的交易（例如早上 6 點，只有價格點 20，但價格點 19 沒有交易）。

當沒有交易時，我想用“0”顯示結果集，而不是像現在這樣不存在。

嘗試在那里使用 LEFT OUTER JOIN。 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;

這將產生以下 output（為簡潔起見截斷）：

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

我實際上想要的是：

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

在我想要的 output 中，值“0”放在給定時間內沒有交易的價格點旁邊。

歡迎您提出建議。 謝謝。

Answer 1

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

我現在沒有時間嘗試這個，所以如果它不起作用，請告訴我，我會嘗試調整。

Answer 2

有人可能有比這更好的解決方案，但我會使用 UNION 來簡化事情：

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

UNION

SELECT hour,0 AS pricepoint,0 AS counter FROM inHour WHERE hour NOT IN (SELECT hour(Stamp) FROM Transactions)