MySQL的選擇與UNION ALL

Question

我（使用UNION ALL，因爲我有萬噸行的）

SELECT id, max(lastdate) AS lastdate, max(san) AS san 
FROM 
    (SELECT `san` AS id, 
      `created` AS lastdate, 
      '' AS san 
    FROM `TableA` a 
    UNION ALL 
    SELECT `san` AS id, 
      cast(null as datetime) AS lastdate, 
      `san-name` AS san 
    FROM `TableB` b 
    ) sq 
GROUP BY id 
ORDER BY san ASC 

了

表A

| id | san | status |  created  | 
------------------------------------------- 
| 1 | 100 | 3 | 2013-11-01 19:26:20 | 
| 2 | 200 | 8 | 2013-11-02 03:19:35 | 
| 3 | 300 | 1 | 2013-11-03 06:13:38 | 
| 4 | 100 | 2 | 2013-11-06 08:28:37 | 
| 5 | 100 | 4 | 2013-11-27 19:00:00 | 
| 6 | 200 | 1 | 2013-11-27 19:16:22 | 
| 7 | 200 | 3 | 2013-11-27 19:33:33 | 
| 8 | 300 | 5 | 2013-11-27 19:50:29 | 

表B

| san | san-name | 
------------------ 
| 100 | xxx1 | 
| 200 | xxx2 | 
| 300 | xxx3 | 

與此查詢

我得到這個結果（一切都運行完美）

| id |  lastdate  | san | 
-------------------------------------- 
| 100 | 2013-11-27 19:00:00 | xxx1 | 
| 200 | 2013-11-27 19:33:33 | xxx2 | 
| 300 | 2013-11-27 19:50:29 | xxx3 | 

，但現在我也想加入的「創造」最新的「ID」，「身份」到我的結果。

從TableA的，我想行（見< - 最新的San .....）

，它應該是這樣的：

| id | status |  lastdate  | san | 
------------------------------------------------ 
| 100 | 4 | 2013-11-27 19:00:00 | xxx1 | 
| 200 | 3 | 2013-11-27 19:33:33 | xxx2 | 
| 300 | 5 | 2013-11-27 19:50:29 | xxx3 | 

我用此代碼試用過

SELECT id, status, max(lastdate) AS lastdate, max(san) AS san 
FROM 
    (SELECT `san` AS id, 
      `created` AS lastdate, 
      '' AS san, 
      status AS status 
    FROM `TableA` a 
    UNION ALL 
    SELECT `san` AS id, 
      cast(null as datetime) AS lastdate, 
      `san-name` AS san, 
      '' AS status 
    FROM `TableB` b 
    ) sq 
GROUP BY id 
ORDER BY san ASC 

，但它給了我這個（錯誤的結果）

| id | status |  lastdate  | san | 
------------------------------------------------ 
| 100 | 3 | 2013-11-27 19:00:00 | xxx1 | 
| 200 | 8 | 2013-11-27 19:33:33 | xxx2 | 
| 300 | 1 | 2013-11-27 19:50:29 | xxx3 | 

，不正確的是這樣

| id | status |  lastdate  | san | 
------------------------------------------------ 
| 100 | 4 | 2013-11-27 19:00:00 | xxx1 | 
| 200 | 3 | 2013-11-27 19:33:33 | xxx2 | 
| 300 | 5 | 2013-11-27 19:50:29 | xxx3 | 

有人能幫助我嗎？ ：O）

我創建了一個sqlfiddle

問候

bernte

Answer 1

我會嘗試與此查詢：

SELECT tA.san AS id, tA.status, tA.created AS lastdate, tB.`san-name` 
FROM 
    tableA tA INNER JOIN (SELECT san, MAX(created) AS max_created 
         FROM tableA 
         GROUP BY san) m 
    ON tA.san=m.san AND tA.created=m.max_created 
    INNER JOIN tableB tB 
    ON tA.san=tB.san 

請參閱小提琴here。

編輯

如果您需要從tableB的返回所有的記錄，只有符合您需要使用正確的記錄JOIN（這是一個有點少見），或者我們可以交換表的順序加入並使用LEFT JOIN，類似這樣的：

SELECT tB.san AS id, tA.status, tA.created AS lastdate, tB.`san-name` 
FROM 
    tableB tB LEFT JOIN 
    (SELECT san, MAX(created) AS max_created 
    FROM tableA 
    GROUP BY san) m 
    ON tB.san=m.san 
    LEFT JOIN TableA tA 
    ON tA.san=m.san AND tA.created=m.max_created 

小提琴是here。