MySQL的选择与UNION ALL

Question

我（使用UNION ALL，因为我有万吨行的）

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 | 

与此查询

我得到这个结果（一切都运行完美）

| 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 | 

，但现在我也想加入的“创造”最新的“ID”，“身份”到我的结果。

从TableA的，我想行（见< - 最新的San .....）

，它应该是这样的：

| 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 | 

我用此代码试用过

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 

，但它给了我这个（错误的结果）

| 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 | 

，不正确的是这样

| 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 | 

有人能帮助我吗？ ：O）

我创建了一个sqlfiddle

问候

bernte

Answer 1

我会尝试与此查询：

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 

请参阅小提琴here。

编辑

如果您需要从tableB的返回所有的记录，只有符合您需要使用正确的记录JOIN（这是一个有点少见），或者我们可以交换表的顺序加入并使用LEFT JOIN，类似这样的：

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。