MySQL选择每个组中第一个、第二个和最后一个值的最佳方法

我尝试使用MySQL获取表中每个组的第一个、第二个和最后一个值。我的数据行如下所示：

userID    purchaseTime
----------------------
1         2018-01-01
1         2018-01-02
1         2018-01-03
1         2018-01-04
2         2018-02-01
2         2018-02-02
3         2018-03-01

预期结果将是：

userID    first         second        last
------------------------------------------------
1         2018-01-01    2018-01-02    2018-01-04
2         2018-02-01    2018-02-02    2018-02-02
3         2018-03-01    null          2018-03-01

在谷歌搜索了半天之后，我只能想出一个愚蠢的方法，分别执行以下两个查询，然后用我的服务器端代码合并结果：

//get 1st, 2nd values 
SELECT userID, purchaseTime
FROM   purchaseLog t1
WHERE 
    (
        SELECT  COUNT(*) 
        FROM    purchaseLog t2
        WHERE t2.userID = t1.userID AND 
              t2.purchaseTime<= t1.purchaseTime
    ) <= 2 order by t1.userID , t1.purchaseTime;

//get last value
SELECT max(purchaseTime) FROM purchaseTime GROUP BY userID 

---

我敢肯定，一定有一种更优雅的方法可以一次性获得结果。有人能帮我达到我的要求吗？谢谢大家!

以下代码未经测试，但应该会给您一个好主意：

SELECT
    t1.userID,
    t1.purchaseTime AS first,
    t2.purchaseTime AS `second`,
    t4.purchaseTime AS last
FROM purchaseLog t1
LEFT JOIN purchaseLog t0 ON t1.userID = t0.userID AND t0.purchaseTime < t1.purchaseTime
LEFT JOIN purchaseLog t2 ON t1.userID = t2.userID AND t1.purchaseTime < t2.purchaseTime
LEFT JOIN purchaseLog t3 ON t1.userID = t3.userID AND t1.purchaseTime < t3.purchaseTime
                        AND t3.purchaseTime < t2.purchaseTime
JOIN purchaseLog t4 ON t1.userID = t4.userID AND t1.purchaseTime <= t4.purchaseTime
LEFT JOIN purchaseLog t5 ON t1.userID = t5.userID AND t4.purchaseTime < t5.purchaseTime
WHERE t0.purchaseTime IS NULL AND t3.purchaseTime IS NULL AND t5.purchaseTime IS NULL

让我一步一步地把它分解：

首先，我获取所有不存在相同userID的较早行的行：

SELECT
    t1.userID,
    t1.purchaseTime AS first
FROM purchaseLog t1
LEFT JOIN purchaseLog t0 ON t1.userID = t0.userID AND t0.purchaseTime < t1.purchaseTime
WHERE t0.purchaseTime IS NULL

接下来，我获取purchaseTime大于第一个purchaseTime的所有行，其中没有purchaseTime介于两者之间的行：

SELECT
    t1.userID,
    t2.purchaseTime AS `second`
FROM purchaseLog t1
LEFT JOIN purchaseLog t2 ON t1.userID = t2.userID AND t1.purchaseTime < t2.purchaseTime
LEFT JOIN purchaseLog t3 ON t1.userID = t3.userID AND t1.purchaseTime < t3.purchaseTime
                        AND t3.purchaseTime < t2.purchaseTime
WHERE t3.purchaseTime IS NULL

最后，我得到purchaseTime大于或等于第一行的行，其中不存在更大的purchaseTime：

SELECT
    t1.userID,
    t4.purchaseTime AS last
FROM purchaseLog t1
JOIN purchaseLog t4 ON t1.userID = t4.userID AND t1.purchaseTime <= t4.purchaseTime
LEFT JOIN purchaseLog t5 ON t1.userID = t5.userID AND t4.purchaseTime < t5.purchaseTime
WHERE t5.purchaseTime IS NULL

---

将它们组合到一个查询中，即可得到上述答案。

好吧，您可能可以这样做：

选择第一个、第二个和最后一个语句，然后将它们连接在一起：

SELECT a.userID, a.purchaseTime 
    FROM fsl a WHERE a.purchaseTime = 
    (SELECT MAX(b.purchaseTime) FROM fsl b WHERE b.userID = a.userID)

SELECT a.userID, a.purchaseTime 
    FROM fsl a WHERE a.purchaseTime = 
    (SELECT MAX(c.purchaseTime) FROM fsl c 
        WHERE c.purchaseTime < (SELECT MAX(b.purchaseTime) 
            FROM fsl b WHERE b.userID = a.userID));

SELECT a.userID, a.purchaseTime 
    FROM fsl a WHERE a.purchaseTime = 
    (SELECT MIN(b.purchaseTime) 
        FROM fsl b WHERE b.userID = a.userID)

并使用连接将其拼接在一起：

SELECT fst.userID as userID, fst.purchaseTime as first, snd.purchaseTime as snd, trd.purchaseTime as last FROM
    (SELECT a.userID, a.purchaseTime 
    FROM fsl a 
    WHERE a.purchaseTime = 
        (SELECT MAX(b.purchaseTime) FROM fsl b WHERE b.userID = a.userID)) fst
    JOIN (SELECT a.userID, a.purchaseTime FROM fsl a 
            WHERE a.purchaseTime = (SELECT MAX(c.purchaseTime) 
            FROM fsl c WHERE c.purchaseTime < (SELECT MAX(b.purchaseTime) 
            FROM fsl b WHERE b.userID = a.userID))) snd
    ON fst.userID = snd.userID
    JOIN (SELECT a.userID, a.purchaseTime 
            FROM fsl a WHERE a.purchaseTime = 
            (SELECT MIN(b.purchaseTime) FROM fsl b WHERE b.userID = a.userID)) trd
    ON trd.userID = snd.userID;

---

然而，我不能保证这对于任何类型的生产使用都足够快

我很困惑，在输入的情况下，你是如何得到期望的结果的。那么第一个应该是各自ID的最早购买时间，第二个应该是中间时间，第三个应该是最晚的购买时间？您的第一行似乎是这样的，但您的其他两行与之不匹配，所以我猜这不是您想要的？可以将上述内容组合到一个查询中，但我认为这不是一个真正有效的查询。@Obsidiage希望这个解释更清楚：在我的要求中，如果用户在表中有3个以上的购买日志，则字段last将是该用户的最后购买时间。如果一个用户只有一个购买日志，最后一个字段将与第一个字段相同，因为只有一个日志属于该用户，所以购买时间是第一个和最后一个。@Obsidian对于我来说似乎很清楚，因为我给了我一个提示，我现在就要研究它！虽然这可能需要我的时间，因为我是mysql的新手，无论如何，谢谢你！这是一个非常聪明的解决方案，给了我一些我一直在寻找的东西。谢谢你给我的帮助。在这一刻，响应时间不是我的首要任务，我会花时间研究你给出的答案，谢谢！