获取每行两列之间的最小值

Question

这是我的数据表：

| uid |   date   | visit | transactionDate |
+-----+----------+-------+-----------------+
|  1  | 6/2/2014 |   1   |     6/9/2014    |
|  1  | 6/2/2014 |   1   |     8/4/2014    |
|  2  | 6/2/2014 |   1   |     8/2/2014    |
|  2  | 6/2/2014 |   1   |     10/17/2014  |
|  2  | 6/2/2014 |   1   |     10/20/2014  |
|  3  | 6/2/2014 |   1   |     6/9/2014    |
|  3  | 6/2/2014 |   1   |     6/10/2014   |
|  3  | 6/2/2014 |   1   |     6/11/2014   | 
|  3  | 6/2/2014 |   1   |     6/12/2014   |
|  3  | 6/2/2014 |   1   |     6/14/2014   |
|  3  | 6/2/2014 |   1   |     6/15/2014   |
|  3  | 6/2/2014 |   1   |     6/17/2014   |
|  3  | 6/2/2014 |   1   |     6/18/2014   |
|  3  | 6/2/2014 |   1   |     6/23/2014   |

我正在尝试编写一个查询来提取两列日期和交易日期的最小值。 有没有办法做MIN（date，transactionDate）这样的事情？ 查询应该选择以下内容：

uid 1 then minimum of date and transaction_dt
uid 2 then min date and transaction_dt

Answer 1

使用CASE条件。

SELECT uid, visit, 
   CASE WHEN date < transactionDate THEN date ELSE transactionDate END AS minDate
FROM table;

Answer 2

如果您正在寻找每行的最小值：

select uid,visit,least(date,transactionDate) as minDate from t;

如果您正在寻找最低每个人群：

select uid,sum(visit) as totalVisits,min(least(date,transactionDate)) as minDate
  from t
  group by uid;

Answer 3

   SELECT UID ,MIN(tdate) FROM 
       (SELECT a.uid, a.date tdate FROM tableA a 
      UNION 
      SELECT a.uid, a.transaction_dt tdate FROM tableA a ) AS tABLE2 T GROUP BY T.UID

Answer 4

使用LEAST（）函数和MIN（）函数。

尝试这个：

SELECT a.uid, MIN(LEAST(a.date, a.transaction_dt)) tdate 
FROM tableA a 
GROUP BY a.uid;

要么

SELECT a.uid, MIN(a.tdate) tdate
FROM (SELECT a.uid, MIN(a.date) tdate FROM tableA a GROUP BY a.uid
      UNION 
      SELECT a.uid, MIN(a.transaction_dt) tdate FROM tableA a GROUP BY a.uid
     ) AS a
GROUP BY a.uid;