问题
This is my data table:
| 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 |
I am trying to write a query to pull the minimum of the two columns date and transaction date. Is there a way to do something like MIN(date, transactionDate)? The query should select something like this:
uid 1 then minimum of date and transaction_dt
uid 2 then min date and transaction_dt
回答1:
Use CASE condition.
SELECT uid, visit,
CASE WHEN date < transactionDate THEN date ELSE transactionDate END AS minDate
FROM table;
回答2:
If you're looking for minimum per row:
select uid,visit,least(date,transactionDate) as minDate from t;
If you're looking for minimum per uid group:
select uid,sum(visit) as totalVisits,min(least(date,transactionDate)) as minDate
from t
group by uid;
回答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
回答4:
Use LEAST() function with MIN() function.
Try this:
SELECT a.uid, MIN(LEAST(a.date, a.transaction_dt)) tdate
FROM tableA a
GROUP BY a.uid;
OR
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;
来源:https://stackoverflow.com/questions/27372259/get-minimum-value-between-two-columns-for-each-row