[英]Get minimum value between two columns for each row
這是我的數據表:
| 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 |
我正在嘗試編寫一個查詢來提取兩列日期和交易日期的最小值。 有沒有辦法做MIN(date,transactionDate)這樣的事情? 查詢應該選擇以下內容:
uid 1 then minimum of date and transaction_dt
uid 2 then min date and transaction_dt
使用CASE
條件。
SELECT uid, visit,
CASE WHEN date < transactionDate THEN date ELSE transactionDate END AS minDate
FROM table;
如果您正在尋找每行的最小值:
select uid,visit,least(date,transactionDate) as minDate from t;
如果您正在尋找最低每個人群:
select uid,sum(visit) as totalVisits,min(least(date,transactionDate)) as minDate
from t
group by uid;
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
使用LEAST()函數和MIN()函數。
嘗試這個:
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;
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