Total_payments是每个玩家支付的所有款项的总和。如何为每个ID只显示playerid和total_payments获得一行。我不想要复印机。
SELECT playerid
, payment_amount, SUM(payment_amount) OVER (PARTITION BY playerid) AS total_payment
FROM DEPOSIT$
| playerid | payment_amount| total_payment |
|:1002 |:10 :| 60 :|
|:1002 |:50 :| 60 :|
|:1003 |:40 :| 50 :|
|:1003 |:10 :| 50 :|
|:1004 |:10 :| 30 :|
|:1004 |:20 :| 30 :|
|:1006 |:100 :| 100 :|
|:1008 |:60 :| 60 :|
谢谢!
发布于 2022-06-18 16:29:49
一般来说,这是正确的方法--使用sum()和group by子句。因为您是在playerid上分组,所以对于每个playerid,它将是唯一的,并且支付金额的总和将被求和。这实际上只是一个标准的sql查询,用于通过某些属性进行求和。
SELECT
playerid,
SUM(payment_amount) AS total_payment
FROM
DEPOSIT$
GROUP BY
playerid
也许最后你忘了按条款分组了?
https://stackoverflow.com/questions/72672174
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