Question

如果考虑汇率，如何选择在2018年12月支付了大笔款项的客户

我有一张桌子：

最后日期-交易日期
数字转帐（20,2）-付款金额
CurrencyRate数字（20,2）-货币汇率

 ID_Client  Trandate    Transum CurrencyRate    Currency
 --------------------------------------------------------
 1          2018.12.01  100     1               UAH
 1          2018.12.02  150     2               USD
 2          2018.12.01  200     1               UAH
 3          2018.12.01  250     3               EUR
 3          2018.12.02  300     1               UAH
 3          2018.12.03  350     2               USD
 7          2019.01.08  600     1               UAH

但是我认为“ max”根本不是我所需要的

 SELECT ID_Client, MAX(Transum*CurrencyRate) 
 FROM `Payment.TotalPayments` 
 WHERE YEAR(Trandate) = 2018
       AND MONTH(Trandate) = 12

我需要这个东西

ID_Client   Transum 
   3         1750

其中1750是“ UAH”，而350USD + 300UAH + 250EUR，USD的汇率是2，EUR的汇率是3。

Answer 1

我认为您想要sum()。然后，您可以order by的结果：

SELECT ID_Client, SUM(Transum*CurrencyRate) as total 
FROM `Payment.TotalPayments` 
WHERE Trandate >= '2018-12-01' AND Transdate < '2019-01-01'
GROUP BY ID_Client
ORDER BY total DESC;

Answer 2

如果您尝试按客户获取2018年和12月的交易金额之和，则可以这样写：

SELECT ID_Client, SUM(Transum*CurrencyRate) as payment_total_converted
FROM `Payment.TotalPayments` 
WHERE YEAR(Trandate) = 2018
and MONTH(Trandate) = 12
group by ID_Client

如果您希望按给定日期范围内的每个客户，年份和月份对事物进行分组，则可以这样编写：

SELECT ID_Client, YEAR(Trandate) as tran_year, MONTH(Trandate) as tran_month, 
SUM(Transum*CurrencyRate) as payment_total_converted
FROM `Payment.TotalPayments` 
WHERE Trandate between '2018-12-01' and '2019-01-01'
group by ID_Client, YEAR(Trandate), MONTH(Trandate)

我为您的计算列添加了列名，以便结果集仍然是相关的（列需要不同的名称）。

我建议阅读SQL'group by'子句（https://www.w3schools.com/sql/sql_groupby.asp）和聚合（https://www.w3schools.com/sql/sql_count_avg_sum.asp，https://www.w3schools.com/sql/sql_min_max.asp）运算符。

金额和汇率的SQL任务

2 个答案: