Question

我需要能够将两列相加并获得另外两列的第一个输入值。

示例表

trans,invoice,email,amount,ticket,date
111,inv-111,pp@gmail.com,15.00,3,"2014-09-13 15:20:12"
125,inv-125,ab@gmail.com,10.00,2,"2014-09-13 15:15:40"
132,inv-132,ab@gmail.com,10.00,2,"2014-09-13 15:17:19"
145,inv-145,ab@gmail.com,5.00,1,"2014-09-13 15:17:52"
199,inv-199,pp@gmail.com,5.00,1,"2014-09-13 15:18:18"

这就是我所得到的：

通过电子邮件从反式组中选择金额（金额）作为金额，金额（票）作为票，反式，发票，电子邮件，最小（日期）;

amount,ticket,trans,invoice,email,min(date)
25.00,5,125,inv-125,ab@gmail.com,"2014-09-13 15:15:40"
20.00,4,111,inv-111,pp@gmail.com,"2014-09-13 15:18:18"

期望的结果：我希望发票和反式列从第一个日期开始。

amount,ticket,trans,invoice,email,min(date)
25.00,5,125,inv-125,ab@gmail.com,"2014-09-13 15:15:40"
20.00,4,199,inv-199,pp@gmail.com,"2014-09-13 15:18:18"

Answer 1

您可以使用substring_index() / group_concat()诀窍来执行此操作：

select sum(amount) as amount, sum(ticket) as ticket,
       substring_index(group_concat(trans order by date asc), ',', 1) as trans,
       substring_index(group_concat(invoice order by date asc), ',', 1) as invoice,
       email,
       min(date)
from trans
group by email;

请注意，这会将值放入长字符串中。这通常在实践中有效，但在中间字符串上有大约1000个字符的默认长度。此外，此结果数据类型是字符串而不是数字，这可能会影响trans列。输出看起来相同，但底层数据类型不同。

Answer 2

您可以简单地对预先排序的数据执行操作。使用子查询按日期排序，以便获取最小值。

SELECT sum( n.amount ) AS amount, sum( n.ticket ) AS ticket, n.trans, n.invoice, n.email, min( n.date ) 
FROM (
SELECT * 
FROM test
ORDER BY date
) AS n
GROUP BY email

输出：

amount  ticket  trans  invoice  email  min(n.date)  
25 5 125 inv-125 ab@gmail.com 2014-09-13 15:15:40 
20 4 199 inv-199 pp@gmail.com 2014-09-13 15:18:18

Answer 3

您也可以使用自我加入

来完成此操作

select 
t1.amount,
t1.ticket,
t.email,
t.trans,
t.invoice,
t.date
from test t
join (
     select min(date) date,
      email,
      sum(amount) as amount, 
      sum(ticket) as ticket
      from test 
      group by email
     ) t1
on(t.email = t1.email and t.date = t1.date)

Demo

Mysql选择查询将两列相加并从另外两列获取第一个值

3 个答案: