Invoice
| id | invoice_date|
123435 2016/02/09
123436 2016/02/09
123437 2016/02/09
Payments
|invoice_id| pmt_type| amount |
123435 1 1000
123435 2 2500
123436 2 5000
...
Payments_legend
|type| description|
1 cash
2 check
3 credit
我对这个问题很满意。我需要提取日常交易并认为这很简单。我遇到的问题是许多嵌套连接,除非有更简单的方法来执行此操作。从表面上看,它看起来确实很简单。
以下是我需要的所有发票......
SELECT id AS Invoice FROM invoice WHERE DATE(invoice_date) = DATE(NOW())
但那是我感到困惑的地方。我需要按发票分组的每个payments.amount列的总和,并将其分为每个payments.pmt_type的列。让人惊讶。
我需要报告如下......
id sum(pmt_type=1) sum(pmt_type=2) sum(pmt_type=3)
看起来像......
Invoice Cash Check Credit
123435 1000 2500 0
123436 0 5000 0
123437 0 0 7500
非常感谢任何帮助。
感谢NECULON的帮助
SELECT invoice.id,
sum(if(payments.pmt_type = 1, payments.amount, 0)) AS Cash,
sum(if(payments.pmt_type = 2, payments.amount, 0)) AS Chek,
sum(if(payments.pmt_type = 3, payments.amount, 0)) AS Credit,
sum(if(payments.pmt_type = 9, payments.amount, 0)) AS Warranty,
sum(if(payments.pmt_type = 10, payments.amount, 0)) AS Paypal,
sum(if(payments.pmt_type = 7, payments.amount, 0)) AS Refund,
(sum(if(payments.pmt_type = 1, payments.amount, 0))+sum(if(payments.pmt_type = 2, payments.amount, 0))+sum(if(payments.pmt_type = 3, payments.amount, 0))+sum(if(payments.pmt_type = 9, payments.amount, 0))+sum(if(payments.pmt_type = 10, payments.amount, 0))-sum(if(payments.pmt_type = 7, payments.amount, 0))) AS Total
FROM invoice
JOIN payments ON invoice.id = payments.invoice_id
WHERE DATE(invoice_date) = DATE(now())
GROUP BY invoice.id WITH ROLLUP
这给了我我需要的东西。再次感谢!
答案 0 :(得分:0)
尝试使用" IF"或" CASE"。像这样:
select
invoice.id,
sum(if(payments.pmt_type = 1, payments.amount, 0)) cash,
sum(if(payments.pmt_type = 2, payments.amount, 0)) check,
sum(if(payments.pmt_type = 3, payments.amount, 0)) credit
from invoice
join payment on invoice.id = payments.invoice_id
where date(invoice_date) = date(now())
group by invoice.id
见这里:http://dev.mysql.com/doc/refman/5.7/en/control-flow-functions.html#function_if