我有2个表,我需要根据日期和2个值加在一起.
这给了我所有信息的清单 - 很好.
$query = (SELECT date, debit, credit , note FROM proj3_cash )
UNION
(SELECT settle, purch, sale, issue FROM proj3_trades)
ORDER BY date";
现在我需要从两个表中对每日总计的信息进行分组.
$query = "(SELECT date, SUM(debit), SUM(credit)FROM proj3_cash GROUP BY date)
UNION
(SELECT settle as date, SUM(purch) as debit, SUM(sale) as credit FROM proj3_trades GROUP BY date)
ORDER BY date";
很好,但是如果每张桌子上的同一天有什么东西,我会得到这个:
date SUM(debit) SUM(credit)
--------------------------------------
2010-12-02 0.00 170.02
2010-12-02 296449.91 233111.10
如何将两者分组到同一天？
如果我在最后添加GROUP BY - 我只会收到错误.或者这应该通过JOIN完成？