我有一个income
表,如下所示:
date income
---------------------------
09/05/13 56000
09/05/13 66600
09/05/13 50000
一个expense
表,如下所示:
date expense
----------------------------
09/05/13 68800
我想编写一个输出如下的查询:
date income expense
---------------------------------------------
09/05/13 56000 68800
09/05/13 66600
09/05/13 50000
答案 0 :(得分:3)
UPDATE2:如果您不关心收入值与费用值匹配的特定订单,您可以通过此类查询获得所需的输出
SELECT date,
MAX(CASE WHEN type = 1 THEN amount END) income,
MAX(CASE WHEN type = 2 THEN amount END) expense
FROM
(
SELECT 1 type, date, income amount, @n := IF(@g = date, @n + 1, 1) rnum, @g := date g
FROM income CROSS JOIN (SELECT @n := 0, @g := NULL) i1
UNION ALL
SELECT 2, date, expense amount, @m := IF(@f = date, @m + 1, 1) rnum, @f := date g
FROM expense CROSS JOIN (SELECT @m := 0, @f := NULL) i2
) q
GROUP BY date, rnum
输出:
| DATE | INCOME | EXPENSE | |--------------------|--------|---------| | September, 05 2013 | 56000 | 68800 | | September, 05 2013 | 66600 | (null) | | September, 05 2013 | 50000 | (null) |
这是 SQLFiddle 演示
UPDATE1:使用像
这样的查询对每日收入和费用进行分组似乎合乎逻辑SELECT a.date, i.income, e.expense
FROM
(
SELECT date
FROM income
UNION
SELECT date
FROM expense
) a LEFT JOIN
(
SELECT date, SUM(income) income
FROM income
GROUP BY date
) i
ON a.date = i.date LEFT JOIN
(
SELECT date, SUM(expense) expense
FROM expense
GROUP BY date
) e
ON a.date = e.date
输出:
| DATE | INCOME | EXPENSE | |--------------------|--------|---------| | September, 05 2013 | 172600 | 68800 |
这是 SQLFiddle 演示
原始问题的原始答案:如果您需要原始问题中所述的FULL OUTER JOIN
(超过date
列),那么这样就可以了
SELECT a.date, i.income, e.expense
FROM
(
SELECT date
FROM income
UNION
SELECT date
FROM expense
) a LEFT JOIN income i
ON a.date = i.date LEFT JOIN expense e
ON a.date = e.date
但是你会得到这个作为输出
| DATE | INCOME | EXPENSE | |---------------|--------|---------| | September, 05 | 56000 | 68800 | | September, 05 | 66600 | 68800 | | September, 05 | 50000 | 68800 |
这是 SQLFiddle 演示