我正在尝试将多行合并为具有多列的单个行,其中数据来自三个表。我遵循了MySQL pivot table中的建议,但问题和答案并未考虑多个联接。
以下是基础表:
table n
+------+----+------+
| name | id | code |
+------+----+------+
| foo | 1 | NULL |
| bar | 2 | z |
| baz | 3 | y |
+------+----+------+
table ac
+------+----+
| code | id |
+------+----+
| h | 1 |
| i | 2 |
+------+----+
table c
+-----+------+-------+
| cid | code | desc |
+-----+------+-------+
| 9 | h | desch |
| 9 | i | desci |
| 8 | z | descz |
| 8 | y | descy |
+-----+------+-------+
以下是预期结果:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| foo | null | desch |
| bar | descz | desci |
| baz | descy | null |
+------+-------+-------+
我可以很接近我想要的结果:
select
n.name,
n.code as type8,
ac.code as type9
from n
left join ac
on ac.id=n.id
但是,正如预期的那样,这仅产生代码:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| foo | null | h |
| bar | z | i |
| baz | y | null |
+------+-------+-------+
并且我有兴趣用表c中的较长描述替换代码。
答案 0 :(得分:0)
第一步是使用pivot tables in mysql中的技术,使用case
从同一表中的多行返回多列数据:
select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join c
on n.code=c.code
对于type8,这将产生正确的结果,但对于type9,则为空:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| foo | null | null |
| bar | descz | null |
| baz | descy | null |
+------+-------+-------+
下一步是获取type9的结果:
select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join ac
on ac.id=n.id
left join c
on c.code=ac.code
这将产生:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| foo | null | desch |
| bar | null | desci |
| baz | null | null |
+------+-------+-------+
如果这两个结果与
结合在一起select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join c
on n.code=c.code
union
select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join ac
on ac.id=n.id
left join c
on c.code=ac.code
行仍需要合并:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| bar | descz | null |
| baz | descy | null |
| foo | null | null |
| foo | null | desch |
| bar | null | desci |
| baz | null | null |
+------+-------+-------+
最后,通过使用removing nulls from a unioned query的聚合技术,我能够达到预期的结果:
select
n.name,
min(type8), min(type9)
from
(select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join c
on n.code=c.code
union
select
n.name,
case when c.cid=8 then c.desc end as type8,
case when c.cid=9 then c.desc end as type9
from n
left join ac
on ac.id=n.id
left join c
on c.code=ac.code) as n
group by n.name
产生预期的结果:
+------+-------+-------+
| name | type8 | type9 |
+------+-------+-------+
| foo | null | desch |
| bar | descz | desci |
| baz | descy | null |
+------+-------+-------+
答案 1 :(得分:0)
获得预期结果的另一种方法是通过将ifnull(.,.)
与2条left join
语句组合在一起,然后聚合cid
来模仿pivot
来创建伪联合。
select name
, min(if(cid = 8, `desc`, null)) Type8
, min(if(cid = 9, `desc`, null)) Type9
from (
select cid, code, `desc`, ifnull(name1, name2) name from (
select c.cid, c.code, c.desc, n1.name name1, n2.name name2
from c
left join n n1 on n1.code = c.code
left join ac on ac.code = c.code
left join n n2 on n2.id = ac.id
) q1
) p1
group by name
order by name
;