MySQL 给两次转置的结果集增加列头

作者: 只是甲 | 来源:发表于2021-03-03 13:47 被阅读0次

备注:测试数据库版本为MySQL 8.0

测试数据:

create table it_research(deptno int, ename varchar(20));

insert into it_research values (100,'HOPKINS');
insert into it_research values (100,'JONES');
insert into it_research values (100,'TONEY');
insert into it_research values (200,'MORALES');
insert into it_research values (200,'P.WHITAKER');
insert into it_research values (200,'MARCIANO');
insert into it_research values (200,'ROBINSON');
insert into it_research values (300,'LACY');
insert into it_research values (300,'WRIGHT');
insert into it_research values (300,'J.TAYLOR');


CREATE TABLE IT_APPS (deptno int,ename varchar(20));

insert into it_apps values (400,'CORRALES');
insert into it_apps values (400,'MAYWEATHER');
insert into it_apps values (400,'CASTILLO');
insert into it_apps values (400,'MARQUEZ');
insert into it_apps values (400,'MOSLEY');
insert into it_apps values (500,'GATTI');
insert into it_apps values (500,'CALZAGHE');
insert into it_apps values (600,'LAMOTTA');
insert into it_apps values (600,'HAGLER');
insert into it_apps values (600,'HEARNS');
insert into it_apps values (600,'FRAZIER');
insert into it_apps values (700,'GUINN');
insert into it_apps values (700,'JUDAH');
insert into it_apps values (700,'MARGARITO');

一.需求

把两个结果集叠在一起，然后把他们转置为两列，另外，还要为每列加一个"标题"。
例如，有两个表，它们包含公司中有关员工的信息，这些员工在不同地区从事开发工作(也即研究和应用):

mysql> select * from it_research;
+--------+------------+
| deptno | ename |
+--------+------------+
| 100 | HOPKINS |
| 100 | JONES |
| 100 | TONEY |
| 200 | MORALES |
| 200 | P.WHITAKER |
| 200 | MARCIANO |
| 200 | ROBINSON |
| 300 | LACY |
| 300 | WRIGHT |
| 300 | J.TAYLOR |
+--------+------------+
10 rows in set (0.00 sec)

mysql> select * from it_apps;
+--------+------------+
| deptno | ename |
+--------+------------+
| 400 | CORRALES |
| 400 | MAYWEATHER |
| 400 | CASTILLO |
| 400 | MARQUEZ |
| 400 | MOSLEY |
| 500 | GATTI |
| 500 | CALZAGHE |
| 600 | LAMOTTA |
| 600 | HAGLER |
| 600 | HEARNS |
| 600 | FRAZIER |
| 700 | GUINN |
| 700 | JUDAH |
| 700 | MARGARITO |
+--------+------------+
14 rows in set (0.00 sec)

要创建一个报表，它分两栏列出两个表中的员工。
对于每一列，都返回deptno和ename。

最后，返回下列结果集:

image.png

二.解决方案

本解决方案只需要一个简单的堆叠及转置(合并后转置)并且再"拧"一次:deptno 一定在每个员工的ename之前。
这里的技巧采用了笛卡尔积为每个deptno生产附加行，这样才有足够的行显示所有员工和deptno。

with tmp1 as
(
select deptno,
       ename,
       count(*) over (partition by deptno) cnt
from it_apps
order by deptno,ename 
),
tmp2 as
(
select 1 id union select 2 
),
tmp3 as
(
select deptno,
       ename,
       cnt,
       row_number() over w1 as 'rn'
  from tmp1,tmp2
window w1 as (partition by deptno order by id,ename)
),
tmp4 as
(
select 1 flag1,
       1 flag2,
       case when rn = 1 then deptno else concat('  ',ename) end it_dept
  from tmp3 
  where rn <= cnt + 1
)
,
tmp5 as
(
select deptno,
       ename,
       count(*) over (partition by deptno) cnt
from it_research
order by deptno,ename 
),
tmp6 as
(
select deptno,
       ename,
       cnt,
       row_number() over w2 as 'rn'
  from tmp5,tmp2
 window w2 as (partition by deptno order by id,ename)
),
tmp7 as
(
select 1 flag1,0 flag2,case when rn = 1 then deptno else concat('  ',ename) end it_dept
  from tmp6
  where rn <= cnt + 1
),
tmp8 as
(
select flag1,
       flag2,
       it_dept
 from tmp7
union all
select flag1,
       flag2,
       it_dept
 from tmp4
),
tmp9 AS
(
select flag1,
       flag2,
       it_dept,
       row_number() over w3 as 'rn2'
  from tmp8
 window w3 as ()
),
tmp10 as
(
select sum(flag1) over (partition by flag2 order by flag1,rn2) flag,
       flag2,
       it_dept
  from tmp9
)
select max(case when flag2 = 0 then it_dept end) research,
       max(case when flag2 = 1 then it_dept end) apps
  from tmp10
 group by flag
 order by flag

测试记录:

mysql> with tmp1 as
    -> (
    -> select deptno,
    ->        ename,
    ->        count(*) over (partition by deptno) cnt
    -> from it_apps
    -> order by deptno,ename
    -> ),
    -> tmp2 as
    -> (
    -> select 1 id union select 2
    -> ),
    -> tmp3 as
    -> (
    -> select deptno,
    ->        ename,
    ->        cnt,
    ->        row_number() over w1 as 'rn'
    ->   from tmp1,tmp2
    -> window w1 as (partition by deptno order by id,ename)
    -> ),
    -> tmp4 as
    -> (
    -> select 1 flag1,
    ->        1 flag2,
    ->        case when rn = 1 then deptno else concat('  ',ename) end it_dept
    ->   from tmp3
    ->   where rn <= cnt + 1
    -> )
    -> ,
    -> tmp5 as
    -> (
    -> select deptno,
    ->        ename,
    ->        count(*) over (partition by deptno) cnt
    -> from it_research
    -> order by deptno,ename
    -> ),
    -> tmp6 as
    -> (
    -> select deptno,
    ->        ename,
    ->        cnt,
    ->        row_number() over w2 as 'rn'
    ->   from tmp5,tmp2
    ->  window w2 as (partition by deptno order by id,ename)
    -> ),
    -> tmp7 as
    -> (
    -> select 1 flag1,0 flag2,case when rn = 1 then deptno else concat('  ',ename) end it_dept
    ->   from tmp6
    ->   where rn <= cnt + 1
    -> ),
    -> tmp8 as
    -> (
    -> select flag1,
    ->        flag2,
    ->        it_dept
    ->  from tmp7
    -> union all
    -> select flag1,
    ->        flag2,
    ->        it_dept
    ->  from tmp4
    -> ),
    -> tmp9 AS
    -> (
    -> select flag1,
    ->        flag2,
    ->        it_dept,
    ->        row_number() over w3 as 'rn2'
    ->   from tmp8
    ->  window w3 as ()
    -> ),
    -> tmp10 as
    -> (
    -> select sum(flag1) over (partition by flag2 order by flag1,rn2) flag,
    ->        flag2,
    ->        it_dept
    ->   from tmp9
    -> )
    -> select max(case when flag2 = 0 then it_dept end) research,
    ->        max(case when flag2 = 1 then it_dept end) apps
    ->   from tmp10
    ->  group by flag
    ->  order by flag ;
+--------------+--------------+
| research     | apps         |
+--------------+--------------+
| 100          | 400          |
|   JONES      |   CORRALES   |
|   TONEY      |   MARQUEZ    |
|   HOPKINS    |   MAYWEATHER |
| 200          |   MOSLEY     |
|   MORALES    |   CASTILLO   |
|   P.WHITAKER | 500          |
|   ROBINSON   |   GATTI      |
|   MARCIANO   |   CALZAGHE   |
| 300          | 600          |
|   LACY       |   HAGLER     |
|   WRIGHT     |   HEARNS     |
|   J.TAYLOR   |   LAMOTTA    |
| NULL         |   FRAZIER    |
| NULL         | 700          |
| NULL         |   JUDAH      |
| NULL         |   MARGARITO  |
| NULL         |   GUINN      |
+--------------+--------------+
18 rows in set (0.00 sec)

这样看起来临时表都10个临时表了，逻辑看起来太复杂了，下面我们拆开来讲解

2.1 分解求出it_apps各部门及员工

我们首先来看看tmp4的结果集:

mysql> with tmp1 as
    -> (
    -> select deptno,
    ->        ename,
    ->        count(*) over (partition by deptno) cnt
    -> from it_apps
    -> order by deptno,ename
    -> ),
    -> tmp2 as
    -> (
    -> select 1 id union select 2
    -> ),
    -> tmp3 as
    -> (
    -> select deptno,
    ->        ename,
    ->        cnt,
    ->        row_number() over w1 as 'rn'
    ->   from tmp1,tmp2
    -> window w1 as (partition by deptno order by id,ename)
    -> ),
    -> tmp4 as
    -> (
    -> select 1 flag1,
    ->        1 flag2,
    ->        case when rn = 1 then deptno else concat('  ',ename) end it_dept
    ->   from tmp3
    ->   where rn <= cnt + 1
    -> )
    -> select * from tmp4;
+-------+-------+--------------+
| flag1 | flag2 | it_dept      |
+-------+-------+--------------+
|     1 |     1 | 400          |
|     1 |     1 |   CORRALES   |
|     1 |     1 |   MARQUEZ    |
|     1 |     1 |   MAYWEATHER |
|     1 |     1 |   MOSLEY     |
|     1 |     1 |   CASTILLO   |
|     1 |     1 | 500          |
|     1 |     1 |   GATTI      |
|     1 |     1 |   CALZAGHE   |
|     1 |     1 | 600          |
|     1 |     1 |   HAGLER     |
|     1 |     1 |   HEARNS     |
|     1 |     1 |   LAMOTTA    |
|     1 |     1 |   FRAZIER    |
|     1 |     1 | 700          |
|     1 |     1 |   JUDAH      |
|     1 |     1 |   MARGARITO  |
|     1 |     1 |   GUINN      |
+-------+-------+--------------+
18 rows in set (0.00 sec)

那么tmp4的结果集是怎么来的呢？
我们来看看tmp1和tmp2以及tmp3
这个地方为什么要有tmp2，一个id为1 和 2 的表，其实因为第一行要显示deptno，后面的才是ename，所以此时应该是n+1，如果只是n，这个地方deptno带不出来。
tmp3就是在tmp1和tmp2的基础上进行了加工，记住order by这个地方一定要正确，不然数据会不准确

tmp3加工完成后，根据每个deptno rn为1的则显示deptno，从2到n+1遍历ename，根据rn <= cnt + 1来进行控制

mysql> select deptno,
    ->        ename,
    ->        count(*) over (partition by deptno) cnt
    -> from it_apps
    -> order by deptno,ename ;
+--------+------------+-----+
| deptno | ename      | cnt |
+--------+------------+-----+
|    400 | CASTILLO   |   5 |
|    400 | CORRALES   |   5 |
|    400 | MARQUEZ    |   5 |
|    400 | MAYWEATHER |   5 |
|    400 | MOSLEY     |   5 |
|    500 | CALZAGHE   |   2 |
|    500 | GATTI      |   2 |
|    600 | FRAZIER    |   4 |
|    600 | HAGLER     |   4 |
|    600 | HEARNS     |   4 |
|    600 | LAMOTTA    |   4 |
|    700 | GUINN      |   3 |
|    700 | JUDAH      |   3 |
|    700 | MARGARITO  |   3 |
+--------+------------+-----+
14 rows in set (0.00 sec)

mysql> select 1 id union select 2 ;
+----+
| id |
+----+
|  1 |
|  2 |
+----+
2 rows in set (0.00 sec)

mysql> with tmp1 as
    -> (
    -> select deptno,
    ->        ename,
    ->        count(*) over (partition by deptno) cnt
    -> from it_apps
    -> order by deptno,ename
    -> ),
    -> tmp2 as
    -> (
    -> select 1 id union select 2
    -> )
    -> select deptno,
    ->        ename,
    ->        cnt,
    ->        row_number() over w1 as 'rn'
    ->   from tmp1,tmp2
    -> window w1 as (partition by deptno order by id,ename)
    -> ;
+--------+------------+-----+----+
| deptno | ename      | cnt | rn |
+--------+------------+-----+----+
|    400 | CASTILLO   |   5 |  1 |
|    400 | CORRALES   |   5 |  2 |
|    400 | MARQUEZ    |   5 |  3 |
|    400 | MAYWEATHER |   5 |  4 |
|    400 | MOSLEY     |   5 |  5 |
|    400 | CASTILLO   |   5 |  6 |
|    400 | CORRALES   |   5 |  7 |
|    400 | MARQUEZ    |   5 |  8 |
|    400 | MAYWEATHER |   5 |  9 |
|    400 | MOSLEY     |   5 | 10 |
|    500 | CALZAGHE   |   2 |  1 |
|    500 | GATTI      |   2 |  2 |
|    500 | CALZAGHE   |   2 |  3 |
|    500 | GATTI      |   2 |  4 |
|    600 | FRAZIER    |   4 |  1 |
|    600 | HAGLER     |   4 |  2 |
|    600 | HEARNS     |   4 |  3 |
|    600 | LAMOTTA    |   4 |  4 |
|    600 | FRAZIER    |   4 |  5 |
|    600 | HAGLER     |   4 |  6 |
|    600 | HEARNS     |   4 |  7 |
|    600 | LAMOTTA    |   4 |  8 |
|    700 | GUINN      |   3 |  1 |
|    700 | JUDAH      |   3 |  2 |
|    700 | MARGARITO  |   3 |  3 |
|    700 | GUINN      |   3 |  4 |
|    700 | JUDAH      |   3 |  5 |
|    700 | MARGARITO  |   3 |  6 |
+--------+------------+-----+----+
28 rows in set (0.00 sec)

2.2 it_apps与it_research进行拼接

tmp9得到如下结果:

image.png

此时就可以知道 flag1和flag2的作用了，flag2用来区别是 apps还是research，flag1根据flag2的区别进行累计求和，然后根据新的flag来进行group by，这样就可以将apps和research进行展示了

tmp10返回如下结果:

image.png

有了tmp10的结果，就很可以根据flag直接分组，通过max函数进行求求值。

MySQL 给两次转置的结果集增加列头

一.需求

二.解决方案

2.1 分解求出it_apps各部门及员工

2.2 it_apps与it_research进行拼接

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