SQL面试题练习-实现pivot行列转换

题目：实现把表t1的数据变成表t2的形式

t1 year month amount 1991 1 1.1 1991 2 1.2 1991 3 1.3 1991 4 1.4 1992 1 2.1 1992 2 2.2 1992 3 2.3 1992 4 2.4

t2 year m1 m2 m3 m4 1991 1.1 1.2 1.3 1.4 1992 2.1 2.2 2.3 2.4

1. 一般方法

抓住修改之后的表的每一列的特征，也就是说第n列month值为n；select后面选取出列是针对某一行来说的，这样对于某一行的各个列之间可以作运算，脑洞大开一下，可以想到下面几种方法：

1. 巧妙的使用sign()函数

createdatabase practice;
usepractice;
createtable t1
     (
         year int not null,
         month int not null,
         amount dec(2,1)
     );
insertinto t1 values('1991','1','1.1');
    insert into t1 values('1991','2','1.2');
    insert into t1 values('1991','3','1.3');
    insert into t1 values('1991','4','1.4');
    insert into t1 values('1992','1','2.1');
    insert into t1 values('1992','2','2.2');
    insert into t1 values('1992','3','2.3');
    insert into t1 values('1992','4','2.4');
  
createtable t2 as
selectyear,
  sum(amount*(1-abs(sign(month-1))))as m1,
  sum(amount*(1-abs(sign(month-2))))as m2,
  sum(amount*(1-abs(sign(month-3))))as m3,
       sum(amount*(1-abs(sign(month-4)))) as m4
fromt1
        group by year;

select*from t2;

可以看到达到想要的目的了：

 

1. 使用case…when…then…else…end结构：

select year,
max(case month when '1' then amount else 0 end ) as m1,
max(casemonth when '2' then amount  else 0end)  as m2,
max(casemonth when '3' then amount else 0 end ) as m3,
max(casemonth when '4' then amount else 0 end )as m4
from t1
group by year;

1. 既然如此还有一种办法：用  if(…,…,) ，而且如果非的情况为0，MAX()可以改为SUM()：

select year,
  sum(if (month ='1',amount,0))asm1,
  sum(if (month ='2',amount,0))asm2,
  sum(if (month ='3',amount,0))asm3,
  sum(if (month ='4',amount,0))asm4
from t1
group by year;

1. 使用pivot()函数

mysql中似乎没有pivot函数。pivot()函数一般形式为：

table_source

PIVOT

(

    <聚合函数>(<要聚合的列>)

FOR

[<包含要成为列标题的值的列>]

    IN ( [第一个透视的列], [第二个透视的列],

    ... [最后一个透视的列])

) AS <透视表的别名>

select year ,
     [1] as m1,
     [2] as m2,
     [3] as m3,
     [4] as m4
from t1
pivot(
      sum(amount)
      for month in ([1],[2],[3],[4])
      )
      as pvt;