有点像行列互换交叉的题目

table1

月份mon 部门dep 业绩yj
-------------------------------
一月份      01      10
一月份      02      10
一月份      03      5
二月份      02      8
二月份      04      9
三月份      03      8

 

 

 table3 （result）

部门dep 一月份      二月份      三月份
--------------------------------------
      01      10        null      null
      02      10         8        null
      03      null       5        8
      04      null      null      9

 

 

 

1)
select a.部门名称dname,b.业绩yj as '一月份',c.业绩yj as '二月份',d.业绩yj as '三月份'
from table1 a,table2 b,table2 c,table2 d
where a.部门dep = b.部门dep and b.月份mon = '一月份' and
a.部门dep = c.部门dep and c.月份mon = '二月份' and
a.部门dep = d.部门dep and d.月份mon = '三月份' and
2)
select a.dep,
sum(case when b.mon=1 then b.yj else 0 end) as '一月份',
sum(case when b.mon=2 then b.yj else 0 end) as '二月份',
sum(case when b.mon=3 then b.yj else 0 end) as '三月份',
sum(case when b.mon=4 then b.yj else 0 end) as '四月份',
sum(case when b.mon=5 then b.yj else 0 end) as '五月份',
sum(case when b.mon=6 then b.yj else 0 end) as '六月份',
sum(case when b.mon=7 then b.yj else 0 end) as '七月份',
sum(case when b.mon=8 then b.yj else 0 end) as '八月份',
sum(case when b.mon=9 then b.yj else 0 end) as '九月份',
sum(case when b.mon=10 then b.yj else 0 end) as '十月份',
sum(case when b.mon=11 then b.yj else 0 end) as '十一月份',
sum(case when b.mon=12 then b.yj else 0 end) as '十二月份',
from table2 a left join table1 b on a.dep=b.dep 

 

 

8.华为一道面试题
一个表中的Id有多个记录，把所有这个id的记录查出来，并显示共有多少条记录数。

select id, Count（*) from tb group by id having count(*)>1
select * from(select count(ID) as count from table group by ID)T where T.count>1