题解 | #考试分数(四)#

考试分数(四)

http://www.nowcoder.com/practice/502fb6e2b1ad4e56aa2e0dd90c6edf3c

# 思路:
# 1、针对每条记录,生成和每个岗位的记录数量
# 2、根据每个岗位记录数量是奇数、偶数两种情况,用case when生成start 和end字段.
select distinct t1.job
,(case when t1.cnt_score % 2 = 0 then round(t1.cnt_score / 2,0) 
 else round((t1.cnt_score+1)/2,0) end ) as start
 ,(case when t1.cnt_score % 2 = 0 then round(t1.cnt_score/2+1,0) 
  else round((t1.cnt_score+1)/2,0) end ) as end from 
  (
select g1.job 
# ,dense_rank() over(partition by g1.job order by g1.score asc) as rank_score
,count(*) over(partition by g1.job) as cnt_score 
from grade as g1 ) t1 order by t1.job 


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后来123321:别着急,我学院本大二,投了1100份,两个面试,其中一个还是我去线下招聘会投的简历,有时候这东西也得看运气
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