蜂巢爆炸/侧面视图表

时间:2019-08-22 06:19:30

标签: hive hiveql

表格:jd1(可比性表格)

表:data1(新值表)

我已在运行中的sql server中编写了此查询,但在蜂巢中却显示了错误

select * from data1;

1   siva    hadoop
1   siva    hive
1   siva    spark
1   siva    hbase
1   siva    mapreduce
1   siva    hdfs
2   kumar   hadoop
2   kumar   hive
2   kumar   python
2   kumar   spark
3   naveen  hive
3   naveen  hadoop
3   naveen  flume
3   naveen  kafka

从jd1中选择*;

1   hadoop
1   hive
1   spark
1   hbase
1   mapreduce
1   hdfs
1   python
1   java  

预期产量

1   siva    6   85.71428571428571
2   kumar   4   57.142857142857146
3   naveen  2   28.571428571428573

我的查询

select id, name, count(*), ((count(*)*100)/(select count(skills)from jd1))avg
from (select n.id, n.name, n.skills
      from data1 n join jd1 t on (n.skills=t.skills))a
group by id,name;

错误

FAILED: ParseException line 1:44 cannot recognize input near 'select' 'count' '(' in expression specification

3 个答案:

答案 0 :(得分:1)

select id, name, count(*) cnt, count(*)*100/skill_cnt cnt_pct
from (select n.id, n.name, n.skills, t.skill_cnt 
      from data1 n 
           inner join (select skills, count(*) over() skill_cnt from jd1) t 
                      on n.skills=t.skills
     ) a
group by id,name;

答案 1 :(得分:0)

您可以在下面的查询中尝试-

SELECT n.id, n.name, COUNT(n.skills), COUNT(n.skills)/skill_cnt.total_skill
FROM data1 n
JOIN jd1 t ON n.skills=t.skills
CROSS JOIN (SELECT COUNT(*) total_skill FROM jd1) skill_cnt
GROUP BY n.id, n.name, total_skill

答案 2 :(得分:0)

为jd技能再创建一个表,作为skill_count并加入该表。

SELECT n.Id, n.Job_Id, n.Name, n.Email, n.Mobile_Number, n.Education, n.Total_Experiance,((count(n.skills)*100)/s.skill_count) Average
FROM new_resume n
JOIN new_jd t ON n.skills=t.skills
JOIN skill_count s ON n.job_id = s.job_id
GROUP BY n.Id, n.Job_Id, n.Name, n.Email, n.Mobile_Number, n.Education, n.Total_Experiance,s.skill_count;