在工作中经常遇到统计占比的需求,有时候还要把没有值得统计为0,如何写sql呢?

下面写一个小例子,作为参考,方便以后查阅.

数据准备:

create table t_group( id NUMBER not null, name VARCHAR2(100));

alter table t_group add primary key (id);

create table t_score( id NUMBER not null, g_id NUMBER not null, score NUMBER);

alter table t_score add primary key (id);

insert into t_group (id, name) values ('1', 'a');

insert into t_group (id, name) values ('2', 'b');

insert into t_group (id, name) values ('3', 'c');

insert into t_group (id, name) values ('4', 'd');

insert into t_group (id, name) values ('5', 'e');

commit;

insert into t_score (id,g_id, score) values ('1','1', '81');

insert into t_score (id,g_id, score) values ('2','1', '73');

insert into t_score (id,g_id, score) values ('3','2', '78');

insert into t_score (id,g_id, score) values ('4','3', '92');

commit;

/* 需求1.求每个gropu的得分在总得分中的占比 */

select tg.name,

sum(ts1.score) / (select sum(ts.score) from t_score ts) * 100 as perc

from t_score ts1 join t_group tg on ts1.g_id = tg.id

group by tg.name

/* 需求2.统计所有group在总得分中的占比,要求得分为0的数据也统计到结果中,显示为数值0 */

--思路一:用case when 转换结果.

--最先想到的是直接在join上做文章,用right join,但这个显示的结果中占比为空而不是数值0

select name,

(case

when perc is null then

0

else

perc

end) as perc

from (select tg.name,

sum(ts1.score) / (select sum(ts.score) from t_score ts) * 100 as perc

from t_score ts1

right join t_group tg

on ts1.g_id = tg.id

group by tg.name)

order by name;

--思路二: 使用union all + 0 补充缺位 + group by

select tg1.name, sum(TT.perc)

from (select ts1.g_id,

sum(ts1.score) / (select sum(ts.score) from t_score ts) * 100 as perc

from t_score ts1

group by g_id

union all

select tg.id as g_id, 0 as perc from t_group tg) TT

join t_group tg1

on TT.g_id = tg1.id

group by tg1.name

order by tg1.name;

/* 需求3.按得分统计出 优良中差等级的分别有多少条数据,没有数据的等级显示为0 */

select TT.lev, sum(TT.cou)

from (select lev, count(lev) cou

from (select (case

when score >= 90 then

'优'

when score >= 80 then

'良'

when score >= 70 then

'中'

else

'差'

end) as lev

from t_score) T

group by lev

union all (select '优' as lev, 0 as cou

from dual

union all

select '良' as lev, 0 as cou

from dual

union all

select '中' as lev, 0 as cou

from dual

union all

select '差' as lev, 0 as cou from dual)) TT

group by TT.lev;