我试图通过使用UNION运算符来获取总计数,但它会给出错误的计数。
select count(*) as companyRatings from (
select count(*) hrs from (
select distinct hrs from companyA
)
union
select count(*) financehrs from (
select distinct finance_hrs from companyB
)
union
select count(*) hrids from (
select regexp_substr(hr_id,'[^/]+',1,3) hrid from companyZ
)
union
select count(*) cities from (
select regexp_substr(city,'[^/]+',1,3) city from companyY
)
);
个人查询工作正常但总计数不匹配。
个人结果:12 19 3 6 目前总人数:31
实际总数:40。 所以有没有UNION运算符的替代解决方案?
答案 0 :(得分:1)
要添加您使用+
的值。 UNION
是添加数据集。
select
(select count(distinct hrs) from companyA)
+
(select count(distinct finance_hrs) from companyB)
+
(select count(regexp_substr(hr_id,'[^/]+',1,3)) from companyZ)
+
(select count(regexp_substr(city,'[^/]+',1,3)) from companyY)
as total
from dual;
但我同意juergen d;你不应该在每个公司都有单独的表格。
答案 1 :(得分:0)
编辑。使用Sum
select sum(cnt) as companyRatings from
(
select count(*) as cnt from (select distinct hrs from companyA)
union all
select count(*) as cnt from (select distinct finance_hrs from companyB)
union all
select count(*) as cnt from (select regexp_substr(hr_id,'[^/]+',1,3) hrid from companyZ)
union all
select count(*) as cnt from (select regexp_substr(city,'[^/]+',1,3) city from companyY)
)
上一个回答:
试试这个
SELECT (
SELECT count(*) hrs
FROM (
SELECT DISTINCT hrs
FROM companyA
)
)
+
(
SELECT count(*) financehrs
FROM (
SELECT DISTINCT finance_hrs
FROM companyB
)
)
+
(
SELECT count(*) hrids
FROM (
SELECT regexp_substr(hr_id, '[^/]+', 1, 3) hrid
FROM companyZ
)
)
+
(
SELECT count(*) cities
FROM (
SELECT regexp_substr(city, '[^/]+', 1, 3) city
FROM companyY
)
)
AS total_count
FROM dual