结合两个CTE块

Question

有没有一种方法可以组合这两个cte块而不会丢失任何记录？我的目的是最大程度地减少此处发生的代码重复。我需要所有记录都显示在结果中，而不管它们是在cte1还是cte2中，还是两者都存在。

CTE 1-

with grouping (num, count) as 
(
SELECT num, count(*)
FROM pay t
where 
code = '1'
group by num
),
total (num, tot_count) as
(
select num, count(*)
from pay
group by num
)
select m.id, round(count/tot_count::decimal, 4), year
FROM total t
join grouping o
on t.num = o.num
join master m
on o.num = m.num;

CTE 2-

with grouping (num, count) as 
(
SELECT num, count(*)
FROM pay t
where 
code = '2'
group by num
),
total (num, tot_count) as
(
select num, count(*)
from pay
group by num
)
select m.id, round(count/tot_count::decimal, 4), year
FROM total t
join grouping o
on t.num = o.num
join master m
on o.num = m.num;

Answer 1

我没有要测试的样本数据，但似乎最大的不同在于您在WHERE CTE中的grouping语句中。

IN子句应该为您做到这一点：

with grouping (num, count) as 
(
SELECT num, count(*)
FROM pay t
where 
code IN ('1', '2')
group by num
),
total (num, tot_count) as
(
select num, count(*)
from pay
group by num
)
select m.id, round(count/tot_count::decimal, 4), year
FROM total t
join grouping o
on t.num = o.num
join master m
on o.num = m.num;

确认这是使用Sublime Text进行查询的唯一区别