Question

我有两个查询，我只需要总记录数，但查询中唯一的区别是一个字段值。

实施例

SELECT COUNT(*) AS group_a
FROM tbl
WHERE category = 'value_a'

SELECT COUNT(*) AS group_b
FROM tbl
WHERE category = 'value_b'

我怎样才能得到这样的东西:(伪）

SELECT COUNT(*) AS group_a, COUNT(*) AS group_b
FROM tbl
WHERE category IN ('value_a', 'value_b')

但结果是这样的

group_a , group_b
56, 101

我在查询中想一个CASE语句来过滤这两个但是我该如何实现呢？还是有更好的方法？

我现在正在做一个UNION，但想知道我是否可以返回一条有两个结果的记录

Answer 1

select sum(case when category = 'value_a' then 1 else 0 end) as group_a,
       sum(case when category = 'value_b' then 1 else 0 end) as group_b
    from tbl
    where category in ('value_a', 'value_b')

Answer 2

select  sum(case when category = 'value_a' then 1 else 0 end) group_a,
        sum(case when category = 'value_b' then 1 else 0 end) group_b
from tbl

Answer 3

SELECT category,COUNT(*) FROM tbl
GROUP BY category;

扩展到更多类别。如果你只想要那些类别

SELECT category,COUNT(*) FROM tbl
WHERE category IN ('value_a', 'value_b')
GROUP BY category;

Answer 4

计数有什么奇怪的答案。这是一个简单的COUNT：

SELECT COUNT(category = 'value_a' OR NULL) AS group_a, COUNT(category = 'value_b' OR NULL) AS group_b FROM tbl;

PostgreSQL中的COUNT聚合允许像我所示的复杂语法。请注意，OR NULL非常重要，因为COUNT仅计算条件category = '...' OR NULL给出非NULL答案的行。

Answer 5

只为了它的乐趣：

SELECT * 
FROM 
(
    SELECT category
    FROM tbl 
) subquery
PIVOT
(
    COUNT(category)
    FOR category IN ([value_a],[value_b])
) AS piv

将两个SELECT查询合并为一个

5 个答案: