得到 (count(column1) + count(column2)) 的总和答案

【问题标题】：Get the sum of (count(column1) + count(column2))得到 (count(column1) + count(column2)) 的总和
【发布时间】：2021-12-14 00:31:37
【问题描述】：

我有一张桌子 A：

entity_id    name
------------------
1            Test1
2            Test2
3            Test3
4            Test4
5            Test5
6            Test6

我有一张桌子 B：

entity_id   value1     value2
-----------------------------
1           10          20
1           15          30
2           10          25
1           9           45
3           null        1
2           45          50
3           20          null

我需要编写一个查询来从表 A 中选择 entity_id 和 name，并计算表 B 中列 value1 和 value2 的 entity_id 的总出现次数，然后计算这些列的总数（null 不计算在内） .

所以我的输出表是：

entity_id    name         value1_count    value2_count     total_count
----------------------------------------------------------------------
1            Test1             3               3              6
2            Test2             1               2              3
3            Test3             1               1              2
4            Test4             0               0              0
5            Test5             0               0              0
6            Test6             0               0              0

我无法将 value1 的计数和 value2 的计数相加，并将该值输出到每个唯一 entity_it 的 total_count 中。

这是我目前的查询：

SELECT DISTINCT a.entity_id, a.name
     , count(b.value1) AS value1_count, count(b.value2) AS value2_count, sum(2) AS total_count
FROM a
LEFT JOIN b ON a.entity_id = b.entity_id
GROUP BY a.entity_id, a.name

我知道sum(2) as total_count 不正确，没有得到我想要的。

【问题讨论】：

value1_count 实体 2 应该是 2，而不是 1 我假设？

标签： sql postgresql count left-join

【解决方案1】：

试试这个：

WITH list AS
(
SELECT b.entity_id
     , count(*) FILTER (WHERE b.value1 IS NOT NULL) OVER () AS value1_count
     , count(*) FILTER (WHERE b.value2 IS NOT NULL) OVER () AS value2_count
FROM Table_B AS b
GROUP BY b.entity_id
)
SELECT a.entity_id, a.name
     , COALESCE(l.value1_count, 0)
     , COALESCE(l.value2_count,0)
     , COALESCE(l.value1_count + l.value2_count, 0) AS total_count
FROM Table_A AS a
LEFT JOIN list AS l
ON a.entity_id = l.entity_id

【讨论】：

【解决方案2】：

SELECT entity_id, a.name
     , COALESCE(b.v1_ct, 0) AS value1_count
     , COALESCE(b.v2_ct, 0) AS value2_count
     , COALESCE(b.v1_ct + b.v2_ct, 0) AS total_count
FROM   a
LEFT   JOIN (
   SELECT entity_id, count(value1) AS v1_ct, count(value2) AS v2_ct
   FROM   b
   GROUP  BY 1
   ) b USING (entity_id);

db小提琴here

先聚合，后加入。这更简单，更快捷。见：

Query with LEFT JOIN not returning rows for count of 0

count() 永远不会产生NULL。只有LEFT JOIN 可以在此查询中引入NULL 计数值，因此v1_ct 和v2_ct 要么都是NULL，要么都是NOT NULL。因此COALESCE(v1_ct + v2_ct, 0) 可以。（否则，一个 NULL 将使加法中的另一个和数无效。）

【讨论】：