SQL

Question

我正在努力解决 https://platform.stratascratch.com/coding/10065-find-whether-the-number-of-seniors-works-at-facebook-is-higher-than-its-number-of-usa-based-employees?python=

这是我尝试编写的查询：

SELECT CASE WHEN COUNT(CASE WHEN location = 'US' THEN 1 ELSE 0 END) >
    COUNT(CASE WHEN is_senior = true THEN 1 ELSE 0 END) THEN 'More USA-based'
    ELSE 'More seniors' END AS what_do_we_have_more_of
FROM facebook_employees

Result: 'More seniors'

然而，当我用翻转的条件重写它时：

SELECT CASE WHEN COUNT(CASE WHEN is_senior = true THEN 1 ELSE 0 END) >
    COUNT(CASE WHEN location = 'US' THEN 1 ELSE 0 END) THEN 'More seniors'
    ELSE 'More USA-based' END AS what_do_we_have_more_of
FROM facebook_employees

Result: 'More USA-based'

有人可以解释一下为什么这里有差异吗？我写的查询有什么问题？

我知道这个问题可以通过子查询来解决，但我想专门尝试 CASE WHEN 方法。这样效率更高吗？

编辑：我用子查询编写的解决方案（适用于相反的条件）

WITH us_employees AS (
    SELECT id, location
    FROM facebook_employees
    WHERE location = 'US'
),

senior_employees AS (
    SELECT id, is_senior
    FROM facebook_employees
    WHERE is_senior = true
)

SELECT CASE WHEN COUNT(location) < COUNT(is_senior) THEN 'More seniors' ELSE 'More US-based' END AS what_do_we_have_more_of
FROM us_employees u
FULL JOIN senior_employees s
ON u.id = s.id

Result: 'More seniors'

Answer 1

您应该使用 SUM 而不是 COUNT。

COUNT 即使您的 CASE return 为 1 或 0，也会算作 +1。

SUM 只有当你的 CASE return 1.

因此假设您的 table 有 1000 行，那么您的两个查询都将是 CASE 1000 > 1000 THEN ... ELSE... END。

Answer 2

您查询中 count() 的使用不正确。但是你如何得到不同的结果呢？因为两个数是一样的。所以 A > B 总是假的，你总是在 ELSE 分支结束。

正确的查询可能如下所示：

SELECT CASE WHEN count(*) FILTER (WHERE location = 'US')
               > count(*) FILTER (WHERE is_senior) THEN 'More USA-based'
            WHEN count(*) FILTER (WHERE location = 'US')
               < count(*) FILTER (WHERE is_senior) THEN 'More seniors'
            ELSE 'US-based and seniors tie' END AS what_do_we_have_more_of
FROM   facebook_employees;

参见：

Aggregate columns with additional (distinct) filters

请注意，这永远不会因 NULL 值而失败，因为 count()（与大多数聚合函数不同）永远不会 returns NULL.

SQL - CASE WHEN 结果输出冲突的结果

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