将两个不同查询的结果分成一个

Question

我有两个查询：

SELECT COUNT(*)
FROM houses
WHERE with_people IS NOT NULL



SELECT COUNT(*)
FROM houses
where color = 'white' AND with_people IS NOT NULL

我先执行两个查询，然后在后端代码中执行（white_houses_with_people / white_houses）。

是否有一种方法可以在SQL中进行该划分，从而不必执行两个查询？

Answer 1

您可以使用条件聚合：

SELECT COUNT(*) AS cnt, COUNT(CASE WHEN color = 'white' THEN 1 END) AS cnt2
FROM houses
WHERE with_people IS NOT NULL;

与比率：

SELECT 
COALESCE((1.0*COUNT(CASE WHEN color = 'white' THEN 1 END)/NULLIF(COUNT(*),0)),0)  
AS ratio
FROM houses
WHERE with_people IS NOT NULL

Answer 2

您可以简单地执行以下操作：

*,
*::before,
*::after {
  box-sizing: border-box; // 1
}

SELECT 1.0 * (SELECT COUNT(*) FROM houses WHERE color = 'white' AND with_people IS NOT NULL) / (SELECT COUNT(*) FROM houses WHERE with_people IS NOT NULL) 可以确保结果为1.0 *而不是decimal(, 1)。