单个SQL查询中的多个计数

Question

我有以下三个不同的查询：

SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Bathtub is  null

SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Bathroom is  null

SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Toilet is  null

我想在一个查询中查看其计数。 我怎样才能做到这一点？ 谢谢。

Answer 1

使用子查询！ 了解SQL的一些基础知识https://blog.sqlauthority.com/

SELECT 
(SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Bathtub is  null) AS BathTub,

(SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Bathroom is  null) AS Bathroom,

(SELECT COUNT(*)
  FROM [myDb].[dbo].[Properties] WHERE Toilet is  null) AS Toilet

警告如果您担心性能会涉及开销

Answer 2

试试这个（这是oracle，我想在mysql中没有什么不同）：

SELECT 
COUNT(CASE WHEN Bathtub IS NULL THEN 1 END) Bathtub, 
COUNT(CASE WHEN Bathroom IS NULL THEN 1 END) Bathroom,
COUNT(CASE WHEN Toilet IS NULL THEN 1 END) Toilet 
from [myDb].[dbo].[Properties];

Answer 3

或这样写：

SELECT total-BT cnt_bt, total-BR cnt_br, total-TL cnt_tl FROM (
  SELECT COUNT(*) total, COUNT(Bathtub) BT, COUNT(Bathroom) BR, COUNT(Toilet) TL
  FROM [myDb].[dbo].[Properties]
) subq

Answer 4

尝试结合自定义描述

SELECT COUNT(*), 'Bathtub is null counts' desc
  FROM [myDb].[dbo].[Properties] WHERE Bathtub is  null
union
SELECT COUNT(*),'Bathroom is null counts' desc
  FROM [myDb].[dbo].[Properties] WHERE Bathroom is  null
union
SELECT COUNT(*), 'Toilet is null counts' desc
  FROM [myDb].[dbo].[Properties] WHERE Toilet is  null

要么

SELECT 
SUM(CASE WHEN Bathtub IS NULL THEN 1 ElSE 0 END) as Bathtub_count, 
SUM(CASE WHEN Bathroom IS NULL THEN 1 ElSE 0 END) as Bathroom_count,
SUM(CASE WHEN Toilet IS NULL THEN 1 ElSE 0 END) as Toilet_count
from [myDb].[dbo].[Properties]

Answer 5

您可以执行此SQL SUM()函数。 如下所示：

SELECT
    SUM(IF(Bathtub IS NULL, 1, 0)) AS Bathtub_count,
    SUM(IF(Bathroom IS NULL, 1, 0)) AS Bathroom_count,
    SUM(IF(Toilet IS NULL, 1, 0)) AS Toilet_count
FROM
    [ myDb ].[ dbo ].[ Properties ]