我有三个不同的查询:
@media screen and (max-device-width: 1024px) and (orientation: portrait)
{#cast1, #cast2, #cast3, #cast4, #cast5, #cast6, #cast7
{
height: 26%;left: 5%;
}
}
@media screen and (max-width: 1024px)
{#cast1, #cast2, #cast3, #cast4, #cast5, #cast6, #cast7
{
height: 45%;left: 5%;
}
}
我想在一个查询中看到他们的计数。我怎样才能做到这一点?感谢。
答案 0 :(得分:3)
使用子查询!学习SQL https://blog.sqlauthority.com/
的一些基础知识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,则涉及案件的开销
答案 1 :(得分:1)
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];
答案 2 :(得分:1)
或者像这样写:
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
答案 3 :(得分:1)
尝试使用自定义说明进行联合
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
或
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]
答案 4 :(得分:1)
您可以执行此SQL SUM()
功能。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 ]