Question

我有一个示例表“ city”，如下所示：

inputHidden

如果我提供的城市不存在，我可以写一个返回id | city ---------------- 1 London 2 Liverpool 3 Birmingham 4 (Other)的查询吗...

(Other)

我知道第二个SELECT语句将不返回任何内容。但是，如何修改它以返回“（其他）”行？概念代码：

SELECT * FROM cities WHERE city = "London"      --> {1, London}
SELECT * FROM cities WHERE city = "Manchester"  --> {4, (Other) }

Answer 1

一种方法是：

select c.*
from cities c
where c.city in (?, '(Other)')
order by c.city = '(Other)'
limit 1;

这将检索可能匹配的两行（?是所需名称的占位符）。如果有两行，则选择一个不是“其他”的行。

Answer 2

考虑：

select * 
from cities 
where 
    city = 'Manchester'
    or (city = '(Other)' and not exists (select 1 from cities where city = 'Manchester'))

Answer 3

select DISTINCT 'Other' AS city
FROM cities WHERE NOT EXISTS (select * from cities where city = 'Brisbane')
UNION ALL
SELECT city from cities where city = 'Brisbane'

Answer 4

使用self join并稍加扭曲

select a.id, a.city
from (select *, city = '(Other)' as other from cities)  a
left join cities b on b.city = 'London'
where a.city = b.city or (b.city is null and a.other = 1)

Demo

如果没有其他行匹配，则返回默认行的MYSQL查询

4 个答案: