对同一个表中的两个SELECT语句执行SQLITE中的JOIN
这就是我在SQLITE中创建示例表的方式对同一个表中的两个SELECT语句执行SQLITE中的JOIN,sqlite,Sqlite,这就是我在SQLITE中创建示例表的方式 ID NAME AGE ADDRESS SALARY 1 Paul 32 California 20000.0 2 Allen 25 Texas 15000.0 3 Teddy 23 Norway 20000.0 4 Mark 25 Rich-Mond 65000.0 5 David 27 Texas 85000.0 6 Kim 22 South-Hall 45
ID NAME AGE ADDRESS SALARY
1 Paul 32 California 20000.0
2 Allen 25 Texas 15000.0
3 Teddy 23 Norway 20000.0
4 Mark 25 Rich-Mond 65000.0
5 David 27 Texas 85000.0
6 Kim 22 South-Hall 45000.0
7 Paul 32 California 20000.0
8 Allen 25 Texas 15000.0
9 Teddy 23 Norway 20000.0
我想要实现的是在我的SQLITE表上连接这两个查询
select AGE, count(*) as SALARYLESSTHAN45 from company where salary < 45000 group by salary
select AGE, count(*) as SALARYMORETHAN45 from company where salary > 45000 group by salary
我尝试了以下方法
select AGE, count(*) as SALARYLESSTHAN45 from company where salary < 45000 group by salary ) T1
INNER JOIN
select AGE, count(*) as SALARYMORETHAN45 from company where salary > 45000 group by salary ) T2
ON T1.AGE = T2.AGE
但是不能让它工作
有人可以分享一个如何在SQLITE中实现这一点的示例吗?两个不同表上的连接如下所示:
SELECT ... FROM Tab1 JOIN Tab2 ON ...
要对查询结果执行联接,必须用子查询替换表名:
select AGE,
SALARYLESSTHAN45,
SALARYMORETHAN45
from (select AGE,
count(*) as SALARYLESSTHAN45
from company
where salary < 45000
group by salary)
join (select AGE,
count(*) as SALARYMORETHAN45
from company
where salary > 45000
group by salary)
using (AGE);
两个不同表上的联接如下所示:
SELECT ... FROM Tab1 JOIN Tab2 ON ...
要对查询结果执行联接,必须用子查询替换表名:
select AGE,
SALARYLESSTHAN45,
SALARYMORETHAN45
from (select AGE,
count(*) as SALARYLESSTHAN45
from company
where salary < 45000
group by salary)
join (select AGE,
count(*) as SALARYMORETHAN45
from company
where salary > 45000
group by salary)
using (AGE);
这些查询没有意义,因为每个组都有一个随机年龄。这些查询没有意义,因为每个组都有一个随机年龄。非常感谢!!!不知道为什么我要这么努力才能完成这件事。下面是一个类似的问题……在这个场景中,我如何做一个完整的外部连接,并用say zero替换null要提问,请单击ask question。非常感谢!!!不知道为什么我要这么努力才能完成这件事。下面是一个类似的问题……在这个场景中,我如何进行完整的外部联接并用say zero替换null要提问,请单击ask question。