Question

将2个表作为输入 - 员工和薪水为

Employee            Salary  
ID  Name            ID  Salary  
1   Anshika         1   1000  
2   Arnav           2   2000  
3   Ishita          3   3000  
4   Iti             4   4000

将输出显示为 -

A       I  
3000    7000

我已将2个查询写为 -

    mysql> select sum(salary) as sumA from salEmp where id in (select id from nameEmp where name LIKE 'A%');  

output  

sumA  
3000

1行（0.00秒）

    mysql> select sum(salary) as sumI from salEmp where id in (select id from nameEmp where name LIKE 'I%');  

output  

sumI  
7000

1行（0.01秒）

但我想编写一个查询，以便我可以在问题中获得所需的输出。

我试过了 -

mysql> select sum(a.salary) as sumA, sum(b.salary) as sumI
->  from salEmp a,salEmp b
->  where a.id in
->          (select id from nameEmp where name LIKE 'A%')
->  AND b.id in
->          (select id from nameEmp where name LIKE 'I%');

但输出如下 -

sumA  sumI  
6000  14000  
    1 row in set (0.00 sec)

价值加倍 PLZ有助于获得原始输出...

Answer 1

你正在做cross join。因此结果表获得重复值。相反，您可以按照以下查询：

SELECT 
    SUM(CASE WHEN ID IN (SELECT ID FROM nameEmp 
                        WHERE name Like 'A%') THEN Salary ELSE 0 END) AS sumA,
    SUM(CASE WHEN ID IN (SELECT ID FROM nameEmp 
                        WHERE name Like 'I%') THEN Salary ELSE 0 END) AS sumI

FROM salEmp

您可以了解＆＃34; CASE WHEN＆＃34;如果您对此不熟悉，请从这里开始。

http://www.java2s.com/Code/SQL/Flow-Control/UseCASEWHENstatementinSELECT.htm

http://dev.mysql.com/doc/refman/5.0/en/case.html

Answer 2

感谢您的帮助!!!我找到了另一个问题的解决方案 -

select a.sumA, b.sumI from   
(select sum(salary) as sumA from salEmp where id in (select id from nameEmp where name LIKE 'A%')) a,
(select sum(salary) as sumI from salEmp where id in (select id from nameEmp where name LIKE 'I%')) b;

使用2进行一个选择查询，查询在同一个表上运行

2 个答案: