结果集中的mysql“和”逻辑
作者:互联网
假设我有一个类似以下的数据集:
table foo
id | employeeType | employeeID
-------------------------
1 | Developer | 1
2 | Developer | 2
3 | Developer | 3
4 | Manager | 1
5 | Manager | 4
6 | Manager | 5
7 | CEO | 1
8 | CEO | 6
我想运行一个查询,该查询将返回所有employeeid(以及employeeTypes),其中所有employeeTypes之间都有一个通用的employee id(即“与”逻辑.只有所有employeeTypes的employeeID都将返回.employeeType= Developer和employeeType =经理,employeeType = CEO).对于上面的数据,示例输出为
result table
id | employeeType | employeeID
-------------------------
1 | Developer | 1
4 | Manager | 1
7 | CEO | 1
当我只有两个员工类型通过自我加入表格时,便能够做到这一点.
select * from foo as fooOne
join foo as fooTwo
on fooOne.employeeID = fooTwo.employeeID
AND
fooOne.employeeType <> fooTwo.employeeType
当“和”逻辑匹配时,该查询返回带有fooTwo值的结果集,但同样,仅适用于两种类型的员工.我的实际用例场景要求我需要能够处理可变数量的employeeType(3、4、5等).
任何对此的想法将不胜感激.
解决方法:
这应该返回所需的行:
SELECT foo.*
FROM
foo
WHERE
employeeID IN (
SELECT employeeID
FROM foo
GROUP BY employeeID
HAVING COUNT(DISTINCT employeeType) =
(SELECT COUNT(DISTINCT employeeType)
FROM foo)
)
请参阅fiddle here.
内部查询将返回不同员工类型的数量:
(SELECT COUNT(DISTINCT employeeType) FROM foo)
中间查询将返回具有最大雇员类型数量的所有雇员ID:
SELECT employeeID
FROM foo
GROUP BY employeeID
HAVING COUNT(DISTINCT employeeType) =
(SELECT COUNT(DISTINCT employeeType) FROM foo)
并且外部查询将返回整个行.
标签:join,conditional,logical-operators,resultset,mysql 来源: https://codeday.me/bug/20191119/2036509.html