Sql 计算所有行数 select empid from employee group by empid having min(case when department in ('Y','Z') then 1 else 0 end) = 1 and count(case when department in ('Y','Z') then 1 end) = 2;

如果一名员工可能被两次分配到同一个部门，那么上述解决方案将不起作用

有最小。。。可以在Postgres中使用聚合bool_和进行简化

当应用标准筛选条件进行条件聚合时，也可以使其适用于员工可以被分配到同一部门两次的情况

select empid
from employee
group by empid
having bool_and(department in ('Y','Z'))
  and count(distinct department) filter (where department in ('Y','Z')) = 2;

bool_和department在“Y”中，仅当组中所有行的条件为true时，“Z”才返回true

标准SQL的另一个解决方案是使用至少拥有这两个部门的员工与分配给两个部门的员工之间的交集：

SELECT empID,
  GROUP_CONCAT(DISTINCT department ORDER BY department ASC) AS depts
FROM emp_dept GROUP BY empID
HAVING depts = 'X,Y'

with depts_to_check (dept) as (
   values ('Z'), ('Y')
)
select empid
from employee
group by empid
having array_agg(department order by department) = array(select dept from depts_to_check order by dept);

-- employees with at least those two departments
select empid
from employee
where department in name in ('Y','Z')
group by empid
having count(distinct department) = 2

intersect

-- employees with exactly two departments
select empid
from employee
group by empid
having count(distinct department) = 2;

---

你为什么不想使用GROUP BY和HAVING？使用JOIN可能会导致性能问题。@weweweshemenance，因为我想使用AND将此条件与另一个条件连接起来。那么为什么不在派生表或cte中使用此sql呢？@MAK这种方法如何从select empID中选择empID，string_aggdepartment，，'dep from employe group by empIDt，其中dep in'Z，Y'？？@MAK yup但您标记了PostgreSQL这就是为什么：D，只标记相关数据库我仍然建议您使用带有GROUP BY的数据库。@WeWesthemenance，实际上我正在为where子句准备一个字符串作为查询。在该字符串中，我不能使用GROUPBY，因为该字符串还有许多其他条件。我只想把这个条件和那个条件连接起来。一个单元格中没有同时有两个不同值的行。永远。部门='Z'和部门='Y'从来都不是真正的好！对于那些没有得到交叉申请的人，使用GROUP_CONCAT或类似的方法。

select empId from employe 
where empId in (select empId from employe 
                where department = 'Z') 
and empId in (select empId from employe 
              where department = 'Y') 
and empId not in (select empId from employe 
                  where department = 'X') ;

select empID from employe 
where empId in (select empId from employe 
where department = 'Z' and department = 'Y') 
and empId not in (select empId from employe 
where department = 'X') ;

SELECT empID 
FROM employe
GROUP BY empID
HAVING SUM(CASE WHEN department= 'Y' THEN 1 ELSE 0 END) > 0
AND SUM(CASE WHEN department= 'Z' THEN 1 ELSE 0 END) > 0
AND SUM(CASE WHEN department NOT IN('Y','Z') THEN 1 ELSE 0 END) = 0

SELECT empID 
FROM employe E1
WHERE (SELECT COUNT(DISTINCT department) FROM employe E2 WHERE E2.empid = E1.empid and  department IN ('Z','Y')) = 2
EXCEPT
SELECT empID 
FROM employe
WHERE department NOT IN ('Z','Y')

;WITH CTE AS 
(

    SELECT empID 
    FROM employe
    GROUP BY empID
    HAVING SUM(CASE WHEN department= 'Y' THEN 1 ELSE 0 END) > 0
    AND SUM(CASE WHEN department= 'Z' THEN 1 ELSE 0 END) > 0
    AND SUM(CASE WHEN department NOT IN('Y','Z') THEN 1 ELSE 0 END) = 0
)
SELECT cols from CTE join othertable on col_cte = col_othertable

SELECT empId
FROM (
    SELECT
        empID, cc = COUNT(DISTINCT department)
    FROM employe
    WHERE department IN('Y', 'Z')
    GROUP BY empID
)t
WHERE
    t.cc = 2
    AND t.cc = (
        SELECT COUNT(*)
        FROM employe
        WHERE empID = t.empID
    )

SELECT e.empId
FROM employe e
WHERE e.department IN('Y', 'Z')
GROUP BY e.empID
HAVING
    COUNT(e.department) = 2
    AND COUNT(e.department) = (SELECT COUNT(*) FROM employe WHERE empID = e.empId)

SELECT DISTINCT e.empID
FROM employe e
WHERE
    EXISTS(
        SELECT 1 FROM employe WHERE department = 'Z' AND empID = e.empID
    )
    AND EXISTS(     
        SELECT 1 FROM employe WHERE department = 'Y' AND empID = e.empID
    )
    AND NOT EXISTS(
        SELECT 1 FROM employe WHERE department NOT IN('Y', 'Z') AND empID = e.empID
    )

SELECT  a.empId
FROM    employe a
        INNER JOIN
        (
            SELECT  empId
            FROM    employe 
            WHERE   department IN ('X', 'Y', 'Z')
            GROUP   BY empId
            HAVING  COUNT(*) = 3
           )b ON a.empId = b.empId
GROUP BY a.empId

   SELECT EMPID FROM EMPLOYE WHERE DEPARTMENT='Z'  AND 
   EMPID IN (SELECT EMPID FROM EMPLOYE WHERE DEPARTMENT ='Y')AND
   EMPID NOT IN(SELECT EMPID FROM EMPLOYE WHERE DEPARTMENT NOT IN ('Z','Y'))

select DISTINCT empID
FROM employe A
CROSS APPLY
            (
                SELECT department + ','
                FROM employe B
                WHERE A.empID = B.empID
                ORDER BY department
                FOR XML PATH ('')
            ) CA(Deps)
WHERE deps = 'Y,Z,'

empID
----------
A103

SELECT empID,
  GROUP_CONCAT(DISTINCT department ORDER BY department ASC) AS depts
FROM emp_dept GROUP BY empID
HAVING depts = 'X,Y'

SELECT empID,
  COUNT(DISTINCT department) AS all_depts,
  COUNT(DISTINCT CASE
    WHEN department IN ('X', 'Y') THEN department ELSE NULL
  END) AS wanted_depts
FROM emp_dept GROUP BY empID
HAVING all_depts = wanted_depts AND wanted_depts = 2

SELECT empID, name, depts
FROM employees
JOIN (
    SELECT empID,
      GROUP_CONCAT(DISTINCT department ORDER BY department ASC) AS depts
    FROM emp_dept GROUP BY empID
    HAVING depts = 'X,Y'
  ) AS tmp USING (empID)
WHERE -- ...add other conditions here...

SELECT ... FROM t1 WHERE t1.a IN (SELECT b FROM t2);

SELECT ... FROM t1 WHERE EXISTS (SELECT 1 FROM t2 WHERE t2.b = t1.a);

SELECT * FROM employee e JOIN employee e2 ON e.empid = e2.empid
WHERE e.department = 'x' AND e2.department ='y'

select distinct e.empID
from employe e
where exists( select * 
              from employe
              where empID = e.empID
              having count(department) = count(case when department in('Y','X','Z') then department end)
                 and count(distinct department) = 3)

select e.empId
from employe e
group by e.empID
having count(department) = count(case when department in('Y','X','Z') then department end)
   and count(distinct department) = 3

having count(case when department not in('Y','X', 'Z') then department end) = 0
   and count(distinct case when department in('Y','X','Z') then department end) = 3

select empid
from employee
group by empid
having array_agg(department order by department)::text[] = array['Y','Z'];

with depts_to_check (dept) as (
   values ('Z'), ('Y')
)
select empid
from employee
group by empid
having array_agg(department order by department) = array(select dept from depts_to_check order by dept);

select empid
from employee
group by empid
having min(case when department in ('Y','Z') then 1 else 0 end) = 1
  and count(case when department in ('Y','Z') then 1 end) = 2;

select empid
from employee
group by empid
having bool_and(department in ('Y','Z'))
  and count(distinct department) filter (where department in ('Y','Z')) = 2;

-- employees with at least those two departments
select empid
from employee
where department in name in ('Y','Z')
group by empid
having count(distinct department) = 2

intersect

-- employees with exactly two departments
select empid
from employee
group by empid
having count(distinct department) = 2;