你可以用简单的 GROUP BY 带的子句 HAVING 子句来过滤出唯一的地方

select place_id, email
from table t
group by place_id, email 
having count(*) = 1;

你可以用开窗的 COUNT(*) 支持现代rdbms。

SELECT *
FROM (SELECT t.*, COUNT(DISTINCT place_id) OVER(PARTITION BY email) AS cnt
      FROM tab t) sub
WHERE cnt > 1

dbfiddle演示
sql服务器/mariadb/mysql 8.0/postgresql：

SELECT *
FROM (SELECT *, COUNT(*) OVER(PARTITION BY email) AS cnt
      FROM (SELECT DISTINCT place_id, email FROM tab) s
      )sub
WHERE cnt > 1;

db-fiddle.com演示

非常感谢，我尝试了gordon linoff的第一个解决方案，结果成功了。但是我有一个小问题，我有一个“where”子句，这个：

select email, count(email) from data where place = 2 or place=3 or place=4 group by email having min(place) <> max(place)

显示与以下相同的结果：

select email, count(email) from data where place = 2 or place=3 group by email having min(place) <> max(place)

因为这是一个or条件，但我不知道如何修复，在第一个查询中，如何只显示所有这些地方的项，而不是其中两个。

如果您只需要电子邮件，可以使用聚合：

select email
from t
group by email
having min(place) <> max(place);

如果您希望这些位置也位于唯一列表中，可以执行以下操作：

select distinct place, email
from t
where exists (select 1
              from t t2
              where t2.email = t.email and t2.place <> t.place
             );

而且，虽然可以使用窗口函数，但解决方案并不那么明显：

select distinct place, email
from (select t.*,
             min(t.place) over (partition by t.email) as min_place,
             max(t.place) over (partition by t.email) as max_place
      from t
     ) t
where min_place <> max_place;