一个简单的方法是 exists . 只有两个地址：

select t.*
from t
where t.city in ('Noida', 'Chennai') and
      exists (select 1
              from t t2
              where t2.customer = t.customer and
                    t2.city in ('Noida', 'Chennai') and
                    t2.city <> t.city
             );

您可以使用windows函数

select * from
(select t.*, count(distinct city) over (partition by customer) as cnt
from your_Table t
where city in ('Noida','Chennai')
) where cnt = 2

SELECT *
FROM your_table 
WHERE Customer IN (
  SELECT Customer 
  FROM your_table 
  WHERE City IN ('Noida', 'Chennai')
  GROUP BY Customer
  HAVING COUNT(DISTINCT City) = 2
)

要获得想要的结果，您需要选择城市为“noida”或“chennai”、客户为“test2”或“test4”的所有数据：

SELECT *
  FROM TEST_DATA
  WHERE CITY IN ('Noida', 'Chennai') AND
        CUSTOMER IN ('Test2', 'Test4')
  ORDER BY Customer

db<>在这里摆弄