您可以用途：

SELECT a.account_id,
       a.customer__id AS CUS_ID,
       b.first_Name AS NAME,
       b.last_time - a.last_time AS LAST_TIME_DIFFERENCE
FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND a.last_time    > b.last_time + INTERVAL '6' HOUR
ORDER BY b.first_name

（无论last_time列是日期还是时间戳，它都有效。）
或者，如果last_time列是时间戳：

FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND a.last_time - b.last_time > INTERVAL '6' HOUR

或者，如果last_time列是DATE s（在Oracle中，它始终具有时间分量）：

FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND a.last_time - b.last_time > 6/24

或者，您可以确保DATE差值被计算为一个区间，使用：

FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND (a.last_time - b.last_time) DAY TO SECOND > INTERVAL '6' HOUR

更新：

如果您希望时间戳在同一天，则：

SELECT a.account_id,
       a.customer__id AS CUS_ID,
       b.first_Name AS NAME,
       b.last_time - a.last_time AS LAST_TIME_DIFFERENCE
FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND a.last_time    > b.last_time + INTERVAL '6' HOUR
          AND a.last_time    < TRUNC(b.last_time) + INTERVAL '1' DAY
ORDER BY b.first_name

或：

FROM   tableA a
       INNER JOIN tableB b
       ON     a.account_id   = b.account_id
          AND a.customer__id = b.customer__id
          AND a.last_time - b.last_time > 6/24
          AND a.last_time < TRUNC(b.last_time) + 1

这是一种技巧，你可以利用使用decode操作。

select a.account_id AS ACCOUNT_ID, a.customer__id AS CUS_ID, b.first_Name AS NAME,
 b.last_time - a.last_time AS LAST_TIME_DIFFERENCE 
from tableA a, tableB b 
where 
  a.account_id = b.account_id 
  AND a.customer__id = b.customer__id 
  AND decode(floor(abs(a.last_time - b.last_time)),0,abs(a.last_time - b.last_time)*24,2) > decode(floor(abs(a.last_time - b.last_time)),0,6,1)
ORDER BY b.first_name