Oracle：如何更新表中的“过时”日期？

Question

Our use case :

For table USERS :

update all rows (set value 'N' for column ACTIVE ) in case when user was inactive (column 'LAST_VISIT' ) more that 60 minutes.

Google helped me to find how to obtain current time in db (I use Oracle ):

SELECT sysdate FROM dual;

Then I found the way how to find the difference in minutes between two dates:

SELECT(date1 - date2)*1440

Looks like a little bit ugly... but ok.

Now, I am trying to combine all together:

UPDATE USERS u 
SET ACTIVE='N' 
WHERE  SELECT((SELECT sysdate FROM dual) - u.LAST_VISIT)*1440 >60;

Could you please review my final query.

Is it ok? Or, maybe, it is possible to optimize it?

Answer 1

You can simplify the query

UPDATE users u
   SET active = 'N'
 WHERE u.last_visit < sysdate - interval '1' hour;

If there is an index on last_visit that can be used for the query (because the predicate is sufficiently selective), this query can potentially use that index.

Answer 2

Justin's answer is very good, but it will probably update too many rows. You should add an extra condition to the where clause:

UPDATE users u
   SET active = 'N'
    WHERE active <> 'N' AND u.last_visit < sysdate - interval '1' hour;

This saves the attempted update for users who are already inactive. (Note: this version assumes that active does not take the value of NULL .)