How to execute cron quartz for the first Saturday and the first Sunday of each month

Question

I have tried with:

0 0 2 ? 1/1 SAT#1,SUN#1 *

but the result is wrong (only appears the sunday):

1. Sunday, April 6, 2014 2:00 AM
2. Sunday, May 4, 2014 2:00 AM
3. Sunday, June 1, 2014 2:00 AM
4. Sunday, July 6, 2014 2:00 AM
5. Sunday, August 3, 2014 2:00 AM

also I have tried with:

0 0 2 1-7 * ? SAT,SUN and a lot of variants but all of them are invalid syntax.

The correct output should be:

1. Saturday, April 5, 2014 2:00 AM
2. Sunday, April 6, 2014 2:00 AM
3. Saturday, May 3, 2014 2:00 AM
4. Sunday, May 4, 2014 2:00 AM
5. Sunday, June 1, 2014 2:00 AM
6. Saturday, June 7, 2014 2:00 AM . . . .

I've been doing all the tests since http://www.cronmaker.com/ (nice page!) unsuccessfully. Regards

Answer 1

Cron Quartz definitely not allowed to do this in a single statement. I get error message: "Support for specifying both a day-of-week AND a day-of-month parameter is not implemented."

Answer 2

I've tried unsuccessfully too. In my opinion, Quartz Cron has some limitations.

By the way, can't you split your cron into two different schedules? For example:

0 0 2 ? 1/1 SAT#1 *
0 0 2 ? 1/1 SUN#1 *

Answer 3

0 and 7 stand for Sunday.

Hence, the numbers from 1 to 6 are fixed to Monday, ..., Saturday.

Graphically:

 +---------------- minute (0 - 59)
 |  +------------- hour (0 - 23)
 |  |  +---------- day of month (1 - 31)
 |  |  |  +------- month (1 - 12)
 |  |  |  |  +---- day of week (0 - 6) (Sunday=0 or 7)
 |  |  |  |  |
 *  *  *  *  *  command to be executed

0 0 1-7 * 0,6 * should work for you.

Cron

Answer 4

You have to split it to 2 cron expressions. From the documentation ( http://www.quartz-scheduler.org/api/2.2.0/org/quartz/CronExpression.html ) :

If the '#' character is used, there can only be one expression in the day-of-week field ("3#1,6#3" is not valid, since there are two expressions).