crontab entry 3

[Mag0007]

Is there an algorithm to schedule something as first business day of the month?

 

[PHV]

man cron
man date

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[Mag0007]

Phv:

Thanks for the quick response

I know how to use crontab and date. Its just getting it so I can use it on a offical first business day of the month.

I was thinking of something like this

0 6 1 * 1 foo > /dev/null

Run foo at 6:00AM everymonth, on the first day?

 

[Chacalinc]

from man 5 crontab
[tt]
field allowed values
----- --------------
minute 0-59
hour 0-23
day of month 1-31
month 1-12 (or names, see below)
day of week 0-7 (0 or 7 is Sun, or use names)
[/tt]

So 1st of every month should be:

0 6 1 * * foo > log.log 2>1&

Cheers.

 

[Mag0007]

Right, thanks.

But if the first day is Sunday, thats no good for me. I need this to run on the first buisness day (Monday). Is that possible?

 

[PHV]

And foo may test with the date command if it has to sleep until a business day ...

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[Mag0007]

PHV:

Now I think I know what you are saying. Thats a very good idea!

 

[vgersh99]

to get the 1st monday of every month in the US locale:

cal 8 2004| nawk 'FNR > 2 && NF >= 6 {print (NF==6) ? $1 : $2; exit}'

vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[SamBones]

Here's a simpler way to run something the 1st Monday of every month...

0 6 1-7 * 1 foo > foo.log 2>1&

The "[tt]1-7[/tt]" limits it to the first 7 days of the month. The "[tt]1[/tt]" limits it to a Monday. This only works if a Monday occurs within the first 7 days of the month.

Now the first business day is just the first Monday through Friday that occurs in a month. You can do this with the following crontab (it takes three entries to do it though)...

0 6 1 * 1-5 foo > foo.log 2>1&
0 6 2 * 1   foo > foo.log 2>1&
0 6 3 * 1   foo > foo.log 2>1&

The first line means if the 1st falls on a Monday through Friday, run it. The second line means if it's the 2nd and it's a Monday, run it. This means the 1st was on a Sunday. The third line means if it's the 3rd and it's a Monday, run it. This is when the 1st is a Saturday.

So, these three lines will run "[tt]foo[/tt]" on the first bsuiness day of the month (assuming a business day is Monday through Friday).

Hope this helps.

 

[PHV]

SamBones, I'm not sure ...

Note that the specification of days may be made by two fields
(day of the month and day of the week). If both are specified as
a list of elements, both are adhered to. For example, 0 0 1,15 *
1 would run a command on the first and fifteenth of each month,
as well as on every Monday. To specify days by only one field,
the other field should be set to `` * '' (for example, 0 0 * * 1
would run a command only on Mondays).

Hope This Helps, PH.
Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884 or FAQ181-2886

 

[SamBones]

Hmmm, you're right! Well that s*cks! How do I return the star I got?

Then it needs to be done in the script. Something like this as the first few lines in the script...

#!/bin/ksh

export DOM=$(date '+%d')
export DOW=$(date '+%a')

(( ${DOM} > 3 )) && exit
[[ "${DOW}" = Sat || "${DOW}" = Sun ]] && exit
[[ "TueWedThuFri" = +(*${DOW}*) && ${DOM} != 01 ]] && exit

# If it reaches here, it's the first business day of the month.
# Your first business day code starts here...

Then, you'd just need to run the script with the following [tt]crontab[/tt] entry...

0 6 1-3 * * first_bus_day.sh > fbd.log 2>&1

That will run it on the 1st, 2nd, and 3rd, but it will only get past those lines if it's the first business day of the month.

Hope this helps.

 

[jstreich]

Heh, what if the first monday is a holiday?
Why not each monday of this month, find out what the first monday is of the next, and [green]at[/green] the job instead?

You could then use this to figgure out the next one:

cal [red]$(($(date +%m)+1))[/red] [red]$(date +%Y)[/red]| nawk 'FNR > 2 && NF >= 6 {print (NF==6) ? $1 : $2; exit}'

And after getting that into $x (not that hard, I'll leave it to you):

at $(($(date +%m)+1))/$x/$(date +%Y) -f nameOfThisScript

[plug=shameless]

http://game-master.us/phpx-3.4.0/

[/plug]

 

[SamBones]

Sorry, everyone works on holidays due to scripting issues!

jstreich, your examples won't work in December. You'll be trying to get a calendar for the 13th month. Try this...

cal $(( ($(date +%m)+1) % 12 )) $(( $(date +%Y) + ($(date +%m)/12) ))

...to get the next month with a correct rollover to the next year.

If Monday is a holiday, the first business day for the month would be the Tuesday immediately following it, not the first Monday of the next month.

I *HATE* problems that seem simple at first, but get messier and messier as you dig into them.

I worked at a place once where we had this issue and the cleanest solution (for us at the time) ended up being a calendar table in an Oracle database that had all the days in the year with holidays and other special processing days flagged.

Hope this helps.

 

[Mag0007]

Sambones:

"I *HATE* problems that seem simple at first, but get messier and messier as you dig into them."

BINGO!!!

I like how PHV was like, "man cron, man date"

 

[PHV]

I like how PHV was like, "man cron, man date"
I was lazzy, the true answer (for me) was:
man cron
man date
man sleep

 

[Mag0007]

Thanks to everyone for helping me

 

[stefanwagner]

@samBones:

Keep the star from me.

I first asked myself 'hey, don't they work ORed (day of month, day of week) but your explanation was so well done, that I forgot my doubt.

But we get so seldom stars, when they would have been appropriate, so we can keep the few, that were spent by accident.

seeking a job as java-programmer in Berlin:

http://home.arcor.de/hirnstrom/bewerbung