I'm new to python and I'm looking to work with datetime. I have some files generated every Sunday and I like to move the furthest Sunday out of the current folder eg: 2020-04-12, 2020-04-19, 2020-04-26.
I have found some examples on getting a specific date from today's date and I was able to modify it a tab bit. Eg. I can go back and get last week's Sunday with a specific date:
from datetime import date
from datetime import timedelta
import datetime
today = datetime.datetime(2020,4,13)
offset = (today.weekday() + 1) % 7
sunday = today - timedelta(days=offset)
#print (offset)
print(sunday)
I am confused by the offset variable. What is (today.weekday() + 1) % 7 doing? I have read the Python doc and not quite wrapping my head around it. With +1, I get the date 2020-04-12, which is a Sunday, great. When I do -1 (the other thing is if I set it to (today.weekday() - 1) % 7), I get 2020-04-07, a Tuesday. How did it jump from Sunday the 12th to Tuesday the 7th?
Additionally, how do I get it to jump back 3 weeks? that's where I'm also stuck.
Alright, so if today's Wednesday, then today.weekday() is 2, because it starts counting from 0 on Monday. Not sure why, but that's life.
So (2 + 1) % 7) = 3. That means that 3 days ago was Sunday. Hence your code:
offset = (today.weekday() + 1) % 7 # How many days ago was sunday
sunday = today - timedelta(days=offset) # Go backwards from today that many days
You'll notice that if you subtract one instead of add one, that means we're going backwards (because we're sutracting the timedelta object) by two fewer days than before (because 2 - 1 is equivalent to (2 + 1) - 2, that is, two fewer days). If you started by going backwards enough days to get to Sunday, and now you're going backwards two fewer days, you'll end up on Tuesday, which is two days later than Sunday.
The easiest way to shift which week you're headed to is to set the weeks argument in timedelta:
n_weeks = 3
sunday = today - timedelta(days=offset, weeks=n_weeks)
that's equivalent to, but much prettier than:
sunday = today - timedelta(days=offset + n_weeks * 7)
I generated a cron to run every Monday at 5am.
0 5 1 * 1
The third number, 1, for day of month, has it set to run on the first of every month as well as Monday.
Do I change that 1 to 0 so it ignores the day of month? Otherwise it will run every Monday as well as run on the 1st of the month.
Have two entries:
to run every Monday at 5 AM : 0 5 * * 1
to run on 1st of every month at 5 AM : 0 5 1 * *
OR, if you want a single entry, the you may have to do something like: https://github.com/xr09/cron-last-sunday/blob/master/run-if-today
To avoid day confusion we can give like this as well -
to run every Monday at 5 AM : 0 5 * * MON
public static ITrigger FireEveryDayAtMidnight()
{
ITrigger trigger = TriggerBuilder.Create()
.WithIdentity("trigger3", "group1")
.WithCronSchedule("0 02 01 * * ?")
.ForJob("myJob", "group1")
.Build();
return trigger;
}
I have a problem thoroughly understanding Cron Trigger Expressions. I intend to schedule the above trigger to fire every day at 2 minutes past midnight. I just want to hear from people much more experienced with cron expressions if my expression above "0 02 01 * * ?", will indeed run as intended i.e run fire every day at 2 minutes past midnight
The cron in your trigger will execute at 02:00 on the 1st of every month.
If you wish to execute it at 00:02 every day using the 7 field Quartz contrab format, use:
0 2 0 * * * * ?
Here's a quick summary of how crons are formed, snipped from the Cron How To on Ubuntu's site:
Each of the sections is separated by a space, with the final section
having one or more spaces in it. No spaces are allowed within Sections
1-7, only between them. Sections 1-7 are used to indicate when and how
often you want the task to be executed. This is how a cron job is laid
out:
second (0-59), minute (0-59), hour (0-23, 0 = midnight), day (1-31), month (1-12),
weekday (0-6, 0 = Sunday), year (empty, 1970-2099) command.
I would like to run a job through cron that will be executed every second Tuesday at given time of day. For every Tuesday is easy:
0 6 * * Tue
But how to make it on "every second Tuesday" (or if you prefer - every second week)?
I would not like to implement any logic in the script it self, but keep the definition only in cron.
Answer
Modify your Tuesday cron logic to execute every other week since the epoch.
Knowing that there are 604800 seconds in a week (ignoring DST changes and leap seconds, thank you), and using GNU date:
0 6 * * Tue expr `date +\%s` / 604800 \% 2 >/dev/null || /scripts/fortnightly.sh
Aside
Calendar arithmetic is frustrating.
#xahtep's answer is terrific but, as #Doppelganger noted in comments, it will fail on certain year boundaries. None of the date utility's "week of year" specifiers can help here. Some Tuesday in early January will inevitably repeat the week parity of the final Tuesday in the preceding year: 2016-01-05 (%V), 2018-01-02 (%U), and 2019-01-01 (%W).
How about this, it does keep it in the crontab even if it isn't exactly defined in the first five fields:
0 6 * * Tue expr `date +\%W` \% 2 > /dev/null || /scripts/fortnightly.sh
pilcrow's answer is great. However, it results in the fortnightly.sh script running every even week (since the epoch). If you need the script to run on odd weeks, you can tweak his answer a little:
0 6 * * Tue expr \( `date +\%s` / 604800 + 1 \) \% 2 > /dev/null || /scripts/fortnightly.sh
Changing the 1 to a 0 will move it back to even weeks.
Something like
0 0 1-7,15-21 * 2
Would hit the first and third Tuesday of the month.
Note: Don't use this with vixie cron (included in RedHat and SLES distros), as it makes an or between the day-of-month and day-of-week fields instead of an and.
Maybe a little dumb, but one could also create two cronjobs, one for every first tuesday and one for every third.
First cronjob:
0 0 8 ? 1/1 TUE#1 *
Second cronjob:
0 0 8 ? 1/1 TUE#3 *
Not sure about the syntax here, I used http://www.cronmaker.com/ to make these.
If you want to do it based on a given start date:
0 6 * * 1 expr \( `date +\%s` / 86400 - `date --date='2018-03-19' +\%s` / 86400 \) \% 14 == 0 > /dev/null && /scripts/fortnightly.sh
Should fire every other Monday beginning with 2018-03-19
Expression reads: Run at 6am on Mondays if ...
1 - Get today's date, in seconds, divided by the number of seconds in a day to convert to days sice epoch
2 - Do the same for the starting date, converting it to the number of days since epoch
3 - Get the difference between the two
4 - divide by 14 and check the remainder
5- If the remainder is zero you are on the two-week cycle
I discovered some additional limitations of above approaches that can fail in some edge cases. For instance, consider:
#xahtep and #Doppelganger discussed issues using %W on certain year boundaries above.
#pilcrow's answer gets around this to some degree, however it too will fail on certain boundaries. Answers in this and or other related topics use the number of seconds in a day (vs. week), which also fail on certain boundaries for the same reasons.
This is because these approaches rely on UTC time (date +%s). Consider a case where we're running a job at 1am and 10pm every 2nd Tuesday.
Suppose GMT-2:
1am local time = 11pm UTC yesterday
10pm local time = 8pm UTC today
If we are only checking a single time each day, this will not be an issue, but if we are checking multiple times -- or if we are close to UTC time and daylight savings occurs, the script wouldn't consider these to be the same day.
To get around this, we need to calculate an offset from UTC based on our local timezone not UTC. I couldn't find a simple way to do this in BASH, so I developed a solution that uses a quick one liner in Perl to compute the offset from UTC in seconds.
This script takes advantage of date +%z, which outputs the local timezone.
Bash script:
TZ_OFFSET=$( date +%z | perl -ne '$_ =~ /([+-])(\d{2})(\d{2})/; print eval($1."60**2") * ($2 + $3/60);' )
DAY_PARITY=$(( ( `date +%s` + ${TZ_OFFSET} ) / 86400 % 2 ))
then, to determine whether the day is even or odd:
if [ ${DAY_PARITY} -eq 1 ]; then
...
else
...
fi
There are many good answers here. Based upon the comments, I see quite a bit of confusion and frustration. My intention with this answer is to not only answer the OPs question How to instruct cron to execute a job every second week?, but also clear up some confusion for folks who may read this in the future.
TL;DR:
The crontab entry look like this:
\<minute\> \<hour\> * * \<Day of Week\> expr \\( $( date+\\%s ) \\/ 604800 \\% 2 \\) > /dev/null && \<command to run on odd weeks\> || \<command to run on even weeks\>
Crontab Entries:
All crontab entries [made with crontab -e] have this format, per the man page:
1: The Minute of the hour to execute [command]
Range: 0-59
2: The Hour of the Day to execute [command]
Range: 0-23
3: The Day of the Month to execute [command]
Range: 1-31
4: The Month to execute [command]
Range: 1-12
5: The Day of Week to execute [command]
Range: Check your man page, could be 0-6, 1-7 or 0-7
6: command to execute
Note, there are also # values, which I will not discuss here.
Some versions of *nix allow for expressions:
*/2 = every even number in range
1 + */2 = every odd number in range
1,15 = First & 15th
2-8 = Second through the Eighth (inclusive)
Some versions allow for the Human Readable words also, such as February or Wednesday.
For Month, */2 will execute on Feb, Apr, Jun, Aug, Oct, Dec.
For Days of Week, */2 will run every even day -- check your man page to see how the Days of Week are numbered. Tue/2 is NOT a valid expression.
Human Readable (Calendar) Woes
Some constants you need to know:
There are 365.2464 days in a year (For convenience, we'll round to 365).
There are 12 months in a year.
There are 7 days in a week.
There 86,400 seconds in a day.
Therefore,
1 month is 4-1/3 weeks
[i.e. 3 months is 13 weeks]
1 year is 52-1/7 weeks
The bane of calendar math:
Semi-Monthly = every half month = 2x/month = 24 times per year.
Bi-weekly = every other week (fortnightly) = 26 times per year.
Note: These terms are often mis-used.
Some years have 51 weeks, some have 53, most have 52. If My cron runs every odd week ( date +%W mod 2), and the year has 51 or 53 weeks, it will also run the following week, which is week 1 of the new year. Conversely, if my cron runs every even week, it will skip 2 weeks. Not what I want.
CRON can support semi-monthly, it cannot support bi-weekly!
Semi-monthly:
The first of the month will always fall between the 1st & 7th. The second half of the week will always occur between the 15th and the 21st.
Semi-monthly would have 2 values, one in the first half and the other in the second half of the month. Such as:
2,16
Unix Time
At a very high level, *nix time is 2 values:
date +%s = Number of Seconds since Epoch (01/01/1970 00:00:00)
date +%N = fractional seconds (in Nano seconds)
Therefore, time in *nix is date +%s.%N
*Nix uses epoch time. The /etc/shadow file contains the date of last password change. It is the integer portion of %s divided by 86,400.
Factino
The GPS satellites measure time as "Number of weeks since epoch time" and "(fractional) seconds into the week.
Note: Epoch weeks are independent of years. It does not matter if the year has 51, 52, or 53 weeks. Epoch weeks never roll over.
Bi-Weekly Time Algorithm
In *Nix date +%W is week number of the year, not epoch week. *nix does not have an epoch week value, but it can be computed.
Epoch Week = Integer( Epoch Seconds / Seconds_per_Day / Days_per_Week )
Fortnightly = Epoch_Week Modulo 2
The Fortnightly value will always be 0 or 1, thus, running the command when Fortnightly = 0 runs on every even week, and 1 = every odd week.
Computing Fortnightly)
The first way (bc):
date +"%s / 604800 % 2" | bc
This will generate "[Epoch Seconds] / 604800 % 2"
That answer is then sent to bc, a basic calculator, which does the math and echoes the answer to STDOUT and any errors to STDERR.
The return code is 0.
The second way (expr):
In this case, send the expression to expr and let it do the math
expr $( date +%s ) / 604800 % 2
The expr command does the math, echoes the answer to STDOUT, errors to STDERR.
The return code is 1 if the answer is 0, otherwise, the return code is 0.
In this manner, I don't care what the answer is, only the return code. Consider this:
$ expr 1 % 2 && echo Odd || echo Even
1
Odd
$ expr 2 % 2 && echo Odd || echo Even
0
Even
Since the result doesn't matter, I can re-direct STDOUT to /dev/null. In this case, I'll leave STDERR incase something unexpected happens, I'll see the error message.
Crontab Entry
In crontab, the percent sign and the parenthesis have special meanings, so they need to be escaped with a backslash (\).
First, I need to decide if I want to run on Even weeks or odd weeks. Or, perhaps, run Command_O on Odd weeks and command_E on even weeks. Then, I need to escape all the special characters.
Since the expr only evaluates WEEKS, it is necessary to specify a particular day of the week (and time) to evaluate, such as every Tuesday at 3:15pm. Thus, the first 5 files (the Time Entry) of my crontab entry will be:
15 3 * * TUE
The first part of my cron command will be the expr command, with STDOUT sent to null:
expr \\( $( date+\\%s ) \\/ 604800 \\% 2 \\) > /dev/null
The second part will be the evaluator, && or ||
The Third part will be the command (or script) I want to run. Which looks like this:
expr \\( $( date+\\%s ) \\/ 604800 \\% 2 \\) > /dev/null && \<command_O\> || \<command_E\>
Hopefully, that will clarify some of the confusion
Try this
0 0 1-7,15-21 * 2
This is run 2 times a month, both on Tuesdays and at least a week apart.
If you want every tuesday of second week only:
00 06 * * 2#2
Cron provides an 'every other' syntax "/2". Just follow the appropriate time unit field with "/2" and it will execute the cronjob 'every other time'. In your case...
0 6 * * Tue/2
The above should execute every other Tuesday.
Syntax "/2" is not goes along with weekday. So my added to the smart above is simply use 1,15 on MonthDay filed.
0 6 1,15 * * /scripts/fornightly.sh > /dev/null 2>&1
0 0 */14 * *
Runs At 00:00 on every 14th day-of-month
Why not something like
0 0 1-7,15-21,29-31 * 5
Is that appropriate?