I have just been reading this about the Cron Expression Format:
The ? Character
The question mark (?) character can be used only in the `dayofmonth` and `dayofweek` fields, but not at the same time. You can think of the `?` character as "I don't care what value is in this field." This is different from the asterisk, which indicates every value for the field. The `?` character says that no value was specified for this field.
The reasons a value can't be specified for both fields are tough to explain and even tougher to understand. Basically, if a value was specified for each, the meaning would become ambiguous: Consider if an expression had the value 11 in a field for the day of the month and a value of WED in the field for the day of the week. Should that trigger fire only on the 11th of the month if it falls on a Wednesday? Or should it fire on both the 11th and every Wednesday? The ambiguity is removed by not allowing a value in both fields at the same time.
Just remember that if you specify a value in one of the two fields, you must put a ? in the other.
Example expression:
0 10,44 14 ? 3 WED
Just wondering: how can I now create a Cron job that does something on Friday 13th (like formatting my data partition) or Saturday 14th (like making a new backup)?
##########################################################
# minute (0-59),
# hour (0-23),
# day of the month (1-31),
# month of the year (1-12),
# day of the week (0-6 with 0=Sunday).
##########################################################
# this is friday the 13th at 0:00 i guess?
# daynr:13 and dayname: friday
0 0 13 * 4 /script.cgi
Related
I have a project where I have to update a field in my database every 1st day of each month and update it again every 10th after.
This cron allow me to open a session and to close it. I would like to know how can I set up the 10th of the month.
1st day of each month I have : 0 0 1 * *
10th after it : ??
Thanks :)
You can create a Cron expression that passes two dates in the day variable. Something like this would work:
0 0 0 1,10 * ? *
Expression description:
At 00:00:00am, on the 1st and 10th day, every month
You can use the comma in order to pass multiple variables in the day parameter in order to set multiple days to run.
A great resource for playing around with Cron statements is the Cron formatter website, which will let you enter statements, then give you the plain English logic that the statement will follow. It also shows which sections of the statement resolve to minutes, hours, days, ect.. It is really helpful for creating new statements in my opinion.
I'm trying to come up with a CRON expression that will allow me to schedule a quartz trigger to run on every Monday in a month except the first one.
References:
http://www.quartz-scheduler.org/documentation/quartz-2.x/tutorials/crontrigger.html
https://docs.oracle.com/cd/E12058_01/doc/doc.1014/e12030/cron_expressions.htm
CRON allows you to specify the nth occurrence of a day of the week easily. An expression for First Monday of the month would be:
0 5 0 ? * 2#1
Where the 2#1 represents the first Monday of the month (2 = day of the week, 1 being the nth occurrence)
However, if I try to do something like
0 5 0 ? * 2#2-2#5
OR
0 5 0 ? * 2#2,2#3,2#4,2#5
It complains with the message
Support for specifying multiple "nth" days is not implemented.
Does anyone know how to achieve this in CRON?
Where cron doesn't give you the expressiveness you desire(a), it's a simple matter to change the command itself to only execute under certain conditions.
For your particular case, you know that the first Monday of a month is between the first and seventh inclusive and subsequent Mondays must be on the eighth or later.
So use cron to select all Mondays but slightly modify the command to exclude the first one in the month:
# mm hh dom mon dow command
0 1 * * 1 [[ $(date +%u) -gt 7 ]] && doSomething
That job will run at 1am every Monday but the actual payload doSomething will only be executed if the day of the month is greater than seven.
Some people often opt for putting the test into the script itself (assuming it even is a script) but I'm not a big fan of that, preferring to keep all scheduling information in the crontab file itself.
(a) Don't be mistaken into thinking you can combine the day of week 1 and day of month 8-31 to do this. As per the man page, those conditions are OR'ed (meaning either will allow the job to run):
Commands are executed by cron when the minute, hour, and month of year fields match the current time, and when at least one of the two day fields (day of month, or day of week) match the current time
Combining those two will run the job on the first Monday and every single day from the eighth onwards.
i have a cron expression-
0 0 12 */2 * ?
If start date is monday and time is 11:40 am, the next trigger date i'm expecting is monday 12:00, followed by wednesday, friday,etc.
But when i give this expression, the first trigger is set to tuesday 12:00, followed by thursday, saturday,etc
i verified this on http://cronmaker.com
Why does this behavior occur for monday?
If the start date is set to any other day it seems to behave the way its supposed to.
So if it was set on Tuesday 11:50 am , the first trigger is on tuesday 12:00.
Please help me understand. Is it a bug or expected behavior? Is there a work around to make it trigger on monday?
Thanks
Your cron schedule doesn't care about the day of the week. It is running simply on every uneven day of the month. This is the expected behaviour.
If you need it to run on Mondays, you should use something like 0 0 12 ? * MON,WED,FRI
First of all you expression only uses ? for the day of the week, so effectively you are not controlling that part.
Second the / character in a Cron expression indicates an increment. And when used next to a *, the star just means the lower bound for that value, 1 for the day of the month.
So indeed you are asking for a fire at noon every uneven day of the month. And the start time of the trigger will only constrain the first instance to be the next uneven day of the month.
You cannot express what you seem to desire with a cron trigger - that is a schedule which is based off the start time of the trigger. You should use s SimpleTrigger for this
I am working on a revenue report that predicts deposit date related to "Revenue Date" based on observed patterns. For example, one stream of revenue is always deposited the next business day, so I would need a formula that adds one day to anything with a "Revenue Day" identified as Monday through Thursday (because these would be deposited Tuesday-Friday, respectively), and adds three days to a "Revenue Day" identified as Friday (because this would be deposited on Monday), and adds two days for Saturday, and one for Sunday.
I managed to get the correct result for Monday-Thursday using this formula, where G1 is the weekday and H1 is the date:
=IF(OR(G1="MONDAY", G1="TUESDAY", G1="WEDNESDAY", G1="THURSDAY"), H1+1)
Unfortunately, I can't figure out how to also get it to add 3 days to Friday, 2 days to Saturday, and 1 day to Sunday. Is this possible?
Your if formula should take 3 parameters IF(condition, do if condition is true, do if condition is false).
Use ; instead of ,
Sundays does the same this as Moday-Thursday, so no need for a special case for it
... So i would guess that you need something like this :
=IF(OR(G1="SUNDAY";G1="MONDAY";G1="TUESDAY";G1="WEDNESDAY";G1="THURSDAY");H1+1;IF(G1="FRIDAY";H1+3;IF(G1="SATURDAY";H1+2;"bad input")))
This is a follow up to an earlier question.
#ForwardED I am trying to convert your original single static formula into a single dynamic formula.
Unfortunately my employer's filters will let me up certain things to a hyperlink, but will not let me download or view from the same site. I am also trying to come up with a formula for floating dates.
Below is a copy of the expanded explanation I gave on the original question. I am not sure if you missed it or not. It deals with holidays that have a set date like Christmas, 25th of December of every year. However if it fall on a Saturday the time of work is the Friday and if it is on a Sunday the day off is the Monday.
Dealing with a holiday falling on a Saturday or Sunday
Again we need to refer to some cell in your spreadsheet with the year so I will again use Q10 as the example and we will assume a date of 2014/10/24.
=IF(WEEKDAY(DATE(YEAR(Q10),12,25))=7,DATE(YEAR(Q10),12,24),IF(WEEKDAY(DATE(YEAR(Q10),12,25))=1,DATE(YEAR(Q10),12,26),DATE(YEAR(Q10),12,25)))
The formula checks first if the weekday is a Saturday. We do this using a function that will return the day of the week See step 2) from the original question. It is this part from the equation above:
WEEKDAY(DATE(YEAR(Q10),12,25))
It will return a single integer 1 through 7 corresponding to the day of the week the date function results in, in this case. If its a 1 we known its Sunday, if its 7 we know its Saturday. So the check for Saturday is:
WEEKDAY(DATE(YEAR(Q10),12,25))=7
If WEEKDAY()=7 is true then we provides the date of the day before which is really just subtracting 1 from the date we were looking at. We use this part of the formula to calculate that:
DATE(YEAR(Q10),12,24)
notice how I changed the day from 25 to 24. An alternate way would be to recycle our date and make the computer do one more calculation using this formula:
DATE(YEAR(Q10),12,25)-1
or
DATE(YEAR(Q10),12,25-1)
That all sits in the TRUE portion of the if statement. so if the date does not fall on Saturday then we wind up in the FLASE portion of the IF statement. Here we check with a second IF for the date falling on a Sunday. we use the same theory and process as we did for the Saturday check.
IF(WEEKDAY(DATE(YEAR(Q10),12,25))=1,DATE(YEAR(Q10),12,26),DATE(YEAR(Q10),12,25))
Placing an IF statement inside an IF statement is commonly referred to as "nesting". This whole IF statement happens in the FALSE portion of the previous IF that checked to see if it was Saturday. This time we checked for Sunday:
WEEKDAY(DATE(YEAR(Q10),12,25))=1
When this is true, then we need to increase the date by 1 day instead of decreasing it like was done for Saturday:
DATE(YEAR(Q10),12,26)
or
DATE(YEAR(Q10),12,25)+1
or
DATE(YEAR(Q10),12,25+1)
So that was the true portion of the Sunday check. Logically speaking the only way to get to the FALSE portion of this nested IF statement is to fail the Saturday check and then fail the Sunday check. Which means you do not need to go through and check if is the WEEKDAY comes out as 2, 3, 4, 5 or 6! Its one of those by the process of eliminating Sunday and Saturday (1 and 7). And if the date falls on Monday-Friday we dont need to change the date and can leave it just as is:
DATE(YEAR(Q10),12,25)
And I realized I did not explain how the date function works, though I think I tried to in one of the previous questions...regardless! DATE(arg1,arg2,arg3) requires three different arguments as integers or other functions that return integers.
arg1 is the year so 2014, 1995, 1965 are all acceptable integers. Also we could use YEAR(Q10), where the cell Q10 holds the date of 2014/10/24. In this case YEAR(Q10) would return 2014.
arg2 is the month and needs to be an integer in the range of 1 to 12. Again you can always use a formula that returns an integer in that range as well such as MONTH(Q10) which from our previous value of Q10 would return 10.
arg3 is day and similar to the above it needs to be an integer. A formula such as DAY(Q10) would return a value of 24.
What this means is if we know what day a holiday is on we can force it to a date by supplying a set month and day, and letting the year be determined by a formula that supplies the year you are interested in. So if you look at the last formula you can see we fixed the month at 12 and the day at 25. They year will be determine from the year of the date supplied in cell Q10.