I am trying to run my job every end of month but i am getting an error every time i add cron expression :
from("timer://ratingTimer?cron=0+52+12+++")
this is i am trying to run my job for specific time.
Caused by: org.apache.camel.ResolveEndpointFailedException: Failed to resolve endpoint: timer://ratingTimer?cron=0+52+12+%3F++* due to: There are 1 parameters that couldn't be set on the endpoint. Check the uri if the parameters are spelt correctly and that they are properties of the endpoint. Unknown parameters=[{cron=0 52 12 ? * *}]
We will schedule cron on 28,28,29 and 31’st of each month. Now find if today is the last day of the month. To find it check if the next day is 01’st of next day and then only execute any command.
Below command will return the date of the next day.
date +%d -d tomorrow
Now check if tomorrow is 01.
[ "$(date +%d -d tomorrow)" = "01" ] && echo "True"
If the next day is 01 then above command will print “True” on screen. Here you can use the above script in crontab and change echo with your command.
59 23 28-31 * * [ “$(date +%d -d tomorrow)” = “01” ] && /root/script.sh
Check more in below url.
CRON job to run on the last day of the month
AND
quartz scheduler: run on last day of the month
from("cron://ratingTimer?schedule=0+0+3+L+*+?")
Related
I want the previous month & previous year in the format: "December2020".
But while executing the below Linux command I'm getting July2020.
echo `date -d "2021-01-08" '+%B' -d 'last month'``date -d '1 year ago' +%Y
The date hardcoded parameter is in format: YYYY-MM-DD (2021-01-08)
date accepts only one -d option. In your command date -d "2021-01-08" '+%B' -d 'last month' the first -d is ignored. Only the -d "last month" applies. Because of that, and since we have August right now, the output is July.
You probably wanted to use
date -d '2021-01-08 - 1 month' +%B%Y
which prints December2020.
If you really wanted to concat the previous month and previous year together, you could use
echo "$(date -d'2021-01-08 - 1 month' +%B)$(date -d'2021-01-08 - 1 year' +%Y)"
but that would give rather strange results:
2021-01-08 → December2020 # 1 month before input date
2021-04-30 → March2020 # 13 months before input date
I have a simple question here.
I have an input date with time of 05:21, but it could be any arbitrary 24 hour time.
I add 1 minute to it using the date command. date then outputs 17:22!!
I need the output format to be 24 hour because that's what the input format is, and what is required.
How do I tell date to stop messing with the times?
I'm very close to just using substr to extract the minutes, and then adding 1, and then adding a check to see if >59 and if so, value = 0 and substr the hour value out, then add 1 to that, because at this point it seems simpler.
end='12-02-2018 17:01'
test=$(echo $end | sed -re 's#(..)-(..)-(....) (..:..)()#\3-\2-\1 \4#') ;
add=$(date +"%d-%m-%Y %H:%M" --date="$test + 1 minute") ;
echo $test --- $add
2018-02-12 17:01 --- 13-02-2018 05:02
I have date and time printed in log files as "14:09:49.922 UTC 12.08.2015"
To analyze the logs in few instances I have to grep for next X seconds from this date and time in the logs.
Note : Time Zone might vary.
I have used grep along with for loop to iterate the seconds or minutes or hours depending upon the give time.
The help I am looking for is the options in the date command to increment seconds so that if I ad x seconds to the given time stamp, then date, month year, hour, minute and second should be updated accordingly.
Eg : 23:59:59 UTC 31.12.2015" + 1 seond should return "00:00:00 UTC 01.01.2016".
Basically I am looking for options in date command instead of me manually checking if seconds crossed 59 then increment minute and so on.
How to achieve this in a shell script using date utility?
Date command doesn't support "14:09:49.922 UTC 12.08.2015" format.
So I converted to "14:09:49.922 UTC 08/12/2015" and then used date utility as below
DATE="14:09:49 UTC 12.08.2015"
NEXT_DATE=`echo $DATE | awk '{ split($3,a,"."); print $1" "$2" "a[2]"/"a[1]"/"a[3]}'`
TIME_ZONE=`echo $NEXT_DATE | awk '{print $2}'` NEXT_DATE=`TZ="$TIME_ZONE" date +"%H:%M:%S %Z %m/%d/%Y" -d "$NEXT_DATE + 1 second"`
GREP_DATE=`echo $NEXT_DATE | awk '{ split($3,a,"/"); print $1" "$2" "a[2]"."a[1]"."a[3]}'`
grep $GREP_DATE logfile
I use the command to get next day relative to specified date as follows:
date -d "20130107 1days" "+%Y%m%d"
but , if the input is 2013010712 , I want to get 2013010711 , how can i do ?
Or, If there is a simple method to do it?
My date isn't new enough to handle the 2013010712 format but, with some helper punctuation added for my date (GNU coreutils 8.13), the method for changing hours is shown:
$ date -d "2013-01-07 12 -1hour" "+%Y%m%d%H"
2013010711
$ date -d "2013-01-07 12 +1hour" "+%Y%m%d%H"
2013010713
I have 2 script in bash, and i have some files:
transaction-2012-01-01.csv.bz2
transaction-2012-01-02.csv.bz2
transaction-2012-01-03.csv.bz2
transaction-2012-01-04.csv.bz2
.
.
transaction-2012-01-31.csv.bz2
transaction-2012-02-01.csv.bz2
.
.
transaction-2012-02-28.csv.bz2
I have a script called script.sh
cat script.sh
YEAR_MONTH=$1
FILEPATH="transaction-$YEAR_MONTH*.csv.bz2"
bzcat $FILEPATH|strings|grep -v "code" >> output
And if you need call the script you can use other script
cat script2.sh
LAST_MONTH=$(date -d -1month +%Y"-"%m)
if [ $# -eq 1 ]; then
DATE=$1
else
DATE=$LAST_MONTH
fi
script.sh $DATE 1>output$DATE.csv 2>> log.txt
And it do cat the files in a month, but now i need call the script with a specific week in a year:
bash script2.sh 2012-01
where 2012 is the year and 01 is the month
Now i need call the script with:
bash script2.sh 2012 13
where 2012 is the year and 13 is the week in a year
Now i need the cat only to the files in the year and week that the user specified, no per month per week
But the format of the files do not help me!!!! because the name is transaction-year-month-day.csv.bz2, not transaction-year-week.csv.bz2
Take a look at the manpage for strftime. These are date format codes. For example:
$ date +"%A, %B %e, %Y at %I:%m:%S %p"
Will print out a date like:
Thursday, May 30, 2013 at 02:05:31 PM
Try to see why this works.
On some systems, the date command will have a -j switch. This means, don't set the date, but reformat the given date. This allows you to convert one date to another:
$ date -f"$input_format" "$string_date" +"$output_format"
The $input_format is the format of your input date. $string_date is the string representation of the date in your $input_format. And, $output_format is the format you want your date in.
The first two fields are easy. Your date is in YY-MM-DD format:
$ date -f"%Y-%m-%d" "$my_date_string"
The question is what can you do for the final format. Fortunately, there is a format for the week in the year. %V which represents the weeks at 01-53 and %W which represents the weeks as 00-53.
What you need to do is find the date string on your file name, then convert that to the year and week number. If that's the same as the input, you need to concatenate this file.
find $dir -type f | while read transaction_file
do
file_date=${transaction_file#transaction-} #Removes the prefix
file_date=${file_date%.csv.bz2} #Removes the suffix
weekdate=$(date -j -f"%Y-%m-%d" "$file_date" +"%Y %W")
[ "$weekdate" -eq "$desired_date" ] || continue
...
done
For example, someone puts in 2013 05 as the desired date, you will go through all of your files and find ones with dates in the range you want. NOTE: That the week of the year is zero filled. You may need to zero fill the input of the week number to match.