I scheduled some jobs in crontab. They are not even running. I also checked with logs along with schedules log file is not even created.
I am getting some EOF error. I think i made some changes in /etc/crontab
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin:/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin
MAILTO=root
HOME=/
# For details see man 4 crontabs
# Example of job definition:
# .---------------- minute (0 - 59)
# | .------------- hour (0 - 23)
# | | .---------- day of month (1 - 31)
# | | | .------- month (1 - 12) OR jan,feb,mar,apr ...
# | | | | .---- day of week (0 - 6) (Sunday=0 or 7) OR sun,mon,tue,wed,thu,fri,sat
# | | | | |
# * * * * * user-name command to be executed
my corntab
35 04 * * * cd /home/ec2-user/cerberus/src/automation_scripts/act/ && sh active.sh >> /home/ec2-user/cerberus/logs/active/`date "+%Y-%m-%d"` 2>&1
Related
On a Debian 10 machine, grep CRON /var/log/syslog logs CRON[455]: (root) CMD (/script.sh)
But crontab -e -u root is empty.
Where can I find the definition of that crontab job for the root user?
first, check if the cron job is defined for other users, then you can check the below file it may include some of the cron jobs
[root~]$ cat /etc/crontab
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root
HOME=/
# For details see man 4 crontabs
# Example of job definition:
# .---------------- minute (0 - 59)
# | .------------- hour (0 - 23)
# | | .---------- day of month (1 - 31)
# | | | .------- month (1 - 12) OR jan,feb,mar,apr ...
# | | | | .---- day of week (0 - 6) (Sunday=0 or 7) OR sun,mon,tue,wed,thu,fri,sat
# | | | | |
# * * * * * user-name command to be executed
I have a cronjob on my server which runs every hour all the days of the week
0 * * * *
I want to run it 09.00 - 21.00 and only from Monday to Friday
Refer this
* * * * * Command to be executed
- - - - -
| | | | |
| | | | +----- Day of week (0-7)
| | | +------- Month (1 - 12)
| | +--------- Day of month (1 - 31)
| +----------- Hour (0 - 23)
+------------- Min (0 - 59)
I want to run it 09.00 - 21.00 and only from Monday to Friday
Answer should be
* 9-21 * * 1-5
I have read the CronFormat to understand the Cron.But I am unable to understand this Cron Format:
According to my understand,the format is
<Minute> <Hour> <Day_of_the_Month> <Month_of_the_Year> <Day_of_the_Week>
But I am unable to understand the below format.
07/10 * 1/1 *?*
My Understanding:
My understanding of the above format is:
After Every 7 Minute,every hour and every month and every year.
Can anyone guide me what is it?
QuestionMark(?) and * I have not understood
This is format of each cron job
# Example of job definition:
# .---------------- minute (0 - 59)
# | .------------- hour (0 - 23)
# | | .---------- day of month (1 - 31)
# | | | .------- month (1 - 12) OR jan,feb,mar,apr ...
# | | | | .---- day of week (0 - 6) (Sunday=0 or 7) OR sun,mon,tue,wed,thu,fri,sat
# | | | | |
# * * * * * user-name command to be executed
so 7/10 means 7th minute of every 10 minute,
next * means every hour,
1/1 also means every so every day.
Unfortunately I am also not aware of "?".
I like this example its easy for me to understand :-
# Minute | Hour | Day of Month | Month | Day of Week | Command
# (0-59) | (0-23) | (1-31) | (1-12 or Jan-Dec) | (0-6 or Sun-Sat)
0 | 2 | 12 | * | * | /usr/bin/find
i have a crontab line and i want to ask some expert what this line will do
10,40 * * * * sh /etc/test/script.sh
please tell me what will 10,40 do in this crontab -e file.i am new to crontab use
* * * * * command
- - - - -
| | | | |
| | | | +----- day of week (0 - 6) (Sunday = 0)
| | | +------- month (1 - 12)
| | +--------- day (1 - 31)
| +----------- hour (0 - 23)
+------------- min (0 - 59)
each position takes a comma-separated list of values.
this will execute sh /etc/test/script.sh at :10 and :40 every hour.
00:10, 00:40, 01:10, 01:40, ...
additionally
you can use / to specify an interval, i.e.
*/5 * * * * sh /etc/test/script.sh
to run it every 5 minutes.
wikipedia cron page
I need to run bash script at 2nd Sat of the month at 11pm.I cant figure out its cronformat.
* * * * * *
| | | | | |
| | | | | +-- Year (range: 1900-3000)
| | | | +---- Day of the Week (range: 1-7, 1 standing for Monday)
| | | +------ Month of the Year (range: 1-12)
| | +-------- Day of the Month (range: 1-31)
| +---------- Hour (range: 0-23)
+------------ Minute (range: 0-59)
This is the cronformat i found from the internet but i am new to this and i think this problem is kind of tough.
Arg i waited so many days for the solution but not a single comment nor an answer. Anyway i think the answer is
0 23 8-14 * Sat
please correct me if i am wrong.
Run the cron every Saturday and only execute the script on the second Saturday:
* * * * 6 * test $(expr $(date +\%d) / 7) -eq 2 && <execute script here>