Ubuntu :: Write A Shell Script To Add Job To Cron?
Apr 1, 2011I write a shell script below to add a job to cron.
#!/bin/sh
touch date.cron
echo '*/3 * * * * /usr/sbin/ntpdate 192.168.2.3' >date.cron
[code]....
I write a shell script below to add a job to cron.
#!/bin/sh
touch date.cron
echo '*/3 * * * * /usr/sbin/ntpdate 192.168.2.3' >date.cron
[code]....
I want to write a shell script which will simultaneously collect OS user information and write in an individual text files.Can anyone tell me the syntax of the script.N.B. The user name will be mentioned in an array within the shell script.
View 8 Replies View RelatedI would like know when it is necessary or advisable to write a shell script instead of shell function ?
View 3 Replies View RelatedI have a script that I would like to run using 'cron'. I want to use 'scp' to transfer files from one machine to another. I have set up the SSH keys on both machines. When I run the script from bash terminal, it works flawlessly. But when I schedule a 'cron' job to run the same script, 'scp' does not transfer the files.
'Return value' is 0 when the script is run from bash directly. But when it runs from 'cron', the 'Return value' is 1. That means, surely, that 'scp' is throwing an error. I don't know which error is being encountered. Could anybody let me know how to make it work?
I'm creating a script all worked fine in the command line. But not work in the cron. Below you could see the script
[Code]...
So far I found when I use corn following part not working, nothing goes to the processedfiles file. ls -l /var/lct/mou2/processed | grep $TODAY | awk '{print " " $8}' > /home/trans/mou/processedfiles ls -l /var/lct/mou2/processed | grep $YESTERDAY | awk '{print " " $8}' >> /home/trans/mou/processedfiles This work perfect in command line. Corn job and command line use by the same user.
I have a cron job that runs a shell script. But it only runs the first line of that shell script and not the rest of the file. I'm a little stumped as to why. If I run the shell script manually, it runs and executes every single line as it should. I think I must need some additional syntax to make this run correctly?
Here is the crontab ...
Code:
root@kchlinux:~/macs# crontab -l
# m h dom mon dow command
[code]....
Setup: 10.04 server with "bash" as /bin/sh
When I run "ls -l" in a shell I get the following format:
Code:
-rw-r----- 1 syslog adm 0 2010-06-13 06:53 /var/log/user.log
Whereas if "ls -l" executes from a cron job the format is:
Code:
-rw-r----- 1 syslog adm 0 Jun 13 06:53 /var/log/user.log
Notice the different time format. Now I could fix this by changing the cron job to
Code:
ls -l --time-style=+%Y-%m-%d %H:%M ...
but I'm interested in knowing why this behavior occurs. What's different between the cron job and the shell?
My backup script works fine after some effort to get it where I wanted it. So the script is fine when I run it.
sudo bash backup.sh
now I have a sym-link to /etc/cron.daily but I do not see any new backups
I'm running my website on a linux server. I'm using a mac. Apparently, mac mail has a bug where it creates too many imap processes.
Since I'm on a shared server with a 25 process limit, with 3 email accts x 4+ processes each = a lot of problems!
I'm trying to write a cron job to delete imap processes every minute.
I've tried two things
1) following this blog's advice [URL]
Code:
killall imap
i realised it was imapd not imap, but i tried another method anyway
2) create a kill-imap.sh file in my server folder.
Code:
#!/bin/bash
/usr/bin/killall imapd
then executing the file from the cron command /kill-imap.sh
both these cases, it didn't seem to work!
I would get a daemon emai reply with
imapd: no process killed
just to give more info, the existing imap processes look like this
/usr/lib/courier-imap/bin/imapd /home/sitename/mail/sitename.com/emailusername
honestly, I'm not too experienced in this, and i've searched the net without luck. can anyone advise what i'm supposed to do to kill these imap processes every 5 min via cron job?
This is regarding the KDE application Basket:
I want to create a shell script that I can run as a cron job to do automated backups. how to proceed. I have the source code and have pulled out the backup source files. Here is the link for the source code. The download is at the bottom of this page. The backup files can be found in the src folder listed as backup.cpp and backup.h [URL]
If someone can do this, I believe it will make a nice addition to the Basket application for all.
Also I am not running Kubuntu I am running Karmic 9.10 64 bit
I've been having a bit of trouble running a shell script with cron. A friend of mine does a community radio show and the station has a live stream but no podcasts, so I've set up a script to record the stream and encode it as an mp3 while I'm away using mplayer and lame -that's what I'm trying to do anyway.
Here's the script, but it doesn't seem to run- at least, I don't see any of the files it should be outputing, would they be in the cron.weekly directory (where I have the script) or in my home directory?
#This is a script to record 'The Unnamed Show'
#it will record the show from the live stream, then convert the output #to an MP3
#Finally, it will delete any files no longer required HOME=/home/byron/
Im having some difficulties with running a shell script from cron which I am unable to resolve for almost 2 days now.
For testing purposes, im trying this with simple shell script test.sh with chmod 777
This script is located at /var
When I type /var/test.sh it runs perfectly and prints asdasdasd
When I type /var/test.sh > /home/log it writes asdasdasd to /home/log - works
The problem occurs whenever I add it as a cron job to var/spool/cron/crontab/root
There is: 11 10 * * * /var/test.sh > /home/log - however, at 10:11 there is no file at /home/log
Cron as a service works, forexample, every day at 4 am it makes this backup sshpass -p xxxx rsync -avz -e ssh root@x.x.10.7:/data/backup/ /home/backups/isp_admin
However,
doesnt makes the folder
Where is the catch?
I am using Shell Script to run my Java program. I have written a Cron job to invoke Shell Script every day at 7pm. Cron job is running every day at 7pm ,but it is not invoking my shellscript and also I am not getting any error message. I am able to run same shell script from cmd propmt using bash, and also I am able to invoke the same script from mainframes Universal command job. Same ShellScript and Same cronjob is working fine in my Dev server. But in my QA server it is not working.
View 7 Replies View RelatedI have an Ubuntu server running Couch Potato, Sick Beard and Sabnzbdplus. Everything "works" pretty well in a sense that CP and SB push the NZB's to Sabnzbdplus, but Sab crashes regularly (haven't found the solution or the cause for this problem, so if you have some advice regarding that, it's welcome).To counter this problem (Sab crashing) I have a script written which checks if Sab is runnning and if it isn't start it:
Code:
bart@Pyro:~$ cat CheckSabRunning.sh
#!/bin/sh
[code]....
What would happen if I place more than one shell script in the cron.weekly directory. would they run at the same time?
View 1 Replies View RelatedHow to write a cron jobs in linux to monitor my project diskspace usage?
check diskspce= du -h .
any clue for me as a beginner?
unable to write a shell script to vi
View 1 Replies View Relatedhow to write own shell in linux?i want to know procedure of shell programming.
View 5 Replies View RelatedI'm trying to write a toy linux shell. For starters this is what I'm trying to do:
1. Start a new process with fork.
2. Execute a program in the new process with execl().
3. Redirect the output from the new process from STDOUT to another file descriptor, using dup2(2).
4. In the parent process, read the output from the child process and write it to the screen.
Creating a new process and executing a program in it is no problems, the problem is that I can't seem to capture the output from it in the parent process.
Code:
//digenv
//C++arl 2011-03-24
#include <stdio.h>
#include <sys/types.h>
[code]...
I have a question, tried to search on the Internet but it is hopeless. I want to write a shell script(bashShell) that will run commands of configuration for vim editor.For example: in the script, it will run ":let", ":set", ":highlight" to configure for vim editor. In addition to, when I searched a pattern and wrote it to file,ed vi to open it automatically. But, I couldn't highlight a word(that is the pattern I'm searching) in vim automatically
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View 8 Replies View RelatedI am new and start learning how to write korn shell, can you someone please translate the below command to common English? code...
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View 4 Replies View RelatedI wish to create my own shell for my operating systems project in college...
My professor has asked me to make sure that my shell can execute at least 30 or 40 commands... I have around a month's time
I have seen endless source codes in the net, I'm not able to understand any of it
How do i get about doing it?
I'm trying to write a simple shell script, its purpose is not important. The script needs to make use of the system $HOSTNAME environment variable. I had a look at this page which provides the following example.
Code:
#!/bin/sh
echo "You are user $UID on $HOSTNAME"
echo "Your home directory is: $HOME"
echo "$HOSTNAME is running $OSTYPE"
[Code]...
I have a number of text files throughout my /home/pjs/Documents directory tree that have execute permissions set. Almost all of my file names have spaces in them. I am trying to write a shell script that will look at each file in my Documents directory, find the ones that have execute permissions set, and run the command chmod 644. Of course, I don't want the command run on the directories.
The following script *doesn't work*, but might serve to illustrate what I am trying to do:
#!/bin/bash
for x in "$(ls -R)" do
if [ -f "$x" ] && [ -x $x ]; then
chmod 644 "$x"
fi
done
I want each file and directory name to be placed, one by one, in the variable $x, and then tested with the "if" conditionals.
The first problem seems to be that, although the command "ls -R" does produce a complete list of the files and directories I need, they are not placed, one by one, in the variable x like I want them to be.
Also, I think I should use the shift command so that the option -R doesn't get included as one of the values of the variable $x, but I can't figure out where to put it.
I need a help regarding writing a function in a shell, what exactly a function does!!
View 3 Replies View RelatedI want to write a shell script contains python one. So,the result of python one is flv file. I want the path of this is copied into a enviroment variable that i have to pass as a flag argument of another program (to convert into mp3). To individuate the result of python script I thought to use (in PWD)
Code:
ls | grep -E '^.*mp3$'
But my question is: How can I copy this result into enviroment variable?
How do i write a shell script to ssh into a remote server with password and run a script in it?
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