I have a script that checks if the PPTP VPN is running, and if not it reconnects the PPTP VPN. When I run the script manually it executes fine, but when I make a cron job, it's not running.
* * * * * /bin/bash /var/scripts/vpn-check.sh
Here is the script:
#!/bin/sh
/bin/ping -c3 192.168.17.27 > /tmp/pingreport
result=`grep "0 received" /tmp/pingreport`
truncresult="`echo "$result" | sed 's/^\(.................................\).*$$'`"
if [[ $truncresult == "3 packets transmitted, 0 received" ]]; then
/usr/sbin/pppd call home
fi
finally i found a solution ... instead of entering the cronjob with
crontab -e
i needed to edit the crontab file directly
nano /etc/crontab
adding e.g. something like
*/5 * * * * root /bin/bash /var/scripts/vpn-check.sh
and its fine now!
Thank you all for your help ... hope my solution will help other people as well.
After a long time getting errors, I just did this:
SHELL=/bin/bash
PATH=/bin:/sbin:/usr/bin:/usr/sbin
* * * * * /bin/bash /home/joaovitordeon/Documentos/test.sh
Where test.sh contains:
#!/bin/bash
/usr/bin/python3 /home/joaovitordeon/Documentos/test.py;
In my case, the issue was that the script wasn't marked as executable. To make sure it is, run the following command:
chmod +x your_script.sh
If you're positive the script runs outside of cron, execute
printf "SHELL=$SHELL\nPATH=$PATH\n* * * * * /bin/bash /var/scripts/vpn-check.sh\n"
Do crontab -e for whichever crontab you're using and replace it with output of the above command. This should mirror most of your environment in case there is some missing path issue or something else. Also check logs for any errors it's getting.
Though it definitly looks like the script has an error or you messed something up when copying it here
sed: -e expression #1, char 44: unterminated `s' command
./bad.sh: 5: ./bad.sh: [[: not found
Simple alternate script
#!/bin/bash
if [[ $(ping -c3 192.168.17.27) == *"0 received"* ]]; then
/usr/sbin/pppd call home
fi
Your script can be corrected and simplified like this:
#!/bin/sh
log=/tmp/vpn-check.log
{ date; ping -c3 192.168.17.27; } > $log
if grep -q '0 received' $log; then
/usr/sbin/pppd call home >>$log 2>&1
fi
Through our discussion in comments we confirmed that the script itself works, but pppd doesn't, when running from cron. This is because something must be different in an interactive shell like your terminal window, and in cron. This kind of problem is very common by the way.
The first thing to do is try to remember what configuration is necessary for pppd. I don't use it so I don't know. Maybe you need to set some environment variables? In which case most probably you set something in a startup file, like .bashrc, which is usually not used in a non-interactive shell, and would explain why pppd doesn't work.
The second thing is to check the logs of pppd. If you cannot find the logs easily, look into its man page, and it's configuration files, and try to find the logs, or how to make it log. Based on the logs, you should be able to find what is missing when running in cron, and resolve your problem.
Was having a similar problem that was resolved when a sh was put before the command in crontab
This did not work :
#reboot ~myhome/mycommand >/tmp/logfile 2>&1
This did :
#reboot sh ~myhome/mycommand >/tmp/logfile 2>&1
my case:
crontab -e
then adding the line:
* * * * * ( cd /directory/of/script/ && /bin/sh /directory/of/script/scriptItself.sh )
in fact, if I added "root" as per the user, it thought "root" was a command, and it didn't work.
As a complement of other's answers, didn't you forget the username in your crontab script ?
Try this :
* * * * * root /bin/bash /var/scripts/vpn-check.sh
EDIT
Here is a patch of your code
#!/bin/sh
/bin/ping -c3 192.168.17.27 > /tmp/pingreport
result=`grep "0 received" /tmp/pingreport`
truncresult=`echo "$result" | /bin/sed 's/^\(.................................\).*$/\1/'`
if [[ $truncresult == "3 packets transmitted, 0 received" ]]; then
/usr/sbin/pppd call home
fi
In my case, it could be solved by using this:
* * * * * root ( cd /directory/of/script/ && /directory/of/script/scriptItself.sh )
I used some ./folder/-references in the script, which didn't work.
The problem statement is script is getting executed when run manually in the shell but when run through cron, it gives "java: command not found" error -
Please try below 2 options and it should fix the issue -
Ensure the script is executable .If it's not, execute below -
chmod a+x your_script_name.sh
The cron job doesn’t run with the same user with which you are executing the script manually - so it doesn't have access to the same $PATH variable as your user which means it can't locate the Java executable to execute the commands in the script. We should first fetch the value of PATH variable as below and then set it(export) in the script -
echo $PATH can be used to fetch the value of PATH variable.
and your script can be modified as below - Please see second line starting with export
#!/bin/sh
export PATH=<provide the value of echo $PATH>
/bin/ping -c3 192.168.17.27 > /tmp/pingreport
result=`grep "0 received" /tmp/pingreport`
truncresult="`echo "$result" | sed 's/^\(.................................\).*$$'`"
if [[ $truncresult == "3 packets transmitted, 0 received" ]]; then
/usr/sbin/pppd call home
fi
First of all, check if cron service is running. You know the first question of the IT helpdesk: "Is the PC plugged in?".
For me, this was happening because the cronjob was executing from /root directory but my shell script (a script to pull the latest code from GitHub and run the tests) were in a different directory. So, I had to edit my script to have a cd to my scripts folder. My debug steps were
Verified that my script run independent of cron job
Checked /var/log/cron to see if the cron jobs are running. Verified that the job is running at the intended time
Added an echo command to the script to log the start and end times to a file. Verified that those were getting logged but not the actual commands
Logged the result of pwd to the same file and saw that the commands were getting executed from /root
Tried adding a cd to my script directory as the first line in the script. When the cron job kicked off this time, the script got executed just like in step 1.
it was timezone in my case. I scheduled cron with my local time but server has different timezone and it does not run at all. so make sure your server has same time by date cmd
first run command env > env.tmp
then run cat env.tmp
copy PATH=.................. Full line and paste into crontab -e, line before your cronjobs.
try this
/home/your site folder name/public_html/gistfile1.sh
set cron path like above
(I'm new in shell script.)
I've been stuck with this issue for a while. I've tried different methods but without luck.
Description:
When my script attempt to run another script (SiebelMessageCreator.sh, which I don't own) it prompts:
-bash: ./SiebelMessageCreator.sh: No such file or directory
But the file exists and has execute permissions:
-rwxr-xr-x 1 owner ownergrp 322 Jun 11 2015 SiebelMessageCreator.sh
The code that is performing the script execution is:
(cd $ScriptPath; su -c './SiebelMessageCreator.sh' - owner; su -c 'nohup sh SiebelMessageSender.sh &' - owner;)
It's within a subshell because I first thought that it was throwing that message because my script was running in my home directory (When I run the script I'm root and I've moved to my non-root home directory to run the script because I can't move my script [ policies ] to the directory where the other script resides).
I've also tried with the sh SCRIPT.sh ./SCRIPT.sh. And changing the shebang from bash to ksh because the SiebelMessageCreator.sh has that shell.
The su -c 'sh SCRIPT.sh' - owner is necessary. If the script runs as root and not as owner it brokes something (?) (that's what my partners told me from their experience executing it as root). So I execute it as the owner.
Another thing that I've found in my research is that It can throw that message if it's a Symbolic link. I'm really not sure if the content of the script it's a symbolic link. Here it is:
#!/bin/ksh
BASEDIRROOT=/path/to/file/cpp-plwsutil-c
ore-runtime.jar (path changed on purpose for this question)
java -classpath $BASEDIRROOT com.hp.cpp.plwsutil.SiebelMessageCreator
exitCode=$?
echo "`date -u '+%Y-%m-%d %H:%M:%S %Z'` - Script execution finished with exit code $exitCode."
exit $exitCode
As you can see it's a very siple script that just call a .jar. But also I can't add it to my script [ policies ].
If I run the ./SiebelMessageCreator.sh manually it works just fine. But not with my script. I suppose that discards the x64 x32 bits issue that I've also found when I googled?
By the way, I'm automating some tasks, the ./SiebelMessageCreator.sh and nohup sh SiebelMessageSender.sh & are just the last steps.
Any ideas :( ?
did you try ?
. ./SiebelMessageCreator.sh
you can also perform which sh or which ksh, then modify the first line #!/bin/ksh
Here is what I'm entering in Terminal:
curl --silent https://raw.githubusercontent.com/githubUser/repoName/master/installer.sh | bash
The WordPress installing bash script contains a "read password" command that is supposed to wait for users to input their MySQL password. But, for some reason, that doesn't happen when I run it with the "curl githubURL | bash" command. When I download the script via wget and run it via "sh installer.sh", it works fine.
What could be the cause of this? Any help is appreciated!
If you want to run a script on a remote server without saving it locally, you can try this.
#!/bin/bash
RunThis=$(lynx -dump http://127.0.0.1/example.sh)
if [ $? = 0 ] ; then
bash -c "$RunThis"
else
echo "There was a problem downloading the script"
exit 1
fi
In order to test it, I wrote an example.sh:
#!/bin/bash
# File /var/www/example.sh
echo "Example read:"
read line
echo "You typed: $line"
When I run Script.sh, the output looks like this.
$ ./Script.sh
Example read:
Hello World!
You typed: Hello World!
Unless you absolutely trust the remote scripts, I would avoid doing this without examining it before executing.
It wouldn't stop for read:
As when you are piping in a way you are forking a child which has been given input from parent shell.
You cannot give the values back to parent(modify parent's env) from child.
and through out this process you are always in parent process.
I want to send an email when the system is going to shutdown to an email ID. I have CentOS 6.4. Below is my Script.
cat /ect/init.d/sendshtmail
#!/bin/bash
EMAIL="example#example.com"
SHUTDOWNSUBJECT="["`hostname`"] - System Shutdown"
SHUTDOWNBODY="This is an automated message to notify you that "`hostname`" is shutting down.
LOCKFILE=/var/lock/subsys/SystemEmail
echo "${SHUTDOWNBODY}" | mutt -s "${SHUTDOWNSUBJECT}" ${EMAIL}
It has the appropriate permission. While running it manually it's working perfectly. I have just symlinked it to /etc/rc0.d/ folder. By issuing below command.
ln -s /etc/init.d/sendshtmail /etc/rc0.d/K00sendshtmail
But the script is not sending any email during shutdown. Thanks in Advance.
Place your shell script in /etc/init.d with executable permission and symlink name should start with K##. If you want to execute your script at first place immediately after shut down then name it with K00scriptname. Script started will K will be executed first based on ascending order then script with S.
ln -s /etc/init.d/script /etc/rc0.d/K00scriptname
Shutdown command will send the stop signal to script, your script (K00scriptname) should have stop function like example
stop()
{
echo "executing scriptname"
"Your script logic"
}
case "$1" in
stop)
stop
;;
esac
Most important, K00scriptname will execute only if there would be lock file present in /var/lock/subsys folder, so do "touch /var/lock/subsys/scriptname" then check by doing shutdown.
Try to set executable permissions for your script. Sometimes you need to do that to activate it.
chmod 755 /etc/init.d/sendshtmail
Also try to use absolute paths for your command, while quoting the other variable as well.
echo "${SHUTDOWNBODY}" | /usr/bin/mutt -s "${SHUTDOWNSUBJECT}" "${EMAIL}"
Another attempt is to switch your user to your current user e.g.
echo "${SHUTDOWNBODY}" | su -l -c "/usr/bin/mutt -s \"${SHUTDOWNSUBJECT}\" \"${EMAIL}\"" yourusername
ln -s /etc/init.d/sendshtmail /etc/rc0.d/S01sendshtmail
The symlink name should begin with a S - for Start (K for Kill)
The two-digit specifies the order of execution for your script, the lowest numbered being execute first.
Lets assume the script i want to write ssh to 1.2.3.4 and then invokes
ls.
The problem is that when the line "ssh 1.2.3.4" is invoked, a password is
Required, hence, the flow is stopped, even when i fill the password,
The script wont continue.
How can i make the script continue after the password is given?
Thx!
You want to do public key authentication. Here are some resources which should get you going.
http://magicmonster.com/kb/net/ssh/auto_login.html
http://www.cs.rpi.edu/research/groups/vision/doc/auto/ssh/ssh_public_key_authentication.html
I would post a couple more links, but I don't have enough reputation points. ;) Just google on "SSH automated login" or "SSH public key authentication" if you need more help.
Actually you're trying to run ls locally but you have an ssh session opened. So it won't run ls until the session is opened. If you want to run ls remotely, you should use
ssh username#host COMMAND
Where command is the command you want to run. Ssh session will finish as soon as the command is invoked and you can capture its output normally.
I would suggest you to use RSA authentication method for script that needs ssh.
I just tried this script:
#!/bin/sh
ssh vps1 ls
mkdir temp
cd temp
echo test > file.txt
And it works. I can connect to my server and list my home. Then, locally, it creates temp dir, cd into it and then creates file.txt with 'test' inside.
write simple login bash script named login_to and give exec permissions (chmod 744 login_to)
#!/bin/bash
if [ $1 = 'srv1' ]; then
echo 'srv1-pass' | pbcopy
ssh root#11.11.11.11
fi
if [ $1 = 'foo' ]; then
echo 'barbaz' | pbcopy
ssh -t dux#22.22.22.22 'cd ~/somedir/someotherdir; bash'
fi
now use it like this
login_to srv1
login_to foo
When asked for password, just pate (ctrl+v or command+v) and you will be logged in.