Very easy question but I really can't find this when I Google. Sorry!
I'm trying to write a script that runs a user's command that he or she enters but I can't run the command that the user enters.
#!/bin/bash
echo -e "Enter a Command: "
read $COMMAND
echo "Output: $COMMAND" # I can't figure how to implement and print the command
Enter a Command: ls
Output: folder1 folder2 folder3 test.txt)
All you need is to delete the dollar sign from the read command
#!/bin/bash
echo "Enter a Command: "
read COMMAND
echo "Output: $COMMAND"
Happy scripting!, please don't forget to marked as answered ;)
Use eval to execute a command from a variable.
Also, you don't put $ before the variable name in the read command. That's only used when you want to get the variable's value.
#!/bin/bash
echo -e "Enter a Command: "
read COMMAND
echo Output:
eval "$COMMAND"
DEMO
Thanks! This answered my question:
#!/bin/bash
echo -e "Enter a Command: "
read COMMAND
echo Output:
eval "$COMMAND"
When I run my script, the .txt file is read, the executable commands are assigned to $eggs, then to execute the commands and redirect the output to a file I use echo $eggs>>eggsfile.txt but when I cat the file, I just see all the commands and not the execution output of those commands.
echo "Hi, $USER"
cd ~/mydata
echo -e "Please enter the name of commands file:/s"
read fname
if [ -z "$fname" ]
then
exit
fi
terminal=`tty`
exec < $fname #exec destroys current shell, opens up new shell process where FD0 (STDIN) input comes from file fname
count=1
while read line
do
echo $count.$line #count line numbers
count=`expr $count + 1`; eggs="${line#[[:digit:]]*}";
touch ~/mydata/eggsfile.txt; echo $eggs>>eggsfile.txt; echo "Reading eggsfile contents: $(cat eggsfile.txt)"
done
exec < $terminal
If you just want to execute the commands, and log the command name before each command, you can use 'sh -x'. You will get '+ command' before each command.
sh -x commands
+pwd
/home/user
+ date
Sat Apr 4 21:15:03 IDT 2020
If you want to build you own (custom formatting, etc), you will have to force execution of each command. Something like:
cd ~/mydata
count=0
while read line ; do
count=$((count+1))
echo "$count.$line"
eggs="${line#[[:digit:]]*}"
echo "$eggs" >> eggsfile.txt
# Execute the line.
($line) >> eggsfile.txt
done < $fname
Note that this approach uses local redirection for the while loop, avoiding having to revert the input back to the terminal.
Can anyone please explain what the following bash command does?
CMD_PATH=${0%/*}
What is the value assigned to the CMD_PATH variable?
It strips anything beyond last occurence of slash character from $0 variable, which is (in most cases, sometimes depending on how the script is run) the folder the script is currently executed from.
It shows the first directory on the working running process. If it is in a script, it shows its name.
From What exactly does "echo $0" return:
$0 is the name of the running process. If you use it inside a shell,
then it will return the name of the shell. If you use it inside a
script, it will be the name of the script.
Let's explain it:
$ echo $0
/bin/bash
is the same as
$ echo ${0}
/bin/bash
Then a bash substitution is done: get text up to last slash:
$ echo ${0%/*}
/bin
This substitution can be understood with this example:
$ a="hello my name is me"
$ echo ${a% *}
hello my name is
Returns the name of the directory from which the currently running script has been started.
To test it:
create directory /tmp/test:
mkdir /tmp/test
create file 't.sh` with such content:
#!/bin/bash
echo $0
echo ${0%/*}
give t.sh execution permission:
chmod +x /tmp/test/t.sh
execute it and you will see:
/tmp/test/s.sh
/tmp/test
Does crontab have an argument for creating cron jobs without using the editor (crontab -e)? If so, what would be the code to create a cron job from a Bash script?
You can add to the crontab as follows:
#write out current crontab
crontab -l > mycron
#echo new cron into cron file
echo "00 09 * * 1-5 echo hello" >> mycron
#install new cron file
crontab mycron
rm mycron
Cron line explaination
* * * * * "command to be executed"
- - - - -
| | | | |
| | | | ----- Day of week (0 - 7) (Sunday=0 or 7)
| | | ------- Month (1 - 12)
| | --------- Day of month (1 - 31)
| ----------- Hour (0 - 23)
------------- Minute (0 - 59)
Source nixCraft.
You may be able to do it on-the-fly
crontab -l | { cat; echo "0 0 0 0 0 some entry"; } | crontab -
crontab -l lists the current crontab jobs, cat prints it, echo prints the new command and crontab - adds all the printed stuff into the crontab file. You can see the effect by doing a new crontab -l.
This shorter one requires no temporary file, it is immune to multiple insertions, and it lets you change the schedule of an existing entry.
Say you have these:
croncmd="/home/me/myfunction myargs > /home/me/myfunction.log 2>&1"
cronjob="0 */15 * * * $croncmd"
To add it to the crontab, with no duplication:
( crontab -l | grep -v -F "$croncmd" ; echo "$cronjob" ) | crontab -
To remove it from the crontab whatever its current schedule:
( crontab -l | grep -v -F "$croncmd" ) | crontab -
Notes:
grep -F matches the string literally, as we do not want to interpret it as a regular expression
We also ignore the time scheduling and only look for the command. This way; the schedule can be changed without the risk of adding a new line to the crontab
Thanks everybody for your help. Piecing together what I found here and elsewhere I came up with this:
The Code
command="php $INSTALL/indefero/scripts/gitcron.php"
job="0 0 * * 0 $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
I couldn't figure out how to eliminate the need for the two variables without repeating myself.
command is obviously the command I want to schedule. job takes $command and adds the scheduling data. I needed both variables separately in the line of code that does the work.
Details
Credit to duckyflip, I use this little redirect thingy (<(*command*)) to turn the output of crontab -l into input for the fgrep command.
fgrep then filters out any matches of $command (-v option), case-insensitive (-i option).
Again, the little redirect thingy (<(*command*)) is used to turn the result back into input for the cat command.
The cat command also receives echo "$job" (self explanatory), again, through use of the redirect thingy (<(*command*)).
So the filtered output from crontab -l and the simple echo "$job", combined, are piped ('|') over to crontab - to finally be written.
And they all lived happily ever after!
In a nutshell:
This line of code filters out any cron jobs that match the command, then writes out the remaining cron jobs with the new one, effectively acting like an "add" or "update" function.
To use this, all you have to do is swap out the values for the command and job variables.
EDIT (fixed overwriting):
cat <(crontab -l) <(echo "1 2 3 4 5 scripty.sh") | crontab -
There have been a lot of good answers around the use of crontab, but no mention of a simpler method, such as using cron.
Using cron would take advantage of system files and directories located at /etc/crontab, /etc/cron.daily,weekly,hourly or /etc/cron.d/:
cat > /etc/cron.d/<job> << EOF
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root HOME=/
01 * * * * <user> <command>
EOF
In this above example, we created a file in /etc/cron.d/, provided the environment variables for the command to execute successfully, and provided the user for the command, and the command itself. This file should not be executable and the name should only contain alpha-numeric and hyphens (more details below).
To give a thorough answer though, let's look at the differences between crontab vs cron/crond:
crontab -- maintain tables for driving cron for individual users
For those who want to run the job in the context of their user on the system, using crontab may make perfect sense.
cron -- daemon to execute scheduled commands
For those who use configuration management or want to manage jobs for other users, in which case we should use cron.
A quick excerpt from the manpages gives you a few examples of what to and not to do:
/etc/crontab and the files in /etc/cron.d must be owned by root, and must not be group- or other-writable. In contrast to the spool area, the files under /etc/cron.d or the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly may also be symlinks, provided that both the symlink and the file it points to are owned by root. The files under /etc/cron.d do not need to be executable, while the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly do, as they are run by run-parts (see run-parts(8) for more information).
Source: http://manpages.ubuntu.com/manpages/trusty/man8/cron.8.html
Managing crons in this manner is easier and more scalable from a system perspective, but will not always be the best solution.
So, in Debian, Ubuntu, and many similar Debian based distros...
There is a cron task concatenation mechanism that takes a config file, bundles them up and adds them to your cron service running.
You can put a file under the /etc/cron.d/somefilename where somefilename is whatever you want.
sudo echo "0,15,30,45 * * * * ntpdate -u time.nist.gov" >> /etc/cron.d/vmclocksync
Let's disassemble this:
sudo - because you need elevated privileges to change cron configs under the /etc directory
echo - a vehicle to create output on std out. printf, cat... would work as well
" - use a doublequote at the beginning of your string, you're a professional
0,15,30,45 * * * * - the standard cron run schedule, this one runs every 15 minutes
ntpdate -u time.nist.gov - the actual command I want to run
" - because my first double quotes needs a buddy to close the line being output
>> - the double redirect appends instead of overwrites*
/etc/cron.d/vmclocksync - vmclocksync is the filename I've chosen, it goes in /etc/cron.d/
* if we used the > redirect, we could guarantee we only had one task entry. But, we would be at risk of blowing away any other rules in an existing file. You can decide for yourself if possible destruction with > is right or possible duplicates with >> are for you. Alternatively, you could do something convoluted or involved to check if the file name exists, if there is anything in it, and whether you are adding any kind of duplicate-- but, I have stuff to do and I can't do that for you right now.
For a nice quick and dirty creation/replacement of a crontab from with a BASH script, I used this notation:
crontab <<EOF
00 09 * * 1-5 echo hello
EOF
Chances are you are automating this, and you don't want a single job added twice.
In that case use:
__cron="1 2 3 4 5 /root/bin/backup.sh"
cat <(crontab -l) |grep -v "${__cron}" <(echo "${__cron}")
This only works if you're using BASH. I'm not aware of the correct DASH (sh) syntax.
Update: This doesn't work if the user doesn't have a crontab yet. A more reliable way would be:
(crontab -l ; echo "1 2 3 4 5 /root/bin/backup.sh") | sort - | uniq - | crontab -
Alternatively, if your distro supports it, you could also use a separate file:
echo "1 2 3 4 5 /root/bin/backup.sh" |sudo tee /etc/crond.d/backup
Found those in another SO question.
echo "0 * * * * docker system prune --force >/dev/null 2>&1" | sudo tee /etc/cron.daily/dockerprune
A variant which only edits crontab if the desired string is not found there:
CMD="/sbin/modprobe fcpci"
JOB="#reboot $CMD"
TMPC="mycron"
grep "$CMD" -q <(crontab -l) || (crontab -l>"$TMPC"; echo "$JOB">>"$TMPC"; crontab "$TMPC")
(2>/dev/null crontab -l ; echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
cat <(crontab -l 2>/dev/null) <(echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
#write out current crontab
crontab -l > mycron 2>/dev/null
#echo new cron into cron file
echo "0 3 * * * /usr/local/bin/certbot-auto renew" >> mycron
#install new cron file
crontab mycron
rm mycron
If you're using the Vixie Cron, e.g. on most Linux distributions, you can just put a file in /etc/cron.d with the individual cronjob.
This only works for root of course. If your system supports this you should see several examples in there. (Note the username included in the line, in the same syntax as the old /etc/crontab)
It's a sad misfeature in cron that there is no way to handle this as a regular user, and that so many cron implementations have no way at all to handle this.
My preferred solution to this would be this:
(crontab -l | grep . ; echo -e "0 4 * * * myscript\n") | crontab -
This will make sure you are handling the blank new line at the bottom correctly. To avoid issues with crontab you should usually end the crontab file with a blank new line. And the script above makes sure it first removes any blank lines with the "grep ." part, and then add in a new blank line at the end with the "\n" in the end of the script. This will also prevent getting a blank line above your new command if your existing crontab file ends with a blank line.
Bash script for adding cron job without the interactive editor.
Below code helps to add a cronjob using linux files.
#!/bin/bash
cron_path=/var/spool/cron/crontabs/root
#cron job to run every 10 min.
echo "*/10 * * * * command to be executed" >> $cron_path
#cron job to run every 1 hour.
echo "0 */1 * * * command to be executed" >> $cron_path
Here is a bash function for adding a command to crontab without duplication
function addtocrontab () {
local frequency=$1
local command=$2
local job="$frequency $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
}
addtocrontab "0 0 1 * *" "echo hello"
CRON="1 2 3 4 5 /root/bin/backup.sh"
cat < (crontab -l) |grep -v "${CRON}" < (echo "${CRON}")
add -w parameter to grep exact command, without -w parameter adding the cronjob "testing" cause deletion of cron job "testing123"
script function to add/remove cronjobs. no duplication entries :
cronjob_editor () {
# usage: cronjob_editor '<interval>' '<command>' <add|remove>
if [[ -z "$1" ]] ;then printf " no interval specified\n" ;fi
if [[ -z "$2" ]] ;then printf " no command specified\n" ;fi
if [[ -z "$3" ]] ;then printf " no action specified\n" ;fi
if [[ "$3" == add ]] ;then
# add cronjob, no duplication:
( crontab -l | grep -v -F -w "$2" ; echo "$1 $2" ) | crontab -
elif [[ "$3" == remove ]] ;then
# remove cronjob:
( crontab -l | grep -v -F -w "$2" ) | crontab -
fi
}
cronjob_editor "$1" "$2" "$3"
tested :
$ ./cronjob_editor.sh '*/10 * * * *' 'echo "this is a test" > export_file' add
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
No, there is no option in crontab to modify the cron files.
You have to: take the current cron file (crontab -l > newfile), change it and put the new file in place (crontab newfile).
If you are familiar with perl, you can use this module Config::Crontab.
LLP, Andrea
script function to add cronjobs. check duplicate entries,useable expressions * > "
cronjob_creator () {
# usage: cronjob_creator '<interval>' '<command>'
if [[ -z $1 ]] ;then
printf " no interval specified\n"
elif [[ -z $2 ]] ;then
printf " no command specified\n"
else
CRONIN="/tmp/cti_tmp"
crontab -l | grep -vw "$1 $2" > "$CRONIN"
echo "$1 $2" >> $CRONIN
crontab "$CRONIN"
rm $CRONIN
fi
}
tested :
$ ./cronjob_creator.sh '*/10 * * * *' 'echo "this is a test" > export_file'
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
source : my brain ;)
Say you're logged in as the user "ubuntu", but you want to add a job to a different user's crontab, like "john", for example. You can do the following:
(sudo crontab -l -u john; echo "* * * * * command") | awk '!x[$0]++' | sudo crontab -u john -
Source for most of this solution: https://www.baeldung.com/linux/create-crontab-script
I was having tons of issues trying to add a job to another user's crontab. It kept duplicating crontabs, or just flat-out deleting them. After some testing, though, I'm confident this line of code will append a new job to a specified user's crontab, non-destructively, including not creating a job that already exists.
I wanted to find an example like this, so maybe it helps:
COMMAND="/var/lib/postgresql/backup.sh"
CRON="0 0 * * *"
USER="postgres"
CRON_FILE="postgres-backup"
# At CRON times, the USER will run the COMMAND
echo "$CRON $USER $COMMAND" | sudo tee /etc/cron.d/$CRON_FILE
echo "Cron job created. Remove /etc/cron.d/$CRON_FILE to stop it."