Check if script is started by cron, rather than invoked manually

Question

Is there any variable that cron sets when it runs a program ? If the script is run by cron, I would like to skip some parts; otherwise invoke those parts.

How can I know if the Bash script is started by cron ?

Answer 1

I'm not aware that cron does anything to its environment by default that can be of use here, but there are a couple of things you could do to get the desired effect.

1) Make a hard or soft link to the script file, so that, for example, myscript and myscript_via_cron point to the same file. You can then test the value of $0 inside the script when you want to conditionally run or omit certain parts of the code. Put the appropriate name in your crontab, and you're set.

2) Add an option to the script, and set that option in the crontab invocation. For example, add an option -c, which tells the script to run or omit the appropriate parts of the code, and add -c to the command name in your crontab.

And of course, cron can set arbitrary environment variables, so you could just put a line like RUN_BY_CRON="TRUE" in your crontab, and check its value in your script.

Answer 2

Scripts run from cron are not run in interactive shells. Neither are startup scripts. The differentiation is that interactive shells have STDIN and STDOUT attached to a tty.

Method 1: check if $- includes the i flag. i is set for interactive shells.

case "$-" in
    *i*)
        interactive=1
        ;;
    *)
        not_interactive=1
        ;;
esac

Method 2: check is $PS1 is empty.

if [ -z "$PS1" ]; then
    not_interactive=1 
else
    interactive=1
fi

references:

• https://www.gnu.org/software/bash/manual/html_node/Special-Parameters.html
• https://www.gnu.org/software/bash/manual/html_node/Is-this-Shell-Interactive_003f.html

Method 3: test your tty. it's not as reliable, but for simple cron jobs you should be ok, as cron does not by default allocate a tty to a script.

if [ -t 0 ]; then
    interactive=1
else
    non_interactive=1
fi

Keep in mind that you can however force an interactive shell using -i, but you'd probably be aware if you were doing this...

Answer 3

First, get cron's PID, then get the current process's parent PID (PPID), and compare them:

CRONPID=$(ps ho %p -C cron)
PPID=$(ps ho %P -p $$)
if [ $CRONPID -eq $PPID ] ; then echo Cron is our parent. ; fi

If your script is started by another process that might have been started by cron, then you can walk your way back up the parent PIDs until you get to either $CRONPID or 1 (init's PID).

something like this, maybe (Untested-But-It-Might-Work<TM>):

PPID=$$   # start from current PID
CRON_IS_PARENT=0
CRONPID=$(ps ho %p -C cron)
while [ $CRON_IS_PARENT -ne 1 ] && [ $PPID -ne 1 ] ; do
  PPID=$(ps ho %P -p $PPID)
  [ $CRONPID -eq $PPID ] && CRON_IS_PARENT=1
done

From Deian: This is a version tested on RedHat Linux

# start from current PID
MYPID=$$
CRON_IS_PARENT=0
# this might return a list of multiple PIDs
CRONPIDS=$(ps ho %p -C crond)

CPID=$MYPID
while [ $CRON_IS_PARENT -ne 1 ] && [ $CPID -ne 1 ] ; do
        CPID_STR=$(ps ho %P -p $CPID)
        # the ParentPID came up as a string with leading spaces
        # this will convert it to int
        CPID=$(($CPID_STR))
        # now loop the CRON PIDs and compare them with the CPID
        for CRONPID in $CRONPIDS ; do
                [ $CRONPID -eq $CPID ] && CRON_IS_PARENT=1
                # we could leave earlier but it's okay like that too
        done
done

# now do whatever you want with the information
if [ "$CRON_IS_PARENT" == "1" ]; then
        CRON_CALL="Y"
else
        CRON_CALL="N"
fi

echo "CRON Call: ${CRON_CALL}"

Answer 4

If your script file is invoked by cron and it contains a shell in the first line like #!/bin/bash you need to find the parent-parent name for your purpose.

1) cron is invoked at the given time in your crontab, executing a shell 2) shell executes your script 3) your script is running

The parent PID is available in bash as variable $PPID. The ps command to get the parent PID of the parent PID is:

PPPID=`ps h -o ppid= $PPID`

but we need the name of the command, not the pid, so we call

P_COMMAND=`ps h -o %c $PPPID`

now we just need to test the result for "cron"

if [ "$P_COMMAND" == "cron" ]; then
  RUNNING_FROM_CRON=1
fi

Now you can test anywhere in your script

if [ "$RUNNING_FROM_CRON" == "1" ]; then
  ## do something when running from cron
else
  ## do something when running from shell
fi

Good luck!

Answer 5

There is no authoritative answer. Some other answers here actually try to match the cron environment, which is a hefty task.

Personally, I just set a variable in my crontab, e.g. CRON='in_cron' at the top. Then you can test it like this:

if [ "$CRON" != "in_cron" ]; then
  echo "This is not a cron job"
fi

This will authoritatively differentiate between interactive/nohup/ssh and cron, but it does require that variable declaration in the relevant user's crontab. For /etc/cron.d scripts, you should be able to put it in /etc/crontab (which works at least on my Debian 12 system).

---

The terminal ($TERM) variable is pretty decent here, as is the shell's "interactive" option flag (special parameter $- containing i). Some systems set TERM=dumb while most leave it empty, so we'll just check for either and also check for the interactive option flag:

if [ "${TERM:-dumb}$-" != "dumb${-#*i}" ]; then
  echo "This is not a cron job"
fi

The above code substitutes the word "dumb" when there is no value for $TERM. Therefore, the conditional fires when there is no $TERM or $TERM is set to "dumb" or if the $PS1 variable is not empty or if $- does not match itself when removing all characters up to the first i (this removes nothing when there is no i).

I've tested this on Debian 9, 11, & 12 (TERM=), CentOS 6.4 & 7.4 (TERM=dumb), and FreeBSD 7.3 && 11.2 (TERM=). This used to additionally include $PS1, but my Debian 12 system's cron now sets that variable somehow.

On at least Debian 12, this will differentiate between interactive/nohup and cron/ssh.

---

You can alternatively check the terminal name. Cron usually (verify this!) doesn't allocate a terminal, so you could run tty and the POSIX standard mandates that (if it really has no terminal) it must say not a tty in its output:

if [ "$(tty)" != "not a tty" ]; then
  echo "This is not a cron job"
fi

On at least Debian 12, this will differentiate between interactive and cron/ssh/nohup.

---

Another answer mentioned ( : > /dev/tty) 2>/dev/null to match interactivity.

On at least Debian 12, this will differentiate between interactive/nohup vs cron/ssh.

---

A fourth way to take advantage of cron not allocating a terminal would simply be to test for whether standard input is open. You can do this with ! [ -t 0 ], but this will give you an improper answer if you're piping data into the script.

Answer 6

A simple echo $TERM | mail [email protected] in cron showed me that on both Linux and AIX, cron seems to set $TERM to 'dumb'.

Now theoretically there may still be actual dumb terminals around, but I suspect that for most occasions, that should suffice...

Answer 7

Works on FreeBSD or on Linux:

if [ "Z$(ps o comm="" -p $(ps o ppid="" -p $$))" == "Zcron" -o \
     "Z$(ps o comm="" -p $(ps o ppid="" -p $(ps o ppid="" -p $$)))" == "Zcron" ]
then
    echo "Called from cron"
else
    echo "Not called from cron"
fi

You can go as far up the process tree as you wish.

Answer 8

This bash function should work on systems with either cron or crond. Tested under Debian bullseye.

## Returns 0 (success) if we are running under Cron
function undercron ()
{
    local cronpid=$(pgrep --uid=root --oldest --exact '^crond?$')
    for ((ppid=PPID; ppid > 1; ppid=$(ps ho %P -p $ppid))); do
        if ((ppid == cronpid)); then
            return 0
        fi
    done
    return 1
}

Answer 9

A generic solution to the question "is my output a terminal or am I running from a script" is:

( : > /dev/tty ) 2>/dev/null && dev_tty_good=y || dev_tty_good=n

Answer 10

There is a problem I do not see anybody mentioning. Most test fail when I run my script from a remote computer using ssh!

ssh [email protected] ./SmartHome.sh

Then (say) [ -t 0 ] and ( : > /dev/tty) and $TERM all indicate wrong (OK, they are all actually right, but are not indicating what we need).

So, the simplest universal solution is to call the script from the cron with an argument, as mentioned in the most upvoted answer under 2). Then, ANY other invocation will be seen as manual run. I use - (but can be anything) as the first argument to cron call and then check for it in my script:

[ "$1" == "-" ] && shift || Run_Manually=1

Answer 11

I've tested several OS (Ubuntu, CentOS, SUSE) and executed a script which includes printenv through SSH and local terminals and it seems these variables are never available if crond executes a script, but available in all other cases:

LESSOPEN
LESSCLOSE

So checking if this variable is not set, should work:

if [[ -z ${LESSOPEN+x} ]]; then
  echo "Script is executed by cron"
else
  echo "Script is executed manually"
fi