How to simulate the environment cron executes a script with?

Cron provides only this environment by default :

• HOME user's home directory
• LOGNAME user's login
• PATH=/usr/bin:/usr/sbin
• SHELL=/usr/bin/sh

If you need more you can source a script where you define your environment before the scheduling table in the crontab.

You can run:

env - your_command arguments

This will run your_command with empty environment.

Create a cron job that runs env and redirects stdout to a file. Use the file alongside "env -" to create the same environment as a cron job.

Don't forget that since cron's parent is init, it runs programs without a controlling terminal. You can simulate that with a tool like this:

http://libslack.org/daemon/

By default, cron executes its jobs using whatever your system's idea of sh is. This could be the actual Bourne shell or dash, ash, ksh or bash (or another one) symlinked to sh (and as a result running in POSIX mode).

The best thing to do is make sure your scripts have what they need and to assume nothing is provided for them. Therefore, you should use full directory specifications and set environment variables such as $PATH yourself.

The accepted answer does give a way to run a script with the environment cron would use. As others pointed out, this is not the only needed criteria for debugging cron jobs.

Indeed, cron also uses a non-interactive terminal, without an attached input, etc.

If that helps, I have written a script that enables painlessly running a command/script as it would be run by cron. Invoke it with your command/script as first argument and you're good.

This script is also hosted (and possibly updated) on Github.

#!/bin/bash
# Run as if it was called from cron, that is to say:
#  * with a modified environment
#  * with a specific shell, which may or may not be bash
#  * without an attached input terminal
#  * in a non-interactive shell

function usage(){
    echo "$0 - Run a script or a command as it would be in a cron job, then display its output"
    echo "Usage:"
    echo "   $0 [command | script]"
}

if [ "$1" == "-h" -o "$1" == "--help" ]; then
    usage
    exit 0
fi

if [ $(whoami) != "root" ]; then
    echo "Only root is supported at the moment"
    exit 1
fi

# This file should contain the cron environment.
cron_env="/root/cron-env"
if [ ! -f "$cron_env" ]; then
    echo "Unable to find $cron_env"
    echo "To generate it, run \"/usr/bin/env > /root/cron-env\" as a cron job"
    exit 0
fi

# It will be a nightmare to expand "$@" inside a shell -c argument.
# Let's rather generate a string where we manually expand-and-quote the arguments
env_string="/usr/bin/env -i "
for envi in $(cat "$cron_env"); do
   env_string="${env_string} $envi "
done

cmd_string=""
for arg in "$@"; do
    cmd_string="${cmd_string} \"${arg}\" "
done

# Which shell should we use?
the_shell=$(grep -E "^SHELL=" /root/cron-env | sed 's/SHELL=//')
echo "Running with $the_shell the following command: $cmd_string"


# Let's route the output in a file
# and do not provide any input (so that the command is executed without an attached terminal)
so=$(mktemp "/tmp/fakecron.out.XXXX")
se=$(mktemp "/tmp/fakecron.err.XXXX")
"$the_shell" -c "$env_string $cmd_string" >"$so" 2>"$se" < /dev/null

echo -e "Done. Here is \033[1mstdout\033[0m:"
cat "$so"
echo -e "Done. Here is \033[1mstderr\033[0m:"
cat "$se"
rm "$so" "$se"

In my case, cron was executing my script using sh, which fail to execute some bash syntax. In my script I added the env variable SHELL:

#!/bin/bash
SHELL=/bin/bash

I don't believe that there is; the only way I know to test a cron job is to set it up to run a minute or two in the future and then wait.