[SOLVED] Cron doesn't accept bash syntax

Issue

I have a bashscript that I'm running with crontab. Unfortunately, a script that works fine when run manually fails with the error:

Syntax error: "(" unexpected (expecting "}")

Where the line in question is line 22 which is:

declare -a PREV_TOTAL=( $(for i in ${range[@]}; do echo 0; done) )

In the larger context:

TOTAL_CPU_USAGE=0
    TOTAL_CPU=$(grep -c ^processor /proc/cpuinfo) #set number of CPUs to check for
    declare -a 'range=({'"0..$TOTAL_CPU"'})'
    let "TOTAL_CPU=$TOTAL_CPU - 1"

    #declare array of size TOTAL_CPU to store values (eg. 8 cpus makes arrays of size 8)
    declare -a PREV_TOTAL=( $(for i in ${range[@]}; do echo 0; done) )
    declare -a PREV_IDLE=( $(for i in ${range[@]}; do echo 0; done) )

This works when manually just fine, but I don't understand what I'm doing wrong that causes cron to give this error? If you know I'd be very appreciative. Thanks.

EDIT: My crontab looks like this:

# m h  dom mon dow   command
SHELL=/bin/bash
@reboot cd /home/ubuntu/waste-cloud-computing/probe && probe.sh >> /var/log/somelogfile.log 2>&1

And I access it with sudo crontab -e. I'm still getting the issue while providing the SHELL variable.

EDIT 1: Thanks to some help I got past the syntax issues by ensuring the shell was using bash. Now I get the error, /bin/bash: probe.bash: command not found. I assume its some kind of PATH issue, but which bash returns /bin/bash so it seems normal to me. Maybe someone knows what's up?

Solution

Thanks to the help of the people here I found my issue was not syntax but rather the use of sh over bash. This was fixed by setting the crontab this way so future users can see:

# m h  dom mon dow   command
SHELL=/bin/bash
@reboot cd /home/ubuntu/waste-cloud-computing/probe && ./probe.sh >> /var/log/somelogfile.log 2>&1

The key points are the SHELL variable being set and the ./ before running the script.

Answered By - Ashton Spina