Bash Code with while, list and crontab command - bash

Question

I have something like this :
all_scripts=()
for i in "${OS_USER[#]}"; do
list_script=$(crontab -u $i -l | grep -o '[^ ]*\.sh' | grep -o '^[^ ]*' | sort -u)
I want to create a list ( all_scripts) which contain all scripts from all users
For example
if in user 1 crontab there are script1.sh , script2.sh
in user 2 there are scrip3.sh, script4.sh, script5.sh
and in user 3 script6.sh
I want a list with all scripts : script1,sh, script2.sh, script3,...script6.sh

Answer 1

What you presented is incomplete. What you have is almost all correct.
You need to include the action ... to build the list ... within the loop.
You also need to initialize list_script as an array.
You probably want to suppress the "no crontab for ..." messaging.
You also need to be selective regarding which fields are "harvested" from the crontab, so ... ignore comments and ignore parameter definitions.
I recognize that the below may not correctly handle parameter definitions that may include spaces, but it is a good start for what you are looking for. So, the end result would look something like this:
#!/bin/bash
## You may have a specific list instead of this next line.
OS_USER=( $(cut -f1 -d\: /etc/passwd) )
all_scripts=()
list_script=()
for i in ${OS_USER[#]}
do
list_script=$(crontab -u $i -l | grep -v '^#' | awk '{
pos=index($0, $6) ;
$0=substr($0, pos) ;
print NF ;
for( i=1 ; i<=NF ; i++){
if( index($i, "=") == 0 ){
print $i ;
break ;
} ;
} ;
}' | grep -o '[^ ]*\.sh' | grep -o '^[^ ]*' | sort -u )
all_scripts=( ${allscripts[#]} ${list_script[#]} )
done
print ${allscripts[#]}