Bash script scenarios to export variables set in a subshell using process substitution
Quick question that's been bugging me - I've been banging my head against this for hours. I tried several approaches but none seem to work. I'm working with an scenario with variable scope in my bash script when using process substitution. I have a scenario where I want to read from a file and set a variable based on the output of a command executed in a subshell. Here's a simplified version of the code I'm working with: ```bash #!/bin/bash # Trying to set a variable from a process substitution result=$(cat <(grep 'pattern' myfile.txt)) # Attempting to export the variable export output="$result" # Check the variable value echo "Output: $output" ``` When I run this script, the `output` variable is empty, and I don't see the expected result. I also tried using `set -x` to debug and found that the value of `result` is being set correctly, but the export seems to be failing silently. I also explored alternatives like using `read` to capture the output directly, but I still face the same scenario when trying to access `output` later in the script. My `myfile.txt` consists of multiple lines, and the `grep` command is supposed to filter some of them. I expected `output` to contain the matched lines but it remains unpopulated. Are there any nuances with variable scoping in bash that I might be missing? Any suggestions on how to properly export the variable in this context would be greatly appreciated! What's the best practice here? This is my first time working with Bash LTS. Cheers for any assistance! I'm using Bash 3.9 in this project. Thanks, I really appreciate it!