Print the content of all zipped files with names containing a regex

	#!/bin/bash
	#
	# The below will print the content of all zipped files with names containing the keyword "application"
	# ./print_zip_files_content.sh app.jar '.*application.*'
	#
	if [ "$#" -ne 2 ]; then
	echo "Usage: $0 <zip file path> <regex containing filename>"
	exit 1
	fi
	ZIP_FILE="$1"
	REGEX="$2"
	# Using unzip -l to list files, then grep to filter by regex. awk to print the 4th column to the end.
	# This handles spaces in filenames more gracefully.
	unzip -l "$ZIP_FILE" | grep -P "$REGEX" | awk '{print $4}' | while read -r file; do
	if [[ ! -z "$file" ]]; then
	echo "File $file:"
	unzip -p "$ZIP_FILE" "$file"
	echo -e "\n"
	fi
	done

view raw print_zip_files_content.sh hosted with ❤ by GitHub

Securely sharing the content of your MAC OSX ~/.zshrc and other profile configuration files

cat ~/.zshrc | sed -E "s/(export[^=]*=).*/\1***/"