bash get one checksum for many files
# Basic syntax:
find /directory/to/search -type f -exec md5sum {} + | awk '{print $1}' | sort | md5sum > single_md5sum.txt
# Where:
# - find -type f returns all files (but not directories) in
# /directory/to/search and its subdirectories
# - -exec md5sum {} obtains the md5 checksums for all files returned
# with find
# - awk '{print $1}' | sort returns the sorted md5 checksums
# - and finally, md5sum is run on the sorted list of md5 checksums to
# return one md5 checksum for all the files
3.89
9
# Basic syntax:
md5sum /path/to/directory/* > list_of_checksums
# Note, md5sum will run on any files that are found with your search
# pattern. For example:
md5sum /path/*/*files_I_want* > list_of_checksums
# This would process all files than contain the substring "files_I_want"
# in all directories in the "path" directory
Thank you!
9
3.89
(9 Votes)
0
0
0
# Basic syntax:
md5sum /path/to/directory/* > list_of_checksums
# Note, md5sum will run on any files that are found with your search
# pattern. For example:
md5sum /path/*/*files_I_want* > list_of_checksums
# This would process all files than contain the substring "files_I_want"
# in all directories in the "path" directory
Thank you!
0
0
0
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