如何在不提取的情况下获取 zip 中所有文件的 md5sum

Question

Is there any way to get md5sum for all/anyone file inside zip without extracting the zip?

I can extract needed files using unzip <.zip>

But i need to get md5sum without extracting the zip.

Answer 1

This may not be exactly what you are looking for but it will get you closer. You wouldn't be extracting the entire zip but extracting a file to pipe it to md5sum to get checksum. Without reading the contents of the file, md5sum won't be able to generate a hash.

Let's say you have 3 files with this MD5:

b1946ac92492d2347c6235b4d2611184  a.txt
591785b794601e212b260e25925636fd  b.txt
6f5902ac237024bdd0c176cb93063dc4  c.txt

You zip them into a single file using zip final.zip a.txt b.txt c.txt

When you list the files, you see there are 3 files.

unzip -l final.zip
Archive:  final.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        6  2021-08-08 17:20   a.txt
        6  2021-08-08 17:20   b.txt
       12  2021-08-08 17:20   c.txt
---------                     -------
       24                     3 files

To get MD5 of each of the files without extracting the entire zip, you can do this:

unzip -p final.zip a.txt | md5sum
b1946ac92492d2347c6235b4d2611184  -

unzip -p final.zip b.txt | md5sum
591785b794601e212b260e25925636fd  -

unzip -p final.zip c.txt | md5sum
6f5902ac237024bdd0c176cb93063dc4  -

Alternative

You can do md5sum *.txt > checksums to get hash of all files and store them in a checksums file. Add that to the zip so you know the md5 of each of the file when the files were added to zipped.

Answer 2

For all files in a zip you can use this;

File='final.zip' ; unzip -lqq $File | while read L ; do unzip -p $File ${L##*[[:space:]]} | md5sum | sed "s/-/${L##*[[:space:]]}/" ; done

Gives;

b1946ac92492d2347c6235b4d2611184  a.txt
591785b794601e212b260e25925636fd  b.txt
6f5902ac237024bdd0c176cb93063dc4  c.txt