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Suppose I have a zip archive file with many files in it. I can get a list of the files in the archive with:

Import["zipfile.zip"]

I can get data out of one of the files inside the archive with:

Import["zipfile.zip","somefile.txt"]

I can get the date the zip file was created with

FileDate["zipfile.zip"]

But I can't figure out how to get the dates of the files inside the zip file.

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    $\begingroup$ I can verify there is no way to do this currently without extracting the file first. This is a good suggestion, we'll look into this for future versions. $\endgroup$
    – GenericAccountName
    Apr 4, 2019 at 2:34

1 Answer 1

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I know you are probably looking for a pure Mathematica solution - looks like it's not supported by Mathematica (yet). However, don't forget that you can run any external program and analyze its output in Mathematica, so e.g. on macOS I can use unzip, which provides data in the following format:

17:06 /tmp $ unzip -l archive.zip 
Archive:  archive.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  04-03-2019 17:06   a.txt
        0  04-03-2019 17:06   b.txt
        0  04-03-2019 17:06   c.txt
---------                     -------
        0                     3 files

17:07 /tmp $

and therefore I can write this Mathematica function to do date processing for me:

zipDates[filename_] := 
 Interpreter["Date"]@
  StringSplit[
    StringSplit[
      RunProcess[{"unzip", "-l", filename}, "StandardOutput"], 
      "\n"][[4 ;; -3]],
    " "][[All, 3]]

result

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  • $\begingroup$ Thanks for the reminder. Using an external function is a great idea. I'm using windows so I'll give this a try with 7-Zip. $\endgroup$
    – Rudy Potter
    Apr 9, 2019 at 20:03

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