Extract List of File Names in a Zip Archive When 'Unzip -L'

Extract list of file names in a zip archive when `unzip -l`

Assuming none of the files have spaces in names:

unzip -l filename.zip | awk '{print $NF}'

My unzip output has both a header and footer, so the awk script becomes:

unzip -l filename.zip | awk '/-----/ {p = ++p % 2; next} p {print $NF}'

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A version that handles filenames with spaces:

unzip -l filename.zip | awk '
    /----/ {p = ++p % 2; next}
    $NF == "Name" {pos = index($0,"Name")}
    p {print substr($0,pos)}
'

How can I list the files in a zip archive without decompressing it?

Perreal's answer is right, but I recommend installing atool (look for it in your distribution's package manager). Then, for any kind of archive file, bzip2, gzip, tar... you have just one command to remember :

als archive_name

How to extract last file added to an zip archive

Try this:

unzip /path/to/archive/dec2020.zip $(unzip -l /path/to/archive/dec2020.zip | sort -k2 | tail -5 | head -1 | awk '{ print $4 }')

$() expands the first command and uses the expansion in the second. We list the archive entries in date order and use tail, head and sort to get the line required. We then use awk to print out the file path (the 4th space delimited field)

Alternatively, this can be done with the awk system function to action the unzip.

 unzip -l /path/to/archive/dec2020.zip | sort -k2 | tail -5 | head -1 | awk '{ system("unzip /path/to/archive/dec2020.zip "$4) }'

List zip file entries with special characters like emojis

It doesn't seem possible to accurately list the files in a zip archive with unzip (tested with unzip 6.00); you'll have to select an other tool.

I chose perl in my answer because it has the required functionality in its core library. Here I used a newline as delimiter (-l) but you should replace it with a NULL-BYTE (-l0) if you want to be able to read and process the outputted paths 100% accurately from bash:

perl -l -e '
    use IO::Uncompress::Unzip;
    $zip = IO::Uncompress::Unzip->new($ARGV[0]);
    while($zip->nextStream()) {
        print $zip->getHeaderInfo()->{Name}
    }
' foo.zip

test/
test/.txt

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remark: Python also have a ZipFile module in its core library. I didn't post any Python solution because of the encoding issues of its stdout. The fixes aren't compatible between Python versions...

Extract certain files from .zip

Thanks to comment from @user20650.

Use two calls to unzip. First with list=TRUE just to get the $Name for the files. Second with files= to extract only the files whose names match the pattern.

  zipped_csv_names <- grep('\\.csv$', unzip('some_archive.zip', list=TRUE)$Name, 
                           ignore.case=TRUE, value=TRUE)
  unzip('some_archive.zip', files=zipped_csv_names)
  comb_tbl <- rbindlist(lapply(zipped_csv_names,  
                               function(x) cbind(fread(x, sep=',', header=TRUE,
                                                       stringsAsFactors=FALSE),
                                                 file_nm=x)), fill=TRUE ) 

Listing the contents of zip files within a tar file

Starting with tar, you can get it to write to stdout with the -O command line option.

Next to the unzip part of the problem. The standard Linux infozip versions of zip/unzip don't support reading input from stdin. The funzip command, which is part of infozip can read from stdin, reads from stdin but it only supports uncompression. It doesn't have an option to list the contents.

Luckily Java jar files are really just zip files with a well-defined structure and the jar command can read from stdin and list the contents of the input file

So, assuming you have a tar file called my.tar that only contains zip files, something like this should list the names of the files in the embedded zip files

tar xOf my.tar | jar -t

Unzip ZIP file and extract unknown folder name's content

I think your approach is a good one. Step 2 could be improved my extracting to a newly created directory (later deleted) so that "detection" is trivial.

# Bash (minimally tested)
tempdest=$(mktemp -d)
unzip -d "$tempdest" TEMPLATE1.ZIP
dir=("$tempdest"/*)
if (( ${#dir[@]} == 1 )) && [[ -d $dir ]]
    # in Bash, etc., scalar $var is the same as ${var[0]}
    mv "$dir"/* /mysite/user_uploaded_templates/myrandomname
else
    echo "rejected"
fi
rm -rf "$tempdest"

How to get a list of directories in a zip?

To list only directories:

unzip -l foo.zip "*/"

Output (e.g.):

Archive:  foo.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
        0  2015-09-10 20:10   work/
        0  2015-08-31 10:45   work/test1/
        0  2015-08-31 10:50   work/test1/cc/
        0  2015-08-31 10:45   work/test1/dd/
        0  2015-08-31 10:45   work/test1/aa/
        0  2015-08-31 10:45   work/test1/bb/
        0  2015-09-09 21:17   work/tmp/
        0  2015-08-23 18:49   work/tmp/work/
        0  2015-09-08 19:33   work/tmp/work/loop/
        0  2015-08-15 16:00   work/tmp/work/1/
        0  2015-08-15 16:00   work/1/
        0  2015-08-24 18:40   work/dir/
        0  2015-09-05 18:07   work/rename/
---------                     -------
        0                     13 files

or use

zipinfo -1 foo.zip "*/"

Output (e.g.):

work/
work/test1/
work/test1/cc/
work/test1/dd/
work/test1/aa/
work/test1/bb/
work/tmp/
work/tmp/work/
work/tmp/work/loop/
work/tmp/work/1/
work/1/
work/dir/
work/rename/

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