How to list files in a zip without extra information in command line

Question

In my bash command line, when I use unzip -l test.zip I get the output like this:

Archive:  test.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
   810000  05-07-2014 15:09   file1.txt
   810000  05-07-2014 15:09   file2.txt
   810000  05-07-2014 15:09   file3.txt
---------                     -------
  2430000                     3 files

But I am interested only by the lines containing the file details.

I tried to make filtering using grep like this:

unzip -l test.zip | grep -v Length | grep -v "\-\-\-\-" | g -v Archive | grep -v " files"

But it is long and prone to error (e.g a file name Archive in this list will be dropped)

Is there any other options with unzip -l (I checked the unzip man page and did not find any) or another tool to do so?

It is important to me to not really unzip the archive but just to look what files are inside.

Answer 1

zipinfo -1 file.zip

Or:

unzip -Z1 file.zip

Would list only the files.

If you still want the extra info for each file names, you could do:

unzip -Zl file.zip | sed '1,2d;$d'

Or:

unzip -l file.zip | sed '1,3d;$d' | sed '$d'

Or (assuming GNU head):

unzip -l file.zip | tail -n +4 | head -n -2

Or you could use libarchive's bsdtar:

$ bsdtar tf test.zip
file1.txt
file2.txt
file3.txt

$ bsdtar tvf test.zip
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file1.txt
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file2.txt
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file3.txt

$ bsdtar tvvf test.zip
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file1.txt
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file2.txt
-rw-rw-r--  0 1000   1000   810000 Jul  5  2014 file3.txt
Archive Format: ZIP 2.0 (deflation),  Compression: none