CODEWITHSUNDEEP
batch-filezip7zip
JHarley1
JHarley1

Reputation: 2112

Listing the contents of a ZIP file

Following an example of tech-recipes I have managed to list the contents of a ZIP file (using 7-Zip:

FOR /F "tokens=* delims=" %%A in ('dir /b /s *.zip') do (7z.exe l -r "%%A" >> listing.txt)

However, this just dumps out the entire directory structure of the ZIP file into a text file (called listing.txt).

I only want to list the directory names of the highest level directories e.g.

Example.Zip 

7-Zip 9.20  Copyright (c) 1999-2010 Igor Pavlov  2010-11-18

Listing archive: C:\Users\Test\Desktop\7zip\Demo.zip

--
Path = C:\Users\Test\Desktop\7zip\Demo.zip
Type = zip
Physical Size = 1252

   Date      Time    Attr         Size   Compressed  Name
------------------- ----- ------------ ------------  ------------------------
2013-04-24 13:12:26 D....            0            0  Directory Three\Sub Folder One
2013-04-24 13:13:00 D....            0            0  Directory Three\Sub Folder Three
2013-04-24 13:12:54 D....            0            0  Directory Three\Sub Folder Two
2013-04-24 13:12:26 D....            0            0  Directory Two\Sub Folder One
2013-04-24 13:13:00 D....            0            0  Directory Two\Sub Folder Three
2013-04-24 13:12:54 D....            0            0  Directory Two\Sub Folder Two
2013-04-24 13:12:26 D....            0            0  Directory One\Sub Folder One
2013-04-24 13:13:00 D....            0            0  Directory One\Sub Folder Three
2013-04-24 13:12:54 D....            0            0  Directory One\Sub Folder Two
------------------- ----- ------------ ------------  ------------------------
                                     0            0  0 files, 9 folders

I would only want the text file to contain:

Can anyone suggest how I could achieve this?

Upvotes: 5

Views: 15144

Answers (2)

ElektroStudios
ElektroStudios

Reputation: 20474

7z.exe l -r "File.7z" | FINDSTR "[0-9].D....\>" | FIND /V "\"

With the first Findstr we filter the output getting a list of only dirs ("D" attribute).

With the second find we delete from the output the paths containing backslashes (Folder\Subfolder).

How to use it with a amount of zips:

@echo off

(For /R %%# in (*.zip) do (

    Echo File: "%%#" | MORE

    For /F "Tokens=5,*" %%A in (
        '7z.exe l -r "%%#" ^| FINDSTR "[0-9].D....\>" ^| FIND /V "\"'
    ) do (
        Set /A "Num+=1"
        Call Echo %%NUM%%.0: %%B
    )

    Set /A "Num=0"
    Echo+

))>"listing.txt"

Pause&Exit

Tested structure:

   Date      Time    Attr         Size   Compressed  Name
------------------- ----- ------------ ------------  ------------------------
2013-04-24 13:02:29 ....A          274          167  ..bat
2013-04-24 13:02:29 ....A          274               1\2\2.bat
2013-04-24 13:02:29 ....A          274               a\b\c\c.bat
2013-04-24 13:12:45 D....            0            0  a\b\c
2013-04-24 13:12:37 D....            0            0  a\b
2013-04-24 13:12:33 D....            0            0  a
2013-04-24 13:12:50 D....            0            0  1\2
2013-04-24 13:12:24 D....            0            0  1
------------------- ----- ------------ ------------  ------------------------
                                   822          167  3 files, 5 folders

Output received:

File: "C:\Users\Administrador\Desktop\Nueva carpeta\x.7z" 

1.0: a
2.0: 1

Upvotes: 0

Nickolay Olshevsky
Nickolay Olshevsky

Reputation: 14160

7z doesn't seem to have builtin key for this, however you can do some batch scripting (this one searches for slash in file name and displays line if slash not found) :

7z.exe l -r archive.zip > lines.txt

@echo off

setlocal ENABLEDELAYEDEXPANSION

for /f "tokens=*" %%a in (lines.txt) do (
  set line=%%a
  set srch=!line:\=!
  if "!line!" == "!srch!" (
     echo !line!
  )
)

Upvotes: 2

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