列出 ZIP 文件的内容 [英] Listing the contents of a ZIP file
问题描述
按照技术食谱的一个例子,我设法列出了一个 ZIP 文件的内容(使用 7-Zip:
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)
然而,这只是将 ZIP 文件的整个目录结构转储到一个文本文件中(称为 listing.txt).
However, this just dumps out the entire directory structure of the ZIP file into a text file (called listing.txt).
我只想列出最高级别目录的目录名称,例如
I only want to list the directory names of the highest level directories e.g.
示例.Zip
7-Zip 9.20 Copyright (c) 1999-2010 Igor Pavlov 2010-11-18
Listing archive: C:UsersTestDesktop7zipDemo.zip
--
Path = C:UsersTestDesktop7zipDemo.zip
Type = zip
Physical Size = 1252
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2013-04-24 13:12:26 D.... 0 0 Directory ThreeSub Folder One
2013-04-24 13:13:00 D.... 0 0 Directory ThreeSub Folder Three
2013-04-24 13:12:54 D.... 0 0 Directory ThreeSub Folder Two
2013-04-24 13:12:26 D.... 0 0 Directory TwoSub Folder One
2013-04-24 13:13:00 D.... 0 0 Directory TwoSub Folder Three
2013-04-24 13:12:54 D.... 0 0 Directory TwoSub Folder Two
2013-04-24 13:12:26 D.... 0 0 Directory OneSub Folder One
2013-04-24 13:13:00 D.... 0 0 Directory OneSub Folder Three
2013-04-24 13:12:54 D.... 0 0 Directory OneSub Folder Two
------------------- ----- ------------ ------------ ------------------------
0 0 0 files, 9 folders
我只希望文本文件包含:
I would only want the text file to contain:
- 目录一
- 目录二
- 目录三
谁能建议我如何实现这一目标?
Can anyone suggest how I could achieve this?
推荐答案
7z 似乎没有为此提供内置密钥,但是您可以执行一些批处理脚本(这个在文件名中搜索斜杠并在斜杠时显示行未找到):
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!
)
)
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