从多个7-zip文件中提取特定的文件扩展名 [英] Extract specific file extensions from multiple 7-zip files

问题描述

我有一个RAR文件和一个ZIP文件。在这两个文件夹中有一个文件夹。文件夹内有几个7-zip（.7z）文件。在每个7z内有多个具有相同扩展名的文件，但其名称有所不同。

I have a RAR file and a ZIP file. Within these two there is a folder. Inside the folder there are several 7-zip (.7z) files. Inside every 7z there are multiple files with the same extension, but whose names vary.

RAR or ZIP file
  |___folder
        |_____Multiple 7z
                  |_____Multiple files with same extension and different name

我只想从数千个文件中提取我需要的文件...
我需要那些名称中包含特定子字符串的文件。例如，如果压缩文件的名称中包含'[！]'或'（U）'或'（J）'是确定要提取文件的标准。

I want to extract just the ones I need from thousands of files... I need those files whose names include a certain substring. For example, if the name of a compressed file includes '[!]' in the name or '(U)' or '(J)' that's the criteria to determine the file to be extracted.

我可以提取文件夹没有问题，因此我具有以下结构：

I can extract the folder without problem so I have this structure:

folder
   |_____Multiple 7z
                |_____Multiple files with same extension and different name

我在Windows环境中，但是我安装了Cygwin。我想知道如何轻松提取我需要的文件？也许使用单个命令行。

I'm in a Windows environment but I have Cygwin installed. I wonder how can I extract the files I need painlessly? Maybe using a single command line line.

该问题有一些改进：

• 内部7z文件及其内的相应文件的名称中可以带有空格。


• 有7z文件中只有一个不符合指定条件的文件。因此，作为唯一可能的文件，它们也必须被提取。

谢谢大家。 bash解决方案是帮助我解决问题的一种方法。我无法测试Python3解决方案，因为尝试使用 pip 安装库时遇到了问题。我不使用Python，所以我必须研究并克服这些解决方案所面临的错误。目前，我已经找到了合适的答案。

Thanks to everyone. The bash solution was the one that helped me out. I wasn't able to test Python3 solutions because I had problems trying to install libraries using pip. I don't use Python so I'll have to study and overcome the errors I face with these solutions. For now, I've found a suitable answer. Thanks to everyone.

推荐答案

该解决方案基于bash，grep和awk，可在Cygwin和Ubuntu上使用。

This solution is based on bash, grep and awk, it works on Cygwin and on Ubuntu.

因为您需要先搜索（X）[！]。ext 文件，如果没有这样的文件，文件，然后查找（X）.ext 文件，我认为不可能编写一些单个表达式来处理这种逻辑。

Since you have the requirement to search for (X) [!].ext files first and if there are no such files then look for (X).ext files, I don't think it is possible to write some single expression to handle this logic.

该解决方案应该具有一些if / else条件逻辑，以测试存档中的文件列表并确定要提取的文件。

The solution should have some if/else conditional logic to test the list of files inside the archive and decide which files to extract.

这里是我在zip / rar存档中测试的脚本的初始结构（我制作了脚本以准备此结构）：

Here is the initial structure inside the zip/rar archive I tested my script on (I made a script to prepare this structure):

folder
├── 7z_1.7z
│   ├── (E).txt
│   ├── (J) [!].txt
│   ├── (J).txt
│   ├── (U) [!].txt
│   └── (U).txt
├── 7z_2.7z
│   ├── (J) [b1].txt
│   ├── (J) [b2].txt
│   ├── (J) [o1].txt
│   └── (J).txt
├── 7z_3.7z
│   ├── (E) [!].txt
│   ├── (J).txt
│   └── (U).txt
└── 7z 4.7z
    └── test.txt

输出是这样的：

output
├── 7z_1.7z           # This is a folder, not an archive
│   ├── (J) [!].txt   # Here we extracted only files with [!]
│   └── (U) [!].txt
├── 7z_2.7z
│   └── (J).txt       # Here there are no [!] files, so we extracted (J)
├── 7z_3.7z
│   └── (E) [!].txt   # We had here both [!] and (J), extracted only file with [!]
└── 7z 4.7z
    └── test.txt      # We had only one file here, extracted it

这是脚本进行提取：

#!/bin/bash

# Remove the output (if it's left from previous runs).
rm -r output
mkdir -p output

# Unzip the zip archive.
unzip data.zip -d output
# For rar use
#  unrar x data.rar output
# OR
#  7z x -ooutput data.rar

for archive in output/folder/*.7z
do
  # See https://stackoverflow.com/questions/7148604
  # Get the list of file names, remove the extra output of "7z l"
  list=$(7z l "$archive" | awk '
      /----/ {p = ++p % 2; next}
      $NF == "Name" {pos = index($0,"Name")}
      p {print substr($0,pos)}
  ')
  # Get the list of files with [!].
  extract_list=$(echo "$list" | grep "[!]")
  if [[ -z $extract_list ]]; then
    # If we don't have files with [!], then look for ([A-Z]) pattern
    # to get files with single letter in brackets.
    extract_list=$(echo "$list" | grep "([A-Z])\.")
  fi
  if [[ -z $extract_list ]]; then
    # If we only have one file - extract it.
    if [[ ${#list[@]} -eq 1 ]]; then
      extract_list=$list
    fi
  fi
  if [[ ! -z $extract_list ]]; then
    # If we have files to extract, then do the extraction.
    # Output path is output/7zip_archive_name/
    out_path=output/$(basename "$archive")
    mkdir -p "$out_path"
    echo "$extract_list" | xargs -I {} 7z x -o"$out_path" "$archive" {}
  fi
done

这里的基本思想是遍历7zip归档文件，并使用 7z l 命令获取每个文件的文件列表（文件列表）

The basic idea here is to go over 7zip archives and get the list of files for each of them using 7z l command (list of files).

该命令的输出非常冗长，因此我们使用 awk 清理并获取列表

The output of the command if quite verbose, so we use awk to clean it up and get the list of file names.

之后，我们使用 grep 过滤此列表以获取<$ c $的列表c> [！] 文件或（X）文件列表。
然后，我们将此列表传递给7zip，以提取所需的文件。

After that we filter this list using grep to get either a list of [!] files or a list of (X) files. Then we just pass this list to 7zip to extract the files we need.

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