7z-extract 中文文档教程

Question

7z-extract

7zip 的包装器，它试图提供提取状态（7z 命令实际上并没有这样做）。

Prerequisites

7z 命令是必需的，需要位于 PATH 或应用程序的根目录中，即 package.json 所在的位置。

# OSX
$ brew update && brew doctor && brew install p7zip
# Ubuntu
$ sudo apt-get install p7zip-full

Install

npm install 7z-extract --save

API

const archive = new Zip(options);

• input String - The input file path
• output String - The output path (optional, if not supplied uses /tmp)
• createFolder Boolean - Creates a random folder name to extract to the files to. (default false)

archive.list(); //- returns a promise, the result of which is an object including array of files in the archive
archive.extract(progress); //- extracts the archive and returns a promise, progress function optionally returns current status as object (index, totalNumberOfFiles, percentage)

Usage

查看 CBZ 文件的信息。

'use strict';

const path = require('path'),
    Zip = require('../index'),
    inputFile = path.join(__dirname, 'archive.zip'),
    outputPath = path.join(__dirname), // optional, if not supplied uses /tmp
    progress = info => {
        console.log(info);
    };

const archive = new Zip({ input: inputFile, output: outputPath, createFolder: true });

  // archive.list()
  //  .then(res => { if (res){ console.log('Success!', res); } })
  //  .catch(err => { console.error(err); });

archive.extract(progress)
    .then(res => { console.log('Success!', res); })
    .catch(err => { console.error(err); });