javascript - 通过 express 将 zip 解压缩到文件夹 Node 中

Question

我尝试找到可以发送 zip 的示例(例如通过 postman )并在我的处理程序中获取此 zip 并 解压缩 它如此指定的文件夹我没有找到太多使用 express 进行压缩的示例我想解压到路径 web/app

我尝试了以下对我不起作用的方法，zip 文件没有解压缩到指定的文件夹中，知道我做错了什么吗？

https://nodejs.org/api/zlib.html#zlib_zlib

var zlib = require('zlib');
var fs = require('fs');
const dir = path.join(__dirname, 'web/app/');

if (req.file.mimetype === 'application/zip') {

    var unzip = zlib.createUnzip();

    var read = fs.createReadStream(req.file);
    var write = fs.createWriteStream(dir);
    //Transform stream which is unzipping the zipped file
    read.pipe(unzip).pipe(write);   
    console.log("unZipped Successfully");

}

任何工作示例都会非常有帮助，或者引用我哪里有问题...

调试时我看到这是代码失败的时候

var read = fs.createReadStream(req.file);

知道为什么吗？

我也试过

var read = fs.createReadStream(req.file.body);

我没有看到错误、原因等的问题。

当我把它改成

var read = fs.createReadStream(req.file.buffer);

程序没有退出，我能够运行它，直到记录器 console.log("unZipped Successfully"); 但什么也没发生...

如果有 https://www.npmjs.com/package/yauzl 的任何示例yauzl 和 multer 在我的上下文中，它会很棒

更新-这是 postman 的请求

Answer 1

首先，zlib不支持解压zip文件。

我推荐formidable用于处理文件，因为

1. 经过实战考验
2. 使用最广泛的
3. 避免编写样板代码，例如从请求中读取文件流、存储和处理错误
4. 易于配置

先决条件
使用 npm i -S extract-zip formidable express 或 yarn add extract-zip formidable express

安装依赖项

使用 formidable 和 extract-zip

解决您的问题的最简单的解决方案

const express = require('express');
const fs = require('fs');
const extract = require('extract-zip')
const formidable = require('formidable');
const path = require('path');
const uploadDir = path.join(__dirname, '/uploads/');
const extractDir = path.join(__dirname, '/app/');
if (!fs.existsSync(uploadDir)) {
  fs.mkdirSync(uploadDir);
}
if (!fs.existsSync(extractDir)) {
  fs.mkdirSync(extractDir);
}

const server = express();

const uploadMedia = (req, res, next) => {
  const form = new formidable.IncomingForm();
  // file size limit 100MB. change according to your needs
  form.maxFileSize = 100 * 1024 * 1024;
  form.keepExtensions = true;
  form.multiples = true;
  form.uploadDir = uploadDir;

  // collect all form files and fileds and pass to its callback
  form.parse(req, (err, fields, files) => {
    // when form parsing fails throw error
    if (err) return res.status(500).json({ error: err });

    if (Object.keys(files).length === 0) return res.status(400).json({ message: "no files uploaded" });
    
    // Iterate all uploaded files and get their path, extension, final extraction path
    const filesInfo = Object.keys(files).map((key) => {
      const file = files[key];
      const filePath = file.path;
      const fileExt = path.extname(file.name);
      const fileName = path.basename(file.name, fileExt);
      const destDir = path.join(extractDir, fileName);

      return { filePath, fileExt, destDir };
    });

    // Check whether uploaded files are zip files
    const validFiles = filesInfo.every(({ fileExt }) => fileExt === '.zip');

    // if uploaded files are not zip files, return error
    if (!validFiles) return res.status(400).json({ message: "unsupported file type" });

    res.status(200).json({ uploaded: true });

    // iterate through each file path and extract them
    filesInfo.forEach(({filePath, destDir}) => {
      // create directory with timestamp to prevent overwrite same directory names
      extract(filePath, { dir: `${destDir}_${new Date().getTime()}` }, (err) => {
        if (err) console.error('extraction failed.');
      });
    });
  });

  // runs when new file detected in upload stream
  form.on('fileBegin', function (name, file) {
    // get the file base name `index.css.zip` => `index.html`
    const fileName = path.basename(file.name, path.extname(file.name));
    const fileExt = path.extname(file.name);
    // create files with timestamp to prevent overwrite same file names
    file.path = path.join(uploadDir, `${fileName}_${new Date().getTime()}${fileExt}`);
  });
}

server.post('/upload', uploadMedia);

server.listen(3000, (err) => {
  if (err) throw err;
});

此解决方案适用于单个/多个文件上传。此解决方案的一个问题是，尽管服务器抛出错误，但错误的文件类型将上传到 uploaded 目录。

用 postman 测试: