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我有一个服务器，用于允许人们上传文件。不幸的是，我的脚本允许人们上传文件名中带有空格的文件。当生成指向这些文件的链接时（也由我的脚本生成），文件名将与这些空格一起输出。然后我的 Python 脚本（也附上）无法下载这些文件。我的问题显然是如何最好地解决这个问题。我不确定是否是：

A：使用某种正则表达式在上传时更改文件名，或者

B：如果有一种方法可以引用我服务器上的实际文件名（我什至不知道，因为当我在其中 ssh 时，看起来正确的文件路径实际上确实有空格）。

我对这个问题的最佳解决方案非常感兴趣。谢谢。

这是我的服务器代码：

var http = require('http'),
    url = require('url'),
    util = require('util'),
    path = require('path'),
    fs = require('fs'),
    qs = require('querystring');

var formidable = require('formidable'),
    mime = require('mime');


function dirTree(filename) {
    var stats = fs.lstatSync(filename),
        info = {
            name: path.basename(filename),
            path: ip + ':' + port + '/uploads/finished/' + path.basename(filename),
            type: mime.lookup(filename).substring(0, 5)
        };
    if (stats.isDirectory()) {
        info.type = "folder";
        info.children = fs.readdirSync(filename).map(function (child) {
            return dirTree(filename + '/' + child);
        });
    }
    return info;
}


http.createServer(function (request, response) {
    if (request.method.toLowerCase() == 'get') {
        var filePath = './content' + request.url;
        if (filePath == './content/') {
            filePath = './content/home.html';
        }
        if (filePath == './content/feed') {
            var a = dirTree('./content/uploads/finished');
            response.end(JSON.stringify(a));
        }
        var extname = path.extname(filePath);
        var contentType = mime.lookup(extname);
        fs.exists(filePath, function (exists) {
            if (exists) {
                fs.readFile(filePath, function (error, content) {
                    if (error) {
                        response.writeHead(500);
                        response.end();
                    }
                    else {
                        response.writeHead(200, {'Content-Type': contentType});
                        response.end(content, 'utf-8');
                    }
                })
            } else {
                response.writeHead(404);
                response.end();
            }
        });
    }

    var form = new formidable.IncomingForm;
    if (request.url == '/upload') {
        var oldPath,
            newPath,
            fileName;
        form.uploadDir = './content/uploads/temp/';
        form.keepExtensions = true;
        form.parse(request, function (err, fields, files) {
            type = files['upload']['type'];
            fileName = files['upload']['name'];
            oldPath = files['upload']['path'];
            newPath = './content/uploads/finished/' + fileName;
        });

        form.on('end', function () {
            fs.rename(oldPath, newPath, function (err) {
                if (err) {
                    response.end('There was an error with your request');
                    console.log('error')
                } else {
                    response.end('<h1>Thanks for uploading ' + fileName + '<h1>');
                }
            });
        });
    }
}).listen(port);

这是有问题的 Python 下载脚本。

import json
import urllib.request

url = '192.241.228.76:9090'
def test():
        response = urllib.request.urlopen('http://192.241.228.76:9090/feed')
        readd = response.read()
        data = json.loads(readd.decode(response.info().get_param('charset', 'utf8')))
        if "children" in data:
            for i in range(len(data['children'])):
                kind = data['children'][i]['type']
                url = 'http://' + data['children'][i]['path']
                name = 'media/' + data['children'][i]['name']
                if kind in ('video', 'image'):
                    try:
                        urllib.request.urlretrieve(url, name)
                    except:
                        try:
                            print('Spaces in ' + url)
                        except:
                            print('weird in file name')
test()
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1 回答 1

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应用程序应该能够处理空格。

在浏览器中，您将空格更改为%20，例如：

http://foo.bar/my%20cat.jpg

虽然 my%20cat 可能看起来不漂亮，但在这种情况下，您可以使用下划线 ( _) 替换空格：

filename.replace(new RegExp(find, ' '), '_');

对于 python 脚本，将其替换为urllib.request.urlretrive....：

urllib.request.urlretrieve(urllib.quote(url + '\' + name))

不要忘记添加import urllib.quote

于 2014-06-26T07:48:06.137 回答