Question

在浏览使用目录存储文件的基本文件夹系统网站时，

yield scrapy.Request(url1, callback=self.parse)

跟踪链接并抓取已爬网链接的所有内容，但我通常会遇到爬网程序通过根目录链接，并且当根目录进入时，它会获取具有不同URL的所有相同文件之间。

http://example.com/root/sub/file
http://example.com/root/sub/../sub/file

任何帮助都将不胜感激。

这是代码示例的片段

class fileSpider(Spider):
    name = 'filespider'
    def __init__(self, filename=None):
        if filename:
            with open(filename, 'r') as f:
                self.start_urls =  [url.strip() for url in f.readlines()]

    def parse(self, response):
        item = Item()
        for url in response.xpath('//a/@href').extract():
            url1 = response.url + url
            if(url1[-4::] in videoext):
                item['name'] = url
                item['url'] = url1
                item['depth'] = response.meta["depth"]
                yield item
            elif(url1[-1]=='/'):
                yield scrapy.Request(url1, callback=self.parse)   
        pass

Answer 1

您可以使用os.path.normpath来规范化所有路径，这样就不会出现重复：

import os
import urlparse
...

    def parse(self, response):
        item = Item()
        for url in response.xpath('//a/@href').extract():
            url1 = response.url + url

            # =======================
            url_parts = list(urlparse.urlparse(url1))
            url_parts[2] = os.path.normpath(url_parts[2])
            url1 = urlparse.urlunparse(url_parts)
            # =======================

            if(url1[-4::] in videoext):
                item['name'] = url
                item['url'] = url1
                item['depth'] = response.meta["depth"]
                yield item
            elif(url1[-1]=='/'):
                yield scrapy.Request(url1, callback=self.parse)   
        pass

如何在抓取文件类型网站时跳过父目录？

1 个答案: