如何编写 python scrapy 代码以提取 url 在网站站点地图中的呈现

Question

我正在尝试使用此代码获取站点地图中的网址列表。当我运行这个时，我在屏幕上看不到任何结果。任何人都可以告诉我问题是什么或建议我更好的一个很好的例子。提前致谢

class MySpider(SitemapSpider):
name = "xyz"
allowed_domains = ["xyz.nl"]
sitemap_urls = ["http://www.xyz.nl/sitemap.xml"] 

def parse(self, response):
    print response.url
    return Request(response.url, callback=self.parse_sitemap_url)

def parse_sitemap_url(self, response):
    # do stuff with your sitemap links

Answer 1

此蜘蛛将从站点地图中获取所有 URL 并将它们保存到列表中。您可以轻松地将其更改为输出到文件或控制台。

# -*- coding: utf-8 -*-
import scrapy
from scrapy.spiders import SitemapSpider
from scrapy.spiders import Spider
from scrapy.http import Request, XmlResponse
from scrapy.utils.sitemap import Sitemap, sitemap_urls_from_robots
from scrapy.utils.gz import gunzip, is_gzipped
import re
import requests

class GetpagesfromsitemapSpider(SitemapSpider):
    name = "test"
    handle_httpstatus_list = [404]

    def parse(self, response):
       print response.url

    def _parse_sitemap(self, response):
        if response.url.endswith('/robots.txt'):
            for url in sitemap_urls_from_robots(response.body):
                yield Request(url, callback=self._parse_sitemap)
        else:
            body = self._get_sitemap_body(response)
            if body is None:
                self.logger.info('Ignoring invalid sitemap: %s', response.url)
                return

            s = Sitemap(body)
            sites = []
            if s.type == 'sitemapindex':
                for loc in iterloc(s, self.sitemap_alternate_links):
                    if any(x.search(loc) for x in self._follow):
                        yield Request(loc, callback=self._parse_sitemap)
            elif s.type == 'urlset':
                for loc in iterloc(s):
                    for r, c in self._cbs:
                        if r.search(loc):
                            sites.append(loc)
                            break
            print sites

    def __init__(self, spider=None, *a, **kw):
            super(GetpagesfromsitemapSpider, self).__init__(*a, **kw)
            self.spider = spider
            l = []
            url = "https://channelstore.roku.com"
            resp = requests.head(url + "/sitemap.xml")
            if (resp.status_code != 404):
                l.append(resp.url)
            else:
                resp = requests.head(url + "/robots.txt")
                if (resp.status_code == 200):
                    l.append(resp.url)
            self.sitemap_urls = l
            print self.sitemap_urls

def iterloc(it, alt=False):
    for d in it:
        yield d['loc']

        # Also consider alternate URLs (xhtml:link rel="alternate")
        if alt and 'alternate' in d:
            for l in d['alternate']:
                yield l

如何编写 python scrapy 代码以提取 url 在网站站点地图中的呈现

How to write python scrapy code for extracting url's present in sitemap of a site

python

scrapy

web-crawler

sitemap