scrapy 无法进行 Request() 回调 [英] scrapy unable to make Request() callback

问题描述

我正在尝试使用 Scrapy 制作递归解析脚本，但是 Request() 函数没有调用回调函数 suppose_to_parse()，也没有在回调值中提供任何函数.我尝试了不同的变体，但它们都不起作用.去哪里挖?

I am trying to make recursive parsing script with Scrapy, but Request() function doesn't call callback function suppose_to_parse(), nor any function provided in callback value. I tried different variations but none of them work. Where to dig ?

from scrapy.http import Request
from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector



class joomler(BaseSpider):
    name = "scrapy"
    allowed_domains = ["scrapy.org"]
    start_urls = ["http://blog.scrapy.org/"]


    def parse(self, response):
        print "Working... "+response.url
        hxs = HtmlXPathSelector(response)
        for link in hxs.select('//a/@href').extract():
            if not link.startswith('http://') and not link.startswith('#'):
               url=""
               url=(self.start_urls[0]+link).replace('//','/')
               print url
               yield Request(url, callback=self.suppose_to_parse)


    def suppose_to_parse(self, response):
        print "asdasd"
        print response.url

推荐答案

将 yield 移到 if 语句之外:

Move the yield outside of the if statement:

for link in hxs.select('//a/@href').extract():
    url = link
    if not link.startswith('http://') and not link.startswith('#'):
        url = (self.start_urls[0] + link).replace('//','/')

    print url
    yield Request(url, callback=self.suppose_to_parse)

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