Question

我一直在寻找解决方案，但我发现的那些都不适合我。在花了两天时间调试之后，我应该向你们寻求帮助。

网址看起来很好。即使我在请求代码之前硬编码了一个url，回调函数仍然无效。

我的代码是：

    def parse_link(self, response):
            print 'lllll', response.url
            print 'bbbbb', len(response.body), response.body

    def parse(self, response):
            hxs = HtmlXPathSelector(response)
            issues = hxs.select('//a//@id').extract()
            for i in range(len(issues)):
                    issue = issues[i]
                    links_2d = hxs.select('//html//body//table[%d+%d]/tr/td//a[contains(text(),"full quotes")]/@href' % (9, i)).extract()
                    links_2d = list(set(links_2d))

                    if len(bb) < 1: continue
                    if len(links_2d) < 1: continue

                    full_link = links_2d[0]

                    yield scrapy.Request(url=full_link, callback = self.parse_link)

Answer 1

试试这个：

def parse(self, response):
        hxs = HtmlXPathSelector(response)
        issues = hxs.select('//a//@id').extract()
        for i in range(len(issues)):
                issue = issues[i]
                links_2d = hxs.select('//html//body//table[%d+%d]/tr/td//a[contains(text(),"full quotes")]/@href' % (9, i)).extract()
                links_2d = list(set(links_2d))

                if len(bb) < 1: continue
                if len(links_2d) < 1: continue

                full_link = links_2d[0]

                yield scrapy.Request(str(full_link), self.parse_link)

def parse_link(self, response):
        print 'lllll', response.url
        print 'bbbbb', len(response.body), response.body

没有输入scrapy回调函数

1 个答案: