在发送请求之前重写scrapy URL

Question

我正在使用scrapy来抓取一个多语言网站.对于每个对象,存在三种不同语言的版本.我正在使用搜索作为起点.不幸的是,搜索包含各种语言的URL,这在解析时会导致问题.

因此,我想在发送之前对URL进行预处理.如果它们包含特定字符串,我想替换URL的那一部分.

我的蜘蛛延伸了CrawlSpider.我查看了文档并找到了make_request_from _url(url)导致此尝试的方法:

def make_requests_from_url(self, url):                                                          
    """                                                                                         
    Override the original function go make sure only german URLs are                            
    being used. If french or italian URLs are detected, they're                                 
    rewritten.                                                                                  

    """                                                                                         
    if '/f/suche' in url:                                                                       
        self.log('French URL was rewritten: %s' % url)                                          
        url = url.replace('/f/suche/pages/', '/d/suche/seiten/')                                
    elif '/i/suche' in url:                                                                     
        self.log('Italian URL was rewritten: %s' % url)                                            
        url = url.replace('/i/suche/pagine/', '/d/suche/seiten/')                                  
    return super(MyMultilingualSpider, self).make_requests_from_url(url)                                                  

但由于某种原因,这不起作用.在请求URL之前重写URL的最佳方法是什么？也许通过规则回调？

Answer 1

可能没什么值得一个例子,因为它花了我大约30分钟来弄明白:

rules = [
    Rule(SgmlLinkExtractor(allow = (all_subdomains,)), callback='parse_item', process_links='process_links')
]

def process_links(self,links):
    for link in links:
        link.url = "something_to_prepend%ssomething_to_append" % link.url
    return links