断断续续的循环取决于草率的响应结果

Question

我有一个爬虫蜘蛛，它在两个for循环中产生请求，是否有一种方法可以根据yield函数的结果使其中断for循环？ （即，在找到parse_results变量x时中断）

def parse(self,response):
    #code here
    offsets = [i for i in range(0,10001,20)]
    for query in all_queries:
        for offset in offsets:
            query = query+f'&offset={offset}'
            yield scrapy.Request(url= query,callback= self.parse_results)
            #break if parse_results gives KeyError (x is True)


def parse_results(self,response):
    try:
        x=response.xpath('//*[@id="alert_bar"]').extract_first()
        if x:
            raise KeyError # condition
    except KeyError:
        rows = response.xpath('//*[@id="sRes"]/div[@class="sResCont"]')
        for row in rows:
            if row.xpath('div[@class="adFrameCnt"]').extract_first():
                continue
            else:
                item = UserItem() # scrapy item
                item['username'] = row.xpath('div/div[@class="sResMain"]/b/a/text()').extract_first()
                item['link'] = response.urljoin(row.xpath('div/div[@class="sResMain"]/b/a/@href').extract_first())
                self.found_items.append(item)