Python 如何从不同的URL获取xpath，由start\u requests方法返回

这是我的密码：

import scrapy
from scrapy.spider import BaseSpider
from scrapy.selector import Selector
import MySQLdb


class AmazonSpider(BaseSpider):
    name = "amazon"
    allowed_domains = ["amazon.com"]
    start_urls = []

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

    def start_requests(self):
        conn = MySQLdb.connect(user='root',passwd='root',db='mydb',host='localhost')
        cursor = conn.cursor()
        cursor.execute(
            'SELECT url FROM products;'
            )
        rows = cursor.fetchall()
        for row in rows:
            yield self.make_requests_from_url(row[0])
        conn.close()

如何获取

start\u请求

函数返回的URL的xpath

---

注意：URL属于不同的域，不尽相同。

yield

使

start\u请求成为生成器。使用
for
循环获取从中返回的每个结果

像这样：

...
my_spider = AmazonSpider()
for my_url in my_spider.start_requests():
    print 'we get URL: %s' % str(my_url)
...