关于python：使用SgmlLinkExtractor进行Scrapy

Scrapy using SgmlLinkExtractor

我正试图对窗体的页进行爬网http://www.wynk.in/music/song/variable_underlined_hydrometic_string.html.我想从笔记本电脑上点击这些URL，但是由于这些URL只在应用程序和waps上工作，所以我给了用户代理settings.py中的'mozilla/5.0(linux；u；android 2.3.4；fr fr；htc desire build/grj22)applewebkit/533.1(khtml，类似gecko)version/4.0 mobile safari/533.1'。我的代码文件读取

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from scrapy import Selector
from wynks.items import WynksItem

from scrapy.contrib.spiders import CrawlSpider, Rule

from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor

class MySpider(CrawlSpider):

name ="wynk"
#allowed_domains = ["wynk.in"]
start_urls = ["http://www.wynk.in/", ]
#start_urls = []
rules = (Rule(SgmlLinkExtractor(allow=[r'/music/song/\w+.html']), callback='parse_item', follow=True),)

def parse_item(self, response):
hxs = Selector(response)
if hxs:
tds = hxs.xpath("//div[@class='songDetails']//tr//td")
if tds:
for td in tds.xpath('.//div'):
titles = td.xpath("a/text()").extract()
if titles:
for title in titles:
print title

我通过运行来启动代码Scrapy Crawl Wynk-o abcd.csv-t csv

但是，我只得到这个结果爬行(200)http://www.wynk.in/>(参考：无)2015-03-23 11:06:04+0530[Wynk]信息：关闭蜘蛛(完成)我做错什么了？