使用python+xpath 获取https://pypi.python.org/pypi/lxml/2.3/的下载链接

使用python+xpath 获取https://pypi.python.org/pypi/lxml/2.3/的下载链接：

使用requests获取html后，分析html中的标签发现所需要的链接在<table class="list" >...</table> 中

然后分别获却<tr class="odd"> 和<tr class="even">中的内容 ，使用xpath时可以写成xpath('//table[@class="list"]/tr[@class="even" or "odd"]/td/span/a[1]/@href')

 

import re

import requests

import urllib2

from lxml import etree

 

url='https://pypi.python.org/pypi/lxml/2.3/'

head={'User-Agent':'Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/31.0.1650.63 Safari/537.36'}

 

def gethtml(url, *args):

    html = requests.get(url, *args).content

    return html

 

def writfile(cont):

    try:

        fd = open('x.txt', 'w')

        try:

            fd.write(cont)

        finally:

            fd.close()

    except IOError:

        print "file not existing!"

 

def readfile():

    try:

        fd = open('x.txt', 'r')

        try:

            all_the_text = fd.read()

        finally:

            fd.close()

    except IOError:

        print "File open error !"

 

    return all_the_text

 

html = gethtml(url, head)

writfile(html)

all_text = readfile()

 

dom = etree.HTML(all_text)

url_list = dom.xpath('//table[@class="list"]/tr[@class="even" or "odd"]/td/span/a[1]/@href')

for url in url_list:

    print url