可能你想要这个：

from urllib import urlopen
import re 

pgno = 2
url = "http://www.eximguru.com/traderesources/pincode.aspx?&GridInfo=Pincode0%s" %str(pgno)
print url +'\n'
sock = urlopen(url)
htmlcode = sock.read()
sock.close()

x = re.search('%;"><a href="javascript:__doPostBack',htmlcode).start()

pat = ('\t\t\t\t<td style="width:\d+%;">(\d+)</td>'
       '<td style="width:\d+%;">(.+?)</td>'
       '<td style="width:\d+%;">(.+?)</td>'
       '<td style="width:30%;">(.+?)</td>\r\n')
regx = re.compile(pat)

print '\n'.join(map(repr,regx.findall(htmlcode,x)))

结果

http://www.eximguru.com/traderesources/pincode.aspx?&GridInfo=Pincode02

('110001', 'New Delhi', 'Delhi', 'Baroda House')
('110001', 'New Delhi', 'Delhi', 'Bengali Market')
('110001', 'New Delhi', 'Delhi', 'Bhagat Singh Market')
('110001', 'New Delhi', 'Delhi', 'Connaught Place')
('110001', 'New Delhi', 'Delhi', 'Constitution House')
('110001', 'New Delhi', 'Delhi', 'Election Commission')
('110001', 'New Delhi', 'Delhi', 'Janpath')
('110001', 'New Delhi', 'Delhi', 'Krishi Bhawan')
('110001', 'New Delhi', 'Delhi', 'Lady Harding Medical College')
('110001', 'New Delhi', 'Delhi', 'New Delhi Gpo')
('110001', 'New Delhi', 'Delhi', 'New Delhi Ho')
('110001', 'New Delhi', 'Delhi', 'North Avenue')
('110001', 'New Delhi', 'Delhi', 'Parliament House')
('110001', 'New Delhi', 'Delhi', 'Patiala House')
('110001', 'New Delhi', 'Delhi', 'Pragati Maidan')
('110001', 'New Delhi', 'Delhi', 'Rail Bhawan')
('110001', 'New Delhi', 'Delhi', 'Sansad Marg Hpo')
('110001', 'New Delhi', 'Delhi', 'Sansadiya Soudh')
('110001', 'New Delhi', 'Delhi', 'Secretariat North')
('110001', 'New Delhi', 'Delhi', 'Shastri Bhawan')
('110001', 'New Delhi', 'Delhi', 'Supreme Court')
('110002', 'New Delhi', 'Delhi', 'Rajghat Power House')
('110002', 'New Delhi', 'Delhi', 'Minto Road')
('110002', 'New Delhi', 'Delhi', 'Indraprastha Hpo')
('110002', 'New Delhi', 'Delhi', 'Darya Ganj')

我在研究后写了这段代码d使用以下代码的HTML源代码的结构（我想你会理解它而不再做任何解释）：

I wrote this code after having studied the structure of the HTML source code with the following code (I think you'll understand it without any more explanations):

from urllib2 import Request,urlopen
import re 

pgno = 2
url = "http://www.eximguru.com/traderesources/pincode.aspx?&GridInfo=Pincode0%s" %str(pgno)
print url +'\n'
sock = urlopen(url)
htmlcode = sock.read()
sock.close()

li = htmlcode.splitlines(True)

print '\n'.join(str(i) + ' ' + repr(line)+'\n' for i,line in enumerate(li) if 275<i<300)


ch = ''.join(li[0:291])
from collections import defaultdict
didi =defaultdict(int)
for c in ch:
    didi[c] += 1

print '\n\n'+repr(li[289])
print '\n'.join('%r -> %s' % (c,didi[c]) for c in li[289] if didi[c]<35)

，则li [289]中c的％s'％（c，didi [c]）

。

.

现在，问题是为pgno的所有值返回相同的HTML。该站点可能检测到它是一个想要连接和获取数据的程序。这个问题必须使用 urllib2 中的工具来处理，但我没有接受过这方面的培训。

Now, the problem is that the same HTML is returned for all the values of pgno. The site may detect it is a program that wants to connect and fetch data. This problem must be treated with the tools in urllib2, but I'm not trained to that.