小编use*_*287的帖子

我试图从发送POST请求获取HTML页面:

import httplib 
import urllib 
import urllib2 
from BeautifulSoup import BeautifulSoup


headers = {
    'Host': 'digitalvita.pitt.edu',
    'Connection': 'keep-alive',
    'Content-Length': '325', 
    'Origin': 'https://digitalvita.pitt.edu',
    'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_7_4) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.89 Safari/537.1',
    'Content-type': 'application/x-www-form-urlencoded; charset=UTF-8',
    'Accept': 'text/javascript, text/html, application/xml, text/xml, */*',
    'Referer': 'https://digitalvita.pitt.edu/index.php',
    'Accept-Encoding': 'gzip,deflate,sdch',
    'Accept-Language': 'en-US,en;q=0.8',
    'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
    'Cookie': 'PHPSESSID=lvetilatpgs9okgrntk1nvn595'
}

data = {
    'action': 'search',
    'xdata': '<search id="1"><context type="all" /><results><ordering>familyName</ordering><pagesize>100000</pagesize><page>1</page></results><terms><name>d</name><school>All</school></terms></search>',
    'request': 'search'
}

data = urllib.urlencode(data) 
print data 
req = urllib2.Request('https://digitalvita.pitt.edu/dispatcher.php', data, headers) 
response …

我无法使用urllib2打开一个特定的URL.同样的方法适用于其他网站,例如"http://www.google.com",但不适用于此网站(在浏览器中也可以正常显示).

我的简单代码:

from BeautifulSoup import BeautifulSoup
import urllib2

url="http://www.experts.scival.com/einstein/"
response=urllib2.urlopen(url)
html=response.read()
soup=BeautifulSoup(html)
print soup

任何人都可以帮助我使它工作吗？

这是我得到的错误:

Traceback (most recent call last):
  File "/Users/jontaotao/Documents/workspace/MedicalSchoolInfo/src/AlbertEinsteinCollegeOfMedicine_SciValExperts/getlink.py", line 12, in <module>
    response=urllib2.urlopen(url);
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
    response = meth(req, response)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
    'http', request, response, code, msg, hdrs)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 432, in error
    result = self._call_chain(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
    result = func(*args)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", …

作为主题，这是代码，没有错误消息，但没有插入数据。这是我的代码，谁能告诉我它有什么问题？

import psycopg2
import sys
import os
import glob 
import csv

#open the csv folder
dictfile='******'
os.chdir(dictfile)
total=[]
for file in glob.glob("*.csv"):
    total.append(file)
con = None
try: 
    con = psycopg2.connect(host='localhost',database='*****',user='postgres', password='*****') 
    cur = con.cursor()
    for i in range(0,1):   
        filename='/Users/Shared'+'/'+total[0]
        print filename
        #better move all files into shared folder
        x="copy public.crossref_sample from "+ "'"+filename+"'"+" DELIMITERS ',' CSV"
        print x
        cur.execute(x)  
except psycopg2.DatabaseError, e:
    print 'Error %s' % e    
    sys.exit(1) 
finally:
    if con:
        con.close()