所以我在函数中有以下几行代码
sock = urllib.urlopen(url)
html = sock.read()
sock.close()
当我手动调用函数时它们工作正常。但是,当我在循环中调用该函数时(使用与之前相同的URL),我收到以下错误:
> Traceback (most recent call last):
File "./headlines.py", line 256, in <module>
main(argv[1:])
File "./headlines.py", line 37, in main
write_articles(headline, output_folder + "articles_" + term +"/")
File "./headlines.py", line 232, in write_articles
print get_blogs(headline, 5)
File "/Users/michaelnussbaum08/Documents/College/Sophmore_Year/Quarter_2/Innovation/Headlines/_code/get_content.py", line 41, in get_blogs
sock = urllib.urlopen(url)
File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib.py", line 87, in urlopen
return opener.open(url)
File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib.py", line 203, in open
return getattr(self, name)(url)
File "/System/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib.py", line 314, in open_http
if not host: raise IOError, ('http error', 'no host given')
IOError: [Errno http error] no host given
有什么想法吗?
修改更多代码:
def get_blogs(term, num_results):
search_term = term.replace(" ", "+")
print "search_term: " + search_term
url = 'http://blogsearch.google.com/blogsearch_feeds?hl=en&q='+search_term+'&ie=utf-8&num=10&output=rss'
print "url: " +url
#error occurs on line below
sock = urllib.urlopen(url)
html = sock.read()
sock.close()
def write_articles(headline, output_folder, num_articles=5):
#calls get_blogs
if not os.path.exists(output_folder):
os.makedirs(output_folder)
output_file = output_folder+headline.strip("\n")+".txt"
f = open(output_file, 'a')
articles = get_articles(headline, num_articles)
blogs = get_blogs(headline, num_articles)
#NEW FUNCTION
#the loop that calls write_articles
for term in trend_list:
if do_find_max == True:
fill_search_term(term, output_folder)
headlines = headline_process(term, output_folder, max_headlines, do_find_max)
for headline in headlines:
try:
write_articles(headline, output_folder + "articles_" + term +"/")
except UnicodeEncodeError:
pass
答案 0 :(得分:6)
当我使用网址连接变量时遇到此问题,在您的情况下search_term
url = 'http://blogsearch.google.com/blogsearch_feeds?hl=en&q='+search_term+'&ie=utf-8&num=10&output=rss'
最后有一个换行符。所以一定要确保
search_term = search_term.strip()
你可能也想做
search_term = urllib2.quote(search_term)
确保您的字符串对于网址是安全的
答案 1 :(得分:1)
在你的函数循环中,就在调用urlopen
之前,也许放一个print语句:
print(url)
sock = urllib.urlopen(url)
这样,当您运行脚本并获得IOError时,您将看到导致问题的url
。如果url
等于'http://'
...
答案 2 :(得分:1)
如果您不想自己处理每个块的读取,请使用urllib2。 这可能就是你所期望的。
import urllib2
req = urllib2.Request(url='http://stackoverflow.com/')
f = urllib2.urlopen(req)
print f.read()