python - 如何使用 url fetch (python) 捕获所有可能的错误？

Answer 1

You can use except without specifying any type to catch all exceptions.

From the python docs http://docs.python.org/tutorial/errors.html:

import sys

try:
    f = open('myfile.txt')
    s = f.readline()
    i = int(s.strip())
except IOError as (errno, strerror):
    print "I/O error({0}): {1}".format(errno, strerror)
except ValueError:
    print "Could not convert data to an integer."
except:
    print "Unexpected error:", sys.exc_info()[0]
    raise

The last except will catch any exception that has not been caught before (i.e. a Exception which is not of IOError or ValueError.)

Answer 2

您可以使用顶级异常类型 Exception，它将捕获之前未捕获的任何异常。

http://docs.python.org/library/exceptions.html#exception-hierarchy

try:

    soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))
    title = str(soup.html.head.title.string)

    if title == "404 Not Found":
        self.redirect("/urlparseerror")
    elif title == "403 - Forbidden":
        self.redirect("/urlparseerror")     
    else:
        title = str(soup.html.head.title.string).lstrip("\r\n").rstrip("\r\n")

except UnicodeDecodeError:    
    self.redirect("/urlparseerror?error=UnicodeDecodeError")

except AttributeError:        
    self.redirect("/urlparseerror?error=AttributeError")

#https url:    
except IOError:        
    self.redirect("/urlparseerror?error=IOError")

except Exception, ex:
    print "Exception caught: %s" % ex.__class__.__name__