Question

我在wing101和python 3.3.2

中写了以下内容

def open_page(url):
    """Open the URL string given and return its contents as file."""
    page = None
    increment_num_calls()
    cont = False
    while not cont:
        try:
            page = urllib2.urlopen(url)
            cont = True
        except urllib2.URLError as e:
            print "Warning: Url load error " + str(e) + " for url " + url
            #if not hasattr(e, "code"):
            #    raise
            if hasattr(e, "code") and e.code == 401:
                return None
            time.sleep(TIME_DELAY_PAGE_RETRY)
            return None
        except httplib.BadStatusLine:
            return None
    return page

任何人都可以帮我弄清楚错误所说的错误

print "Warning: Url load error " + str(e) + " for url " + url

如果你想看到更多的节目让我知道，我可以发布。

Answer 1

打印在Python 3中的工作方式不同

尝试更改

print "Warning: Url load error " + str(e) + " for url " + url

到

print('Warning: Url load error {} for url {}'.format(e, url))

urllib2.urlopen - 错误打开网址

1 个答案: