Question

Im与红gle安全管线和以下法典合作:

def getlist(self, type):
    dlurl = "safebrowsing.clients.google.com/safebrowsing/downloads?client=api&apikey=" + api_key + "&appver=1.0&pver=2.2"
    phish = "googpub-phish-shavar"
    mal = "goog-malware-shavar"
    self.type = type
    if self.type == "phish":
        req = urllib.urlopen(dlurl, phish )
        data = req.read()
        print(data)

Produc:

File "./test.py", line 39, in getlist
  req = urllib.urlopen(dlurl, phish )
File "/usr/lib/python2.6/urllib.py", line 88, in urlopen
return opener.open(url, data)
File "/usr/lib/python2.6/urllib.py", line 209, in open
return getattr(self, name)(url, data)
TypeError: open_file() takes exactly 2 arguments (3 given)

What am I doing wrong here? I cant spot where 3 arguments are being passed. BTW, I m calling this with

x = class()
x.getlist("phish")

Answer 1

基本上,你提供陶尔的一种方法,因此认为这是一种URL的档案,并试图把它作为做工的档案(在失败过程中提供混淆错误的信息)。

Try:

dlurl = "http://safebrowsing.clients.google.com/safebrowsing/downloads?client=api&apikey=" + api_key + "&appver=1.0&pver=2.2"

Answer 2

该功能打开了一个网络,这个网络的标注是URL,供阅读。如果URL没有办法识别码,则可以立案。

适当的开户人按第88条,到209处开放。

如果你看一下这一职能:

  def open_file(self, url):
        """Use local file or FTP depending on form of URL."""

答复:请提供一种办法,例如:http://...

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