python:function just exactly 1 argument(2 given) [英] python: function takes exactly 1 argument (2 given)
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问题描述
我在类中有此方法
class CatList:
lista = codecs.open('googlecat.txt', 'r', encoding='utf-8').read()
soup = BeautifulSoup(lista)
# parse the list through BeautifulSoup
def parseList(tag):
if tag.name == 'ul':
return [parseList(item)
for item in tag.findAll('li', recursive=False)]
elif tag.name == 'li':
if tag.ul is None:
return tag.text
else:
return (tag.contents[0].string.strip(), parseList(tag.ul))
但是当我尝试调用它:
myCL = CatList()
myList = myCL.parseList(myCL.soup.ul)
我有以下错误:
parseList() takes exactly 1 argument (2 given)
作为方法的参数,但是当我这样做时,我得到的错误是以下:
I tried to add self as an argument to the method but when I do that the error I get is the following:
global name 'parseList' is not defined
我不清楚这是如何工作的。
not very clear to me how this actually works.
任何提示?
感谢
推荐答案
忘记了 self
参数。
您需要更改此行:
def parseList(tag):
:
def parseList(self, tag):
你也有一个全局名错误,因为你试图访问 parseList
没有 self
。
虽然你应该这样做:
You also got a global name error, since you're trying to access parseList
without self
.
While you should to do something like:
self.parseList(item)
。
具体来说,你需要在代码的两行中这样做:
To be specific, you need to do that in two lines of your code:
return [self.parseList(item)
和
return (tag.contents[0].string.strip(), self.parseList(tag.ul))
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