Python 3.5 아름다운 스프 4 Error UserWarning : 파서가 명시 적으로 지정되지 않았습니다

Question

BeautifulSoup 4를 사용하려고했습니다. 성공적으로 설치 한 후에는 항상 약간의 오류가 있으며 "soup = BeautifulSoup (html)"

나는 다음 코드를 사용하십시오

: 그럼

//anaconda/lib/python3.5/site-packages/bs4/__init__.py:166: UserWarning: No parser was explicitly specified, so I'm using the best available HTML parser for this system ("lxml"). This usually isn't a problem, but if you run this code on another system, or in a different virtual environment, it may use a different parser and behave differently. 

To get rid of this warning, change this: 

BeautifulSoup([your markup]) 

to this: 

    BeautifulSoup([your markup], "lxml") 

    markup_type=markup_type)) 
Traceback (most recent call last): 

    File "<ipython-input-13-d4b16f497b1d>", line 1, in <module> 
runfile('/Users/beckswu/Desktop/coursera/using python access web data/class 2.py', wdir='/Users/beckswu/Desktop/coursera/using python access web data') 

    File "//anaconda/lib/python3.5/site-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 699, in runfile 
execfile(filename, namespace) 

    File "//anaconda/lib/python3.5/site-packages/spyderlib/widgets/externalshell/sitecustomize.py", line 88, in execfile 
exec(compile(open(filename, 'rb').read(), filename, 'exec'), namespace) 

    File "/Users/beckswu/Desktop/coursera/using python access web data/class 2.py", line 37, in <module> 
soup = BeautifulSoup(html) 

    File "//anaconda/lib/python3.5/site-packages/bs4/__init__.py", line 212, in __init__ 
markup, from_encoding, exclude_encodings=exclude_encodings)): 

    File "//anaconda/lib/python3.5/site-packages/bs4/builder/_lxml.py", line 108, in prepare_markup 
markup, try_encodings, is_html, exclude_encodings) 

TypeError: __init__() takes from 2 to 4 positional arguments but 5 were given 

내가

012에 코드를 변경 :

from bs4 import BeautifulSoup 
soup = BeautifulSoup(html) 

그것은 오류를 보여줍니다 3,516,

from bs4 import BeautifulSoup 
soup = BeautifulSoup(html,"lxml") 
markup_type=markup_type)) 

는 또한

markup_type=markup_type)) 
        ^
SyntaxError: invalid syntax 

가 어떻게 그 문제를 해결 할 수있는 오류를 보여줍니다? 나는 누구의 도움에 감사드립니다. 대신 HTML의

Answer 1

, 당신은 당신이`MARKUP_TYPE = markup_type`가하고있는 생각 거라고 이것이 :)