How to pass parameters to Python's HTMLParser init method

Question

I am learning python and have a problem passing a parameter. According to the documentation, the HTMLParser class is defines as follows:

class html.parser.HTMLParser(strict=False, *, convert_charrefs=False)

In my derived class I like to pass True in the convert_charrefs parameter. How can I do that? Here is my code so far

from html.parser import HTMLParser

class MyParser( HTMLParser ):
    def __init(self)__:
        super(MyParser, self).__init__( ??? )

So, what should i write instead of ??? to pass a True value in convert_charrfs?

Update: The obvious solution is to call __init__(convert_charrefs=True) I did try that, but it did not work for me because I was using python 3.3.4 instead of 3.4

def __init__(self) not def __init(self)__

Nicholas Flees
– Nicholas Flees

2014-04-11 13:11:05 +00:00
Commented Apr 11, 2014 at 13:11 — Nicholas Flees
– Nicholas Flees, Commented Apr 11, 2014 at 13:11

Selcuk · Accepted Answer · 2014-04-11 13:12:07Z

3

That would be

super(MyParser, self).__init__(convert_charrefs=True)

answered Apr 11, 2014 at 13:12

Selcuk

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2 Comments

ragnarius Over a year ago

I have tried that but I get the error message: TypeError: __init__() got an unexpected keyword argument 'convert_charrefs'

Selcuk Over a year ago

Are you sure you are using Python version 3.4 or higher? The convert_charrefs keyword was added in 3.4.

Hugh Bothwell · Accepted Answer · 2014-04-11 13:12:49Z

2

from html.parser import HTMLParser

class MyParser(HTMLParser):
    def __init__(self):
        super(MyParser, self).__init__(convert_charrefs=True)

answered Apr 11, 2014 at 13:12

Hugh Bothwell

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