Python 2.7, ValueError when dealing with HTMLParser

Question

First time working with the HTMLParser module. Trying to use standard string formatting on the ouput, but it's giving me an error. The following code:

import urllib2
from HTMLParser import HTMLParser

class LinksParser(HTMLParser):
    def __init__(self, url):
        HTMLParser.__init__(self)
        req = urllib2.urlopen(url)
        self.feed(req.read())

    def handle_starttag(self, tag, attrs):
        if tag != 'a': return
        for name, value in attrs:
        print("Found Link --> {]".format(value))


if __name__ == "__main__":
    LinksParser("http://www.facebook.com"

Produces the following error:

File "C:\Users\workspace\test\src\test.py", line 15, in handle_starttag  
print("Found Link --> {]".format(value))  
ValueError: unmatched '{' in format

Coding District · Accepted Answer · 2010-09-01 06:24:26Z

2

print("Found Link --> {]".format(value))

Should instead be:

print("Found Link --> {}".format(value))

You used a square bracket instead of a brace.

answered Sep 1, 2010 at 6:24

Coding District

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1 Comment

Stev0 Over a year ago

Wow, I feel silly. Good eye catching that. Lol, perhaps I should increase the font size of my editor. :) Working fine now.

Manoj Govindan · Accepted Answer · 2010-09-01 06:14:53Z

0

This format string looks broken: print("Found Link --> {]".format(value)). You need to change this to print("Found Link --> {key}".format(key = value)).

answered Sep 1, 2010 at 6:14

Manoj Govindan

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Comments

Andre Holzner · Accepted Answer · 2010-09-01 06:21:26Z

0

There are several problems

the print statement in handle_starttag should be indented
in the last line you're missing the closing parenthesis
in the print statement in handle_starttag you should use {0} instead of {]

answered Sep 1, 2010 at 6:21

Andre Holzner

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