CODEWITHSUNDEEP
pythonhtmlregex
Andrew Stef
Andrew Stef

Reputation: 47

Python regular expression match in html file

I am trying to match inside an html file. This is the html:

<td>
<b>BBcode</b><br />
<textarea onclick='this.select();' style='width:300px;     height:200px;' />
[URL=http://someimage.com/LwraZS1]          [IMG]http://t1.someimage.com/LwraZS1.jpg[/IMG][    [/URL] [URL=http://someimage.com/CDnuiST]   [IMG]http://t1.someimage.com/CDnuiST.jpg[/IMG]   [/URL] [URL=http://someimage.com/Y0oZKPb][IMG]http://t1.someimage.com/Y0oZKPb.jpg[/IMG][/URL] [URL=http://someimage.com/W2RMAOR][IMG]http://t1.someimage.com/W2RMAOR.jpg[/IMG][/URL] [URL=http://someimage.com/5e5AYUz][IMG]http://t1.someimage.com/5e5AYUz.jpg[/IMG][/URL] [URL=http://someimage.com/EWDQErN][IMG]http://t1.someimage.com/EWDQErN.jpg[/IMG][/URL]
</textarea>
</td>

I want to extract all the BB code from [ to ] included.

And this is my code:

import re
x = open('/xxx/xxx/file.html', 'r').read
y = re.compile(r"""<td> <b>BBcode</b><br /><textarea onclick='this.select();' style='width:300px; height:200px;' />. (. *) </textarea> </td>""") 
z  = y.search(str(x())
print z

But when i run this i get None object... Where is the mistake?

Upvotes: 2

Views: 2846

Answers (3)

James Doepp - pihentagyu
James Doepp - pihentagyu

Reputation: 1308

I would use a parser for this:

from html import HTMLParser

class MyHtmlParser(HTMLParser):
    def __init__(self):
        self.reset()
        self.convert_charrefs = True
        self.dat = []
    def handle_data(self, d):
        self.dat.append(d.strip())
    def return_data(self):
        return self.dat
>>> with open('sample.html') as htmltext:
        htmldata = htmltext.read()
>>> parser = MyHtmlParser()
>>> parser.feed(htmldata)
>>> res = parser.return_data()
>>> res = [item for item in filter(None, res)]
>>> res[0]
'BBcode'
>>>

Upvotes: 0

C Panda
C Panda

Reputation: 3415

import re
x = open('/xxx/xxx/file.html', 'rt').read()
r1 = r'<textarea.*?>(.*?)</textarea>'
s1 = re.findall(r1, s, re.DOTALL)[1] # just by inspection
r2 = r'\[(.*?)\]'
s2 = re.findall(r2, s1)
for u in s2:
    print(u)

Upvotes: 1

coralvanda
coralvanda

Reputation: 6616

I think you need to add something like z.group() in order to pull out of the regex object, right? So, just changing your last line to

print z.group()

might do it.

Upvotes: 0

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