(I don't know if it is the right place. So if I am wrong, please point
me the right direction.
If this post is read by you masters, I'm honoured. If I am getting a
mere response, I'm blessed!)
Hi,
I'm a newbie regular expression user. I use regex in my Python
programs. I have a strange
(sometimes not strange, but please bear in mind; I'm a newbie ;)
problem using regex. That I want
a particular tag value of one of my HTML files.
ie: I want only the value after 'href=' in the tag >>
'<link href="mystylesh eet.css" rel="stylesheet " type="text/css">'
here it would be 'mystylesheet.c ss'. I used the following regex to get
this value(I dont know if it
is good).
_"<link\s+hr ef=["]?(.*?)["]?\s+rel=["]?stylesheet["]?\s+type=["]?text/css["]?>"_
I thought I was doing fine until I got stuck by this tag >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css" : same
tag but with 'href=' part
at a different place. I think you got the point!
So What should I do to get the exact value(here the value after
'href=') in any case even if the
tags are like these? >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css">
-OR-
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
-OR-
<link type="text/css" href="mystylesh eet.css" rel="stylesheet ">
5  1593 
Hey,
I'm new with regex's as well but here is my idea. Since you don't know
which attribute will come first why don't structure your regex like
this
(first off, I'll assume that \s == ' ', actually now that I think of
it, isn't \s any whitespace character? anyways \s == ' ' for now)
'<link\s*((\s*a ttribute1\s*)|( \s*attribute2\s *)|(\s*attribut e3\s*))+>'
I think that should just about do it.
Hope this helped,
Colin
John Blogger wrote:
(I don't know if it is the right place. So if I am wrong, please point
me the right direction.
If this post is read by you masters, I'm honoured. If I am getting a
mere response, I'm blessed!)
Hi,
I'm a newbie regular expression user. I use regex in my Python
programs. I have a strange
(sometimes not strange, but please bear in mind; I'm a newbie ;)
problem using regex. That I want
a particular tag value of one of my HTML files.
ie: I want only the value after 'href=' in the tag >>
'<link href="mystylesh eet.css" rel="stylesheet " type="text/css">'
here it would be 'mystylesheet.c ss'. I used the following regex to get
this value(I dont know if it
is good).
_"<link\s+hr ef=["]?(.*?)["]?\s+rel=["]?stylesheet["]?\s+type=["]?text/css["]?>"_
I thought I was doing fine until I got stuck by this tag >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css" : same
tag but with 'href=' part
at a different place. I think you got the point!
So What should I do to get the exact value(here the value after
'href=') in any case even if the
tags are like these? >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css">
-OR-
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
-OR-
<link type="text/css" href="mystylesh eet.css" rel="stylesheet ">
John Blogger wrote:
That I want a particular tag value of one of my HTML files.
ie: I want only the value after 'href=' in the tag >>
'<link href="mystylesh eet.css" rel="stylesheet " type="text/css">'
here it would be 'mystylesheet.c ss'. I used the following regex to get
this value(I dont know if it is good).
No matter how good it is you should still use something that
understands html:
>>from BeautifulSoup import BeautifulSoup
html='<link href="mystylesh eet.css" rel="stylesheet " type="text/css">'
page=Beautifu lSoup(html)
page.link.get ('href')
'mystylesheet.c ss'
--
- Justin
Justin Azoff wrote:
>from BeautifulSoup import BeautifulSoup
html='<link href="mystylesh eet.css" rel="stylesheet " type="text/css">'
page=Beautiful Soup(html)
page.link.get( 'href')
'mystylesheet.c ss'
On second thought, you will probably want something like
>>[link.get('href' ) for link in page.fetch('lin k',{'type':'tex t/css'})]
['mystylesheet.c ss']
which will properly handle multiple link tags.
--
- Justin
So What should I do to get the exact value(here the value after
'href=') in any case even if the
tags are like these? >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css">
-OR-
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
-OR-
<link type="text/css" href="mystylesh eet.css" rel="stylesheet ">
The following should do it:
expr = r'<link .*?href="(.*?)" '
or if single quotes might have been used:
expr = r'''<link .*?href=["'](.*?)['"]'''
But like the others have said, beautiful soup is very good for things
like this.
Pyparsing is also good for recognizing basic HTML tags and their
attributes, regardless of the order of the attributes.
-- Paul
testText = """sldkjflsa;fa j
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
here it would be 'mystylesheet.c ss'. I used the following regex to get
this value(I dont know if it
I thought I was doing fine until I got stuck by this tag >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css" : same
tag but with 'href=' part
tags are like these? >>
<link rel="stylesheet " href="mystylesh eet.css" type="text/css">
-OR-
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
-OR-
<link type="text/css" href="mystylesh eet.css" rel="stylesheet ">
"""
from pyparsing import makeHTMLTags,li ne
linkTag = makeHTMLTags("l ink")[0]
for toks,s,e in linkTag.scanStr ing(testText):
print toks.href
print line(s,testText )
print
Prints out:
mystylesheet.cs s
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
mystylesheet.cs s
<link rel="stylesheet " href="mystylesh eet.css" type="text/css" : same
mystylesheet.cs s
<link rel="stylesheet " href="mystylesh eet.css" type="text/css">
mystylesheet.cs s
<link href="mystylesh eet.css" rel="stylesheet " type="text/css">
mystylesheet.cs s
<link type="text/css" href="mystylesh eet.css" rel="stylesheet ">
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