Your improvements work great, thank you. And thank you for the very
detailed explanations!
On Tuesday, March 8, 2016 at 9:41:11 AM UTC+1, Michal Petrucha wrote:
>
> On Mon, Mar 07, 2016 at 05:44:08PM -0800, jorr...@gmail.com <javascript:>
> wrote:
> > I'm trying to replace *[URL]www.link.com[/URL]* with HTML with this
> regexp:
> >
> > topic.text = re.sub("(\[URL\])(.*)(\[\/URL\])", '<a href="$2">$2</a>',
> topic
> > .text, flags=re.I)
> >
> > But it's giving me the following problems:
> >
> > 1. The $2 capture group is only able to be repeated once, so I get
> > <a href="www.link.com">$2</a>
> > instead of
> > <a href="www.link.com">www.link.com</a>
>
> I have my doubts – if you use the standard Python re library, then the
> way to refer to captured groups is "\1", "\2", etc., not "$1". When I
> try the code you posted above, I get the following result (i.e., not
> even the first occurrence of "$2" gets substituted)::
>
> >>> re.sub("(\[URL\])(.*)(\[\/URL\])", '<a href="$2">$2</a>', '[URL]
> www.link.com[/URL]', flags=re.I)
> '<a href="$2">$2</a>'
>
> In order to make the substitution work for a single occurrence of
> [URL]...[/URL], you can use the following, which uses "\2" (Also, when
> writing regular expressions, or other strings that are supposed to
> contain the backslash character, it is a good idea to write them as
> raw string literals, i.e. prefix them with a "r", which I've done
> below; that way, Python won't try to interpret the backslashes as
> special characters – otherwise, "\2" would become a character with an
> ASCII value of 2)::
>
> >>> re.sub(r"(\[URL\])(.*)(\[\/URL\])", r'<a href="\2">\2</a>', '[URL]
> www.link.com[/URL]', flags=re.I)
> '<a href="www.link.com">www.link.com</a>'
>
> > 2. Only the first *[URL]* is matched. Everything after the first
> *[/URL]*
> > is simply deleted...
>
> The solution above gets you halfway there – re.sub will replace all
> matches by default, the problem here is that the "(.*)" part of your
> regex will matches everything between the first "[URL]", and the last
> "[/URL]"::
>
> >>> re.sub(r"(\[URL\])(.*)(\[\/URL\])", r'<a href="\2">\2</a>', '[URL]
> www.link1.com[/URL][URL]www.link2.com[/URL][URL]www.link3.com[/URL]',
> flags=re.I)
> '<a href="www.link1.com[/URL][URL]www.link2.com[/URL][URL]
> www.link3.com">www.link1.com[/URL][URL]www.link2.com[/URL][URL]
> www.link3.com</a>'
>
> The reason is that the asterisk operator in a regex is greedy, which
> means a ".*" will try to match as much as possible. When you use the
> non-greedy version of the operator (which you get by putting a
> question mark after the asterisk), you get the result you want::
>
> >>> re.sub(r"(\[URL\])(.*?)(\[\/URL\])", r'<a href="\2">\2</a>', '[URL]
> www.link1.com[/URL][URL]www.link2.com[/URL][URL]www.link3.com[/URL]',
> flags=re.I)
> '<a href="www.link1.com">www.link1.com</a><a href="www.link2.com">
> www.link2.com</a><a href="www.link3.com">www.link3.com</a>'
>
>
> You can read an explanation of the difference between greedy and
> non-greedy regular expressions in the Python docs:
> https://docs.python.org/2/howto/regex.html#greedy-versus-non-greedy
>
> Good luck,
>
> Michal
>
> >
> > I hope someone can help me with this. I'm using Python 2.7 if it makes a
> > difference.
> >
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