Regex back reference in gsub

ruby-talk
1

I am converting URLs in a text file to hyperlinks with the following regex. But the results only show up after the gsub is run a second time.

str = "A link to http://ruby-lang.org"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to<a href='' target='_blank'></a>"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to <a href='http://ruby-lang.org' target='_blank'>ruby-lang.org</a>"

Is there another way to do this?

2

Dan Fitzpatrick wrote:

I am converting URLs in a text file to hyperlinks with the following
regex. But the results only show up after the gsub is run a second time.

str = "A link to http://ruby-lang.org"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to<a href='' target='_blank'></a>"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to <a href='http://ruby-lang.org'
target='_blank'>ruby-lang.org</a>"

Is there another way to do this?

Hi,

Just a small change.

You need to use the block form of gsub, here, because the $n backrefs
are not set until the regex has completed. When using the non-block
form, you can normally use \1, \2 etc. instead of $1, $2 etc. but,
because you're using interpolation with #{...} inside your replacement
string, it gets a bit tricky doubling up the backslashes.

That's really not clear at all, sorry :slight_smile:

The reason why your example took two runs, is because the first run
sets the $n backrefs exactly the way you want them for the second run.

···

#--------------------------------------
str = "A link to http://ruby-lang.org"

res = str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i) do
    "#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>"
end

puts res #-> A link to <a href='http://ruby-lang.org' target='_blank'>ruby-lang.org</a>
#--------------------------------------

daz

3

--snip--

> #--------------------------------------
> str = "A link to http://ruby-lang.org"
>
> res = str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i) do
> "#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>"
> end
>
> puts res #-> A link to <a href='http://ruby-lang.org' target='_blank'>ruby-lang.org</a>
> #--------------------------------------

daz,

Thanks for the explanation. That works great.

Dan

4

Hi --

···

On Thu, 14 Jul 2005, daz wrote:

Dan Fitzpatrick wrote:

I am converting URLs in a text file to hyperlinks with the following
regex. But the results only show up after the gsub is run a second time.

str = "A link to http://ruby-lang.org"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to<a href='' target='_blank'></a>"
str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
"#{$1}<a href='#{$3}#{$5}#{$6}' target='_blank'>#{$5}#{$6}</a>")
#=> "A link to <a href='http://ruby-lang.org'
target='_blank'>ruby-lang.org</a>"

Is there another way to do this?

Hi,

Just a small change.

You need to use the block form of gsub, here, because the $n backrefs
are not set until the regex has completed. When using the non-block
form, you can normally use \1, \2 etc. instead of $1, $2 etc. but,
because you're using interpolation with #{...} inside your replacement
string, it gets a bit tricky doubling up the backslashes.

The #{...} usage isn't actually necessary, though. You could do:

   '\1<a href=\'\3\5\6\' target=\'_blank\'>\5\6</a>'

or

   %q{\1<a href='\3\5\6' target='_blank'>\5\6</a>')}

(to avoid the '-escaping).

David

--
David A. Black
dblack@wobblini.net

5

Dan Fitzpatrick wrote:

[...]

This is reworked with the backslash backrefs I mentioned:

res = str.gsub(/([\s|\>|\[|\(])((ftp:\/\/|http(s?):\/\/))([\w\.\?\/&=\-~:%]+)\b/i,
                '\1<a href="\3\5\6" target="_blank">\5\6</a>')

Thanks for the explanation. That works great.

Dan

Y'welcome,

daz