in reply to Substitution Problem

Instead of 
  @links1="<a href=\"www\.testurl\.com\">";
  @links1=~s/.*/href\s*\=\s*\".*?\"/is;

[download]
try 
  $links1 = '<a href="www.testurl.com">';
  $links1 =~ s/<a\s+(href=".*?")\s*>/$1/is;

[download]
You're escaping more than you need to in the regex, and you're not capturing the the left-hand side the subset of the string that you want to make the replacement with on the right-hand side.

Replies are listed 'Best First'.
Re: Re: Substitution Problem
by Wassercrats (Initiate) on Sep 10, 2002 at 05:01 UTC
    I figured Perl's greedyness would allow .* to capture the entire string. Your suggestion worked (I changed a few things to make it work like I really need it to work):
    $links1 =~ s/.*<\s*a\s*(href\s*=\s*".*?").*>.*/$1/is;
    but out of curiosity, isn't there a simple regex to indicate the entire string that I can put on the left so I can specify what I want replaced on the right without memory brackets?
[reply]
[d/l]

      Don't confuse matching with capturing:

      s/$a/$b/     # $a matches and is replaced by $b
s/($a)/$1$b/ # $a matches and is captured in $1
             #   $b is appended

      [download]
      Note also, that you can't use character classes like "." or "\s" within the right side of the substitute operator.

      ~Django
      "Why don't we ever challenge the spherical earth theory?"

[reply]
[d/l]
        It crossed my mind that maybe I can't use character classes on the right side. I guess that was my only problem, but I think that should be allowed. If want delete all attributes from an anchor tag that's contained in @links1, except for href, it would be nice if the following (from my original message) would work:
        @links1=~s/.*/href\s*\=\s*\".*?\"/is;

        [download]

        with a slight adjustment if I want the angle brackets.
[reply]
[d/l]
      ...meant that I'd like to put the replacement string on the right. I want to replace the entire string.
[reply]
        I want to replace the entire string.
        Then don't use s///, use a plain match plus an assignment instead. 
        $links1 =~ /<\s*a\s*(href\s*=\s*".*?").*>/is and $links1 = $1;

        [download]
[reply]
[d/l]

In Section Seekers of Perl Wisdom