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

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;
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with a slight adjustment if I want the angle brackets.

That wouldn't make much sense, because a character class can match any of its characters, but how should it be interpreted in the right side of s///? Should that mean all the characters it represents? Or a random one?
The following code does what you want:

for (@links1) {
  s/<\s*a\s.*?\b(href\s*=\s*".*?").*?>/<a $1>/i;
}
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Now let me explain your errors:

@links1=~   # that would work like scalar(@array)=~
s/ .*       # replace complete string with
 / href     # 'href'
   \s*      # as many spacey chars as possible
   \=       # no need to escape '='
   \s*      # as many spacey chars as possible
   \"       # no need to escape '"'
   .*?      # as few as possible (=no) anythings
   \"       # no need to escape '"'
 /isx;      # 's' is not needed, I added 'x'
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^~Django
"Why don't we ever challenge the spherical earth theory?"

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;
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