in reply to Perl RE; how to capture, and replace based on a block?

How about you post actual perl code, you know, stuff ready to run?
  • Comment on Re: Perl RE; how to capture, and replace based on a block?

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Re^2: Perl RE; how to capture, and replace based on a block?
by taint (Chaplain) on Dec 18, 2013 at 00:26 UTC
    Um. I did that;
    Code to replace 
    </div>
</body>

    [download]
    RE I'm using, that doesn't work 
    \<\/div\>\n\<\/body\>

    [download]
    As stated in my OP; my RE (shown) matches one, or the other, not both. I had hoped to match both (</div></body>). Is it clearer?

    --Chris

    Yes. What say about me, is true.
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[d/l]
[select]

      Um. I did that;

      Sorry but you didn't. You posted some data and a pattern you say you want to match the data but it doesn't match -- great, now show your code that uses the pattern with this data that fails to match

      it works "perfectly" as expected 

      Compiling REx "\<\/div\>\n\<\/body\>"
Final program:
   1: EXACT <</div>\n</body>> (6)
   6: END (0)
anchored "</div>%n</body>" at 0 (checking anchored isall) minlen 14
Guessing start of match in sv for REx "\<\/div\>\n\<\/body\>" against 
+"</div>%n</body>"
Found anchored substr "</div>%n</body>" at offset 0...
Guessed: match at offset 0
Freeing REx: "\<\/div\>\n\<\/body\>"

      [download]
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[d/l]
        OK. To be more concise; I used cat to feed the file to Perl using an inline regex. 
        Yes. What say about me, is true.
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