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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 ( #1067556=note: print w/ replies, xml ) Need Help??


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

Um. I did that;
Code to replace

</div> </body>
RE I'm using, that doesn't work
\<\/div\>\n\<\/body\>
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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Re^3: Perl RE; how to capture, and replace based on a block?
by Anonymous Monk on Dec 18, 2013 at 00:30 UTC

    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\>"
      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.
      

        That's concise, but I have no idea what it means.

        OK. To be more concise; I used cat to feed the file to Perl using an inline regex.

        And then what happened? How is that a program that shows the problem, a program I can run on my computer through copy/paste?

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