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

by Anonymous Monk
on Dec 18, 2013 at 00:12 UTC ( #1067554=note: print w/ replies, xml ) Need Help??


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?
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>
    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.
    

      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.
        

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