Re^3: Regex find and replace involving new line

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Re^4: Regex find and replace involving new line by PRA007 (Initiate) on Dec 14, 2015 at 11:12 UTC
My data files are exactly similar to what I have shown in the initial file. I have just installed strawberry perl so don't know how to code. I want to perform find and replace in text file using perl regular expression. I am fan of PCRE and have some knowledge. While asking for command line alternative to notepad++, I was suggested, perl is good choice. I tried for following file 1 2 3 4 dog 1 2 3 4 5 puppy 7 8 I want to replace dog with puppy. Sometimes I want preserve lines between them too sometimes and wan't to remove the lines other times. While using this one liner `perl -i.bak -pe "BEGIN{undef $/;} s/(dog)((.[\r\n]+){6})(.)/$1/smg" 4.txt` it gives 1 2 3 4 dog while notepad pcre gives desired result wile using find `(dog)((.[\r\n]+){6})(.)` and replace with `\1` 1 2 3 4 7 dog 8 like wise when I replace it with \2 it gave all the lines without dog and puppy. and while \3 gave me puppy in place of dog in above mentioned result. all behaves perfectly. So the main thing that I want is to make grouping as follow. group 1= dog, Group 2 = 6 lines after found string i.e.dog. group 3 = whatever present on line 7. I want to be able to put back the groups as required. Questions un answered How to use grouping in perl while using regex that matches new line? Am I using rightly grouping in perl in my above code? How to specify substitution? does substitution formula is correct as a whole in my code? How to specified remembered groups? Am I making any mistake in substitution position while specifying group like $1, $2, $3 or sometimes $& (what is found) I don't have sufficient knowledge to understand what you have suggested so I tried `perl -i.bak -pe "BEGIN{undef $/;} s/(dog)(.([^\n]\r?\n){6})(.*)/$1/smg" 4.txt` gave same results as with earlier oneliner	[reply] [d/l] [select]
Re^5: Regex find and replace involving new line by Corion (Patriarch) on Dec 14, 2015 at 11:54 UTC
When I run your regular expression `(dog)((.[\r\n]+){6})(.)` [download] in Notepad++ with the following text: `1 2 3 4 dog 1 2 3 4 5 puppy 7 8` [download] I also get `1 2 3 4 dog 7 8` [download] Which is what I would expect, because the replacement only specifies `\1`, not `\2` or `\3`. To your questions: How to use grouping in perl while using regex that matches new line? Am I using rightly grouping in perl in my above code? Captured groups are referenced in Perl using `$1` for the first opening parenthesis, `$2` for the second opening parenthesis and so on. You've used `$1` in your existing Perl code, and from your description I think you would also want to use `$2` and `$3` maybe. How to specify substitution? does substitution formula is correct as a whole in my code? You use the `s///` operator, as you already do. How to specified remembered groups? Am I making any mistake in substitution position while specifying group like $1, $2, $3 or sometimes $& (what is found) You use `$1` and its siblings. Assuming that the output text you want (which you do not show us, even though I had recommended this) is the following: `1 2 3 4 dog=puppy 7 8` [download] the following works for me: `s/(dog)(([^\n][\n]){6})(.)/$1=$4/g;` [download] If my interpretation of what you want is wrong, maybe now is a good time to be more specific and show concrete examples of what you want, and the exact cases when you want the specific results. "Sometimes this and sometimes that" is not a specific explanation.	[reply] [d/l] [select]
Re^6: Regex find and replace involving new line by PRA007 (Initiate) on Dec 15, 2015 at 08:44 UTC
Does `(([^\n][\n]){6})(.)` always matches 6 lines? It behaves notoriously sometimes. I have string perl -i.bak -pe "BEGIN{undef $/;} s/\\cellx10464\\pard\\plain\\intbl\\s0\\ql\\fi0\\li0\\ri0\\sl320\\plain\\f4\\fs20\\b\\cf0 Patent Information\\b0([^\r?\n][\r?\n]){88}.(EP \d{5,7})(([^\r?\n][\r?\n]){52}).$2[\r\n]+\\cell\\pard\\plain\\intbl\\s0\\ql\\fi0\\li0\\ri0\\sl320\\plain\\f1\\fs20\\cf0 \\f1\\fs20\\cf0 (B\d?)[\r\n]+\\cell\\pard\\plain\\intbl\\s0\\ql\\fi0\\li0\\ri0\\sl320\\plain\\f1\\fs20\\cf0 [a-zA-Z]{3} [0-9,]{3} [0-9]{4}[\r\n]+\\cell\\pard\\plain\\intbl\\s0\\ql\\fi0\\li0\\ri0\\sl320\\plain\\f1\\fs20\\cf0 \\f1\\fs20\\cf0 [\r\n]+\\cell\\pard\\plain\\intbl\\s0\\ql\\fi0\\li0\\ri0\\plain/tttttt$2 $4/smg;" 1.rtf My rtf contains `somestring(88 paragraphs mached as $1)(string $2)(52 paragraphs mached as string 3).*$2{means found string}[\r\n]+some string(string $4)somestring` It doesn't give the desired result. I thing it matches first occurrence of first found and last occurence of last found and removes all the lines between that are iportant one. I am using windows strawberry perl. where am I making mistake?	[reply] [d/l] [select]
Re^7: Regex find and replace involving new line by AnomalousMonk (Archbishop) on Dec 15, 2015 at 13:17 UTC
Re^7: Regex find and replace involving new line by Corion (Patriarch) on Dec 15, 2015 at 09:21 UTC
Re^8: Regex find and replace involving new line by PRA007 (Initiate) on Dec 15, 2015 at 12:13 UTC
Re^6: Regex find and replace involving new line by PRA007 (Initiate) on Dec 15, 2015 at 06:16 UTC
That's the answer I wanted. Currently its doing job for me. I will make entire list working and then I will come back to you with specific problem. My find string is very big containing approximately 200 lines. Puting it here I guess will not be feasible. I am very new to the forum. Please suggest me how to give that string by any means	[reply]
Re^6: Regex find and replace involving new line by PRA007 (Initiate) on Dec 15, 2015 at 07:19 UTC
It take too much time to test perl regex with command line. Is there any tool to test perl regex for windows?	[reply]