Perl regexp replace? - regex

I'm checking to see if I can replace an include like this:
#include <pathFoo/whatever.H>
with:
#include "whatever.H"
to do so I would use the -i switch, but to check I am doing it correctly I am simply using the -p switch without -i. I have the following command:
perl -p -e 's/<pathFoo\/\(.*\)>/"$1"' thefile
but this isn't quite working and i'm not exactly sure which part is off?


You don't want to escape the parens, i.e.:
perl -p -e 's/<pathFoo\/(.*)>/"$1"/' thefile
should work for you.
Also note the ending /.


perl -i~ -pe's!<pathFoo/(.*)>!"$1"!' file
The following is safer:
perl -i~ -pe's!#include\s+\K<pathFoo/(.*)>!"$1"!' file


Since TIMTOWTDI, here's a double substitution. Or rather a substitution and a transliteration:
perl -pe 's|^#include\s+<\KpathFoo/|| && tr/<>/""/'
So, just remove the pathFoo/ part first, and if that succeeds, then transliterate the <> characters to quotes.


I think this will get you there:
perl -pi -e 's{#include[ ]+<.+?/([^/]+)>}{#include "$1"}g' file

Related

General solutions to replace string regex preceded and followed by '\n'

I have a file in CentOS which looks like following
[root#localhost nn]# cat -A excel.log
real1$
0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I$
real2$
0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I$
real3$
0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I$
real4$
0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I1^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I$
real5$
0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I1^I0.5^I0.5^I0.5^I0.5^I$
real6$
I would like to replace \nreal[2-6]\n with \t\t\t' and have tried unsuccessfully the following
sed -i 's/\nreal[2-6]\n/\t\t\t/g' file
It seems that sed has difficulty to deal with line break. Any idea to fulfill the regex in CentOS?
Much appreciated!

If you want to consider perl then use:
perl -i -0777 -pe 's/\n(?:51[23]real|real[2-6])(?:\n|\z)/\t\t\t/g' file
If you want to avoid last real\d+ line to be replaced with \t\t\t then use:
perl -i -0777 -pe 's/\n(?:51[23]real|real[2-6])\n(?!\z)/\t\t\t/g' file
(?!\z) is negative lookahead to fail the match when we have line end just ahead of us.

With GNU sed, you need to use the -z option:
sed -i -z 's/\nreal[2-6]\n/\t\t\t/g' file
# ^^
Now, that you also want to handle specific alternations, you need to enable the POSIX ERE syntax, either with -r or -E option:
sed -i -Ez 's/\n(51[23]real|real[2-6])\n/\t\t\t/g' file

BASH: replacing PERL with SED for in-place substitution

Would like to replace this statement with perl:
perl -pe "s|(?<=://).+?(?=/)|$2:80|"
with
sed -e "s|<regex>|$2:80|"
Since sed has a much less powerful regex engine (for example it does not support look-arounds) the task boils down to writing a sed compatible regex to match only a domain name in a fully qualitied URL. Examples:
http://php2-mindaugasb.c9.io/Testing/JS/displayName.js
http://php2-mindaugasb.c9.io?a=Testing.js
http://www.google.com?a=Testing.js
Should become:
http://$2:80/Testing/JS/displayName.js
http://$2:80?a=Testing.js
http://$2:80?a=Testing.js
A solution like this would be ok:
sed -e "s|<regex>|http://$2:80|"
Thanks :)

Use the below sed command.
$ sed "s~//[^/?]\+\([?/]\)~//\$2:80\1~g" file
http://$2:80/Testing/JS/displayName.js
http://$2:80?a=Testing.js
http://$2:80?a=Testing.js
You must need to escape the $ at the replacement part.

sed 's|http://[^/?]*|http://$2:80|' file
Output:
http://$2:80/Testing/JS/displayName.js
http://$2:80?a=Testing.js
http://$2:80?a=Testing.js

unmatched parenthesis in regex - Linux

I want to replace (whole string)
$(TOPDIR)/$(OSSCHEMASDIRNAME)
with
/udir/makesh/$(OSSCHEMASDIRNAME)
in a makefile
I tried with
perl -pi.bak -e "s/\$\(TOPDIR\)\/\$\(OSSCHEMASDIRNAME\)/\/udir\/makesh\/\$\(OSSCHEMASDIRNAME\)/g " makefile
but i am getting unmatched parentheses error

You have to "double" escape the dollar sign. Like this:
echo "\$(TOPDIR)/\$(OSSCHEMASDIRNAME)" | perl -p -e "s/\\$\(TOPDIR\)\/\\$\(OSSCHEMASDIRNAME\)/\/udir\/makesh\/\\$\(OSSCHEMASDIRNAME\)/g"

First off, you don't need to use / for regular expressions. They're just canonical. You can use pretty much anything. Thus your code can become (simplify away some \):
perl -pi.bak -e "s|\$\(TOPDIR\)/\$\(OSSCHEMASDIRNAME\)|/udir/makesh/\$\(OSSCHEMASDIRNAME\)|g " makefile
Now to actually address your issue, because you're using " instead of ', the shell attempts to figure out what $\ means which is then replaced with (presumably) nothing. So what you really want is:
perl -p -i.bak -e 's|\$\(TOPDIR\)/\$\(OSSCHEMASDIRNAME\)|/udir/makesh/\$\(OSSCHEMASDIRNAME\)|g' makefile

When in doubt about escaping, you can simply use quotemeta or \Q ... \E.
perl -pe 's#\Q$(TOPDIR)\E(?=/\Q$(OSSCHEMASDIRNAME)\E)#/udir/makesh#;'
Note the use of a look-ahead assertion to save us the trouble of repeating the trailing part in the substitution.
A quotemeta solution would be something like:
perl -pe 'BEGIN { $dir = quotemeta(q#$(TOPDIR)/$(OSSCHEMASDIRNAME)#); }
s#$dir#/udir/makesh/$(OSSCHEMASDIRNAME)#;'
Of course, you don't need to use an actual one-liner. When the shell quoting is causing troubles, the simplest option of them all is to write a small source file for your script:
s#\Q$(TOPDIR)\E(?=/\Q$(OSSCHEMASDIRNAME)\E)#/udir/makesh#;
And run with:
perl -p source.pl inputfile

Using sed to replace <? with <?php

I'm trying to programmatically replace <? with <?php in a bunch of file, but my sed regex isn't behaving like I expected. Can you tell me what's wrong with it?
I'm testing it on the command line here:
$ sed -e 's/<\?/<\?php/g'
<?
<?php?<?php
d
<?phpd<?php

I don't think you need the escapes on the ?:
sed -e 's/<?/<?php/g'

You don't need to escape the back reference in the replacement.
sed 's#<\?#<?php#'
In a pipe, to correct for doubling the php:
sed 's#<\?#<?php#g' | sed 's#phpphp#php#g'

Do you really have to use sed, or can you use perl as well?
perl -pi.tmp -e 's,^<\?(?!php),<?php,' *.php *.inc
rm *.tmp
I am using a negative look-ahead to avoid generating <?phpphp in cases where the files already start with the correct characters.

This will also skip matching with <?=$variable:
sed -e 's/<?\([^=p]\)/<?php\1/g' -e 's/<?$/<?php/g'

Perl regex: remove everything (including line breaks) until a match is found

Apologies for the simple question. I don't clean text or use regex often.
I have a large number of text files in which I want to remove every line until my regex finds a match. There's usually about 15 lines of fluff before I find a match. I was hoping for a perl one-liner that would look like this:
perl -p -i -e "s/.*By.unanimous.vote//g" *.txt
But this doesn't work.
Thanks

Solution using the flip-flop operator:
perl -pi -e '$_="" unless /By.unanimous.vote/ .. 1' input-files
Shorter solution that also uses the x=!! pseudo operator:
per -pi -e '$_ x=!! (/By.unanimous.vote/ .. 1)' input-files

Have a try with:
If you want to get rid until the last By.unanimous.vote
perl -00 -pe "s/.*By.unanimous.vote//s" inputfile > outputfile
If you want to get rid until the first By.unanimous.vote
perl -00 -pe "s/.*?By.unanimous.vote//s" inputfile > outputfile

Try something like:
perl -pi -e "$a=1 if !$a && /By\.unanimous\.vote/i; s/.*//s if !$a" *.txt
Should remove the lines before the matched line. If you want to remove the matching line also you can do something like:
perl -pi -e "$a=1 if !$a && s/.*By\.unanimous\.vote.*//is; s/.*//s if !$a" *.txt
Shorter versions:
perl -pi -e "$a++if/By\.unanimous\.vote/i;$a||s/.*//s" *.txt
perl -pi -e "$a++if s/.*By\.unanimous\.vote.*//si;$a||s/.*//s" *.txt

You haven't said whether you want to keep the By.unanimous.vote part, but it sounds to me like you want:
s/[\s\S]*?(?=By\.unanimous\.vote)//
Note the missing g flag and the lazy *? quantifier, because you want to stop matching once you hit that string. This should preserve By.unanimous.vote and everything after it. The [\s\S] matches newlines. In Perl, you can also do this with:
s/.*?(?=By\.unanimous\.vote)//s

Solution using awk
awk '/.*By.unanimous.vote/{a=1} a==1{print}' input > output