Monster perl regex - regex

Question

I'm trying to change strings like this:
<a href='../Example/case23.html'><img src='Blablabla.jpg'
To this:
<a href='../Example/case23.html'><img src='<?php imgname('case23'); ?>'
And I've got this monster of a regular expression:
find . -type f | xargs perl -pi -e \
's/<a href=\'(.\.\.\/Example\/)(case\d\d)(.\.html\'><img src=\')*\'/\1\2\3<\?php imgname\(\'\2\'); \?>\'/'
But it isn't working. In fact, I think it's a problem with Bash, which could probably be pointed out rather quickly.
r: line 4: syntax error near unexpected token `('
r: line 4: ` 's/<a href=\'(.\.\.\/Example\/)(case\d\d)(.\.html\'><img src=\')*\'/\1\2\3<\?php imgname\(\'\2\'); \?>\'/''
But if you want to help me with the regular expression that'd be cool, too!

Answer 1

Teaching you how to fish:
s/…/…/
Use a separator other than / for the s operator because / already occurs in the expression.
s{…}{…}
Cut down on backslash quoting, prefer [.] over \. because we'll shellquote later. Let's keep backslashes only for the necessary or important parts, namely here the digits character class.
s{<a href='[.][.]/Example/case(\d\d)[.]html'>…
Capture only the variable part. No need to reassemble the string later if the most part is static.
s{<a href='[.][.]/Example/case(\d\d)[.]html'><img src='[^']*'}{<a href='../Example/case$1.html'><img src='<?php imgname('case$1'); ?>'}
Use $1 instead of \1 to denote backreferences. [^']* means everything until the next '.
To serve now as the argument for the Perl -e option, this program needs to be shellquoted. Employ the following helper program, you can also use an alias or shell function instead:
> cat `which shellquote`
#!/usr/bin/env perl
use String::ShellQuote qw(shell_quote); undef $/; print shell_quote <>
Run it and paste the program body, terminate input with Ctrl+d, you receive:
's{<a href='\''[.][.]/Example/case(\d\d)[.]html'\''><img src='\''[^'\'']*'\''}{<a href='\''../Example/case$1.html'\''><img src='\''<?php imgname('\''case$1'\''); ?>'\''}'
Put this together with shell pipeline.
find . -type f | xargs perl -pi -e 's{<a href='\''[.][.]/Example/case(\d\d)[.]html'\''><img src='\''[^'\'']*'\''}{<a href='\''../Example/case$1.html'\''><img src='\''<?php imgname('\''case$1'\''); ?>'\''}'

Answer 2

Bash single-quotes do not permit any escapes.
Try this at a bash prompt and you'll see what I mean:
FOO='\'foo'
will cause it to prompt you looking for the fourth single-quote. If you satisfy it, you'll find FOO's value is
\foo
You'll need to use double-quotes around your expression. Although in truth, your HTML should be using double-quotes in the first place.

Answer 3

Single quotes within single quotes in Bash:
set -xv
echo ''"'"''
echo $'\''

Answer 4

I wouldn't use a one-liner. Put your Perl code in a script, which makes it much easier to get the regex right without wondering about escaping quotes and such.
I'd use a script like this:
#!/usr/bin/perl -pi
use strict;
use warnings;
s{
( <a \b [^>]* \b href=['"] [^'"]*/case(\d+)\.html ['"] [^>]* > \s*
<img \b [^>]* \b src=['"] ) [^'"<] [^'"]*
}{$1<?php imgname('case$2'); ?>}gix;
and then do something like:
find . -type f | xargs fiximgs
– Michael

Answer 5

if you install the package mysql, it comes with a command called replace.
With the replace command you can:
while read line
do
X=`echo $line| replace "<a href='../Example/" ""|replace ".html'><" " "|awk '{print $1}'`
echo "<a href='../Example/$X.html'><img src='<?php imgname('$X'); ?>'">NewFile
done < myfile
same can be done with sed. sed s/'my string'/'replace string'/g.. replace is just easier to work with special characters.