Substitute first character before match - regex

Question

For each line I need to add a semicolon exactly one character before the first match of an alphanumeric sign but only for the alphanumeric sign after the first appearance of a semicolon.
Example:
Input:
00000001;Root;;
00000002; Documents;;
00000003; oracle-advanced_plsql.zip;file;
00000004; Public;;
00000005; backup;;
00000006; 20110323-JM-F.7z.001;file;
00000007; 20110426-JM-F.7z.001;file;
00000008; 20110603-JM-F.7z.001;file;
00000009; 20110701-JM-F-via-summer_school;;
00000010; 20110701-JM-F-via-summer_school.7z.001;file;
Desired output:
00000001;;Root;;
00000002; ;Documents;;
00000003; ;oracle-advanced_plsql.zip;file;
00000004; ;Public;;
00000005; ;backup;;
00000006; ;20110323-JM-F.7z.001;file;
00000007; ;20110426-JM-F.7z.001;file;
00000008; ;20110603-JM-F.7z.001;file;
00000009; ;20110701-JM-F-via-summer_school;;
00000010; ;20110701-JM-F-via-summer_school.7z.001;file;
Could someone helps me please to create Perl regex for that? I'd need it in a program, not as a oneliner.

Answer 1

This is a way to insert a semi-colon after the first semi-colon and whitespace, but before the first non-whitespace.
s/;\s*\K(?=\S)/;/
If you feel the need, you can use \w instead of \S, but I felt with this input it was an unnecessary specification.
The \K (keep) escape is similar to a lookbehind assertion in that it does not remove what it matches. The same goes for the lookahead assertion, so all this substitution does is insert a semi-colon in the designated spot.

Answer 2

First of all, here is a program that seems to match your requirements:
#/usr/bin/perl -w
while(<>) {
s/^(.*?;.*?)(\w)/$1;$2/;
print $_;
}
Store it in a file 'program.pl', make it executable with 'chmod u+x program.pl' and run it on your input data like this:
program.pl input-data.txt
Here is an explanation of the regular expression:
s/ # start search-and-replace regexp
^ # start at the beginning of this line
( # save the matched characters until ')' in $1
.*?; # go forward until finding the first semicolon
.*? # go forward until finding... (to be continued below)
)
( # save the matched characters until ')' in $2
\w # ... the next alphanumeric character.
)
/ # continue with the replace part
$1;$2 # write all characters found above, but insert a ; before $2
/ # finish the search-and-replace regexp.
Based on your sample input, I would use a more specific regular expression:
s/^(\d*; *)(\w)/$1;$2/;
This expression starts at the beginning of the line, skips over numbers (\d*) followed by the first semicolon and space. Before the following word character, it inserts a semicolon.
Take what fits best to your needs!

Answer 3

First of all thank you for your really great answers!
Actually my code snippet looks like this:
our $seperator=";" # at the beginning of the file
#...
sub insert {
my ( $seperator, $line, #all_lines, $count, #all_out );
$count = 0;
#all_lines = read_file($filename);
foreach $line (#all_lines) {
$count = sprintf( "%08d", $count );
chomp $line;
$line =~ s/\:/$seperator/; # works
$line =~ s/\ file/file/; # works
#$line=~s/;\s*\K(?=\S)/;/; # doesn't work
$line =~ s/^(.*?$seperator.*?)(\w)/$1$seperator$2/; # doesn't work
say $count . $seperator . $line . $seperator;
$count++; # btw, is there maybe a hidden index variable in a foreach-loop I could us instead of a new variable??
push( #all_out, $count . $seperator . $line . $seperator . "\n" );
}
write_file( $csvfile, #all_out ); # using File::Slurp
}
In order to get the input which I presented you, I made already some small substitutions, as you can see in the beginning of the foreach-loop.
I am curious, why the regular expressions presented by TLP and Yaakov do not work in my code. In general they work, but only when written like in the example which Yaakov gave:
while(<>) {
s/^(.*?;.*?)(\w)/$1;$2/;
print $_;
}