I need a regex to match \' <---- literally backslash apostrophe.
my $line = '\'this';
$line =~ s/(\o{134})(\o{047})/\\\\'/g;
$line =~ s/\\'/\\\\'/g;
$line =~ s/[\\][']/\\\\'/g;
printf('%s',$line);
print "\n";
All I get out of this is
'this
When what I want is
\\'this
This occurs whether the string is declared using ' or ". This was a test script for tracking down a file parsing bug. I wanted to confirm that the regex was working as expected.
I don't know if when the backslash apostrophe is parsed by the regex it is not treated as 2 characters, but is instead treated as an escaped apostrophe.
Either way. what is the best way to match \' and print out \\'? I don't want to escape any other back slashes or apostrophes and I can't change the text I am parsing, just the way it is handled and outputted.
s/\\'/\\\\'/g
All three of your patterns match a backslash followed by a quote, the above being the simplest.
Your testing was in vain because your string doesn't contain any backslashes. Both string literals "\'this" (from earlier edit) and '\'this' (from later edit) produce the string 'this.
say "\'this"; # 'this
say '\'this'; # 'this
To produce the string \'this, you could use either of the following string literals (among others):
"\\'this"
'\\\'this'
say "\\'this"; # \'this
say '\\\'this'; # \'this
The answer is, of course
s/[\\][']/\\\\'/g
This will match
\'this
And substitute with this
\\'this
This was the only way I could get it to work.
Perl
Too much "regexing" in your snippet. Try:
my $line = '\'this';
$line =~ s/'/\\\\\'/g;
printf('%s',$line);
print "\n";
# \\'this
or... if you want another mode:
my $line = '\'this';
$line =~ s/'/\\'/g;
printf('%s',$line);
print "\n";
# \'this
I need to increment a numeric value in a string:
my $str = "tool_v01.zip";
(my $newstr = $str) =~ s/\_v(\d+)\.zip$/ ($1++);/eri;
#(my $newstr = $str) =~ s/\_v(\d+)\.zip$/ ($1+1);/eri;
#(my $newstr = $str) =~ s/\_v(\d+)\.zip$/ $1=~s{(\d+)}{$1+1}/r; /eri;
print $newstr;
Expected output is tool_v02.zip
Note: the version number 01 may contain any number of leading zeroes
I don't think this question has anything to do with the /r modifier, but rather how to properly format the output. For that, I'd suggest sprintf:
my $newstr = $str =~ s{ _v (\d+) \.zip$ }
{ sprintf("_v%0*d.zip", length($1), $1+1 ) }xeri;
Or, replacing just the number with zero-width Lookaround Assertions:
my $newstr = $str =~ s{ (?<= _v ) (\d+) (?= \.zip$ ) }
{ sprintf("%0*d", length($1), $1+1 ) }xeri;
Note: With either of these solutions, something like tool_v99.zip would be altered to tool_v100.zip because the new sequence number cannot be expressed in two characters. If that's not what you want then you need to specify what alternative behaviour you require.
The bit you're missing is sprintf which works the same way as printf except rather than outputting the formatted string to stdout or a file handle, it returns it as a string. Example:
sprintf("%02d",3)
generates a string 03
Putting this into your regex you can do this. Rather than using /r you can use do a zero-width look ahead ((?=...)) to match the file suffix and just replace the matched number with the new value
s/(\d+)(?=.zip$)/sprintf("%02d",$1+1)/ei
I am after some help trying to convert the following log I have to plain text.
This is a URL so there maybe %20 = 'space' and other but the main bit I am trying convert is the char(1,2,3,4,5,6) to text.
Below is an example of what I am trying to convert.
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
What I have tried so far is the following while trying to added into the char(in here) to convert with the chr($2)
perl -pe "s/(char())/chr($2)/ge"
All this has manage to do is remove the char but now I am trying to convert the number to text and remove the commas and brackets.
I maybe way off with how I am doing as I am fairly new to to perl.
perl -pe "s/word to remove/word to change it to/ge"
"s/(char(what goes in here))/chr($2)/ge"
Output try to achieve is
select -x1-Q-,-x2-Q-,-x3-Q-
Or
select%20-x1-Q-,-x2-Q-,-x3-Q-
Thanks for any help
There's too much to do here for a reasonable one-liner. Also, a script is easier to adjust later
use warnings;
use strict;
use feature 'say';
use URI::Escape 'uri_unescape';
my $string = q{select%20}
. q{char(45,120,49,45,81,45),char(45,120,50,45,81,45),}
. q{char(45,120,51,45,81,45)};
my $new_string = uri_unescape($string); # convert %20 and such
my #parts = $new_string =~ /(.*?)(char.*)/;
$parts[1] = join ',', map { chr( (/([0-9]+)/)[0] ) } split /,/, $parts[1];
$new_string = join '', #parts;
say $new_string;
this prints
select -x1-Q-,-x2-Q-,-x3-Q-
Comments
Module URI::Escape is used to convert percent-encoded characters, per RFC 3986
It is unspecified whether anything can follow the part with char(...)s, and what that might be. If there can be more after last char(...) adjust the splitting into #parts, or clarify
In the part with char(...)s only the numbers are needed, what regex in map uses
If you are going to use regex you should read up on it. See
perlretut, a tutorial
perlrequick, a quick-start introduction
perlre, the full account of syntax
perlreref, a quick reference (its See Also section is useful on its own)
Alright, this is going to be a messy "one-liner". Assuming your text is in a variable called $text.
$text =~ s{char\( ( (?: (?:\d+,)* \d+ )? ) \)}{
my #arr = split /,/, $1;
my $temp = join('', map { chr($_) } #arr);
$temp =~ s/^|$/"/g;
$temp
}xeg;
The regular expression matches char(, followed by a comma-separated list of sequences of digits, followed by ). We capture the digits in capture group $1. In the substitution, we split $1 on the comma (since chr only works on one character, not a whole list of them). Then we map chr over each number and concatenate the result into a string. The next line simply puts quotation marks at the start and end of the string (presumably you want the output quoted) and then returns the new string.
Input:
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
Output:
select%20"-x1-Q-","-x2-Q-","-x3-Q-"
If you want to replace the % escape sequences as well, I suggest doing that in a separate line. Trying to integrate both substitutions into one statement is going to get very hairy.
This will do as you ask. It performs the decoding in two stages: first the URI-encoding is decoded using chr hex $1, and then each char() function is translated to the string corresponding to the character equivalents of its decimal parameters
use strict;
use warnings 'all';
use feature 'say';
my $s = 'select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)';
$s =~ s/%(\d+)/ chr hex $1 /eg;
$s =~ s{ char \s* \( ( [^()]+ ) \) }{ join '', map chr, $1 =~ /\d+/g }xge;
say $s;
output
select -x1-Q-,-x2-Q-,-x3-Q-
I want to be able to be able to replace all of the line returns (\n's) in a single string (not an entire file, just one string in the program) with spaces and all commas in the same string with semicolons.
Here is my code:
$str =~ s/"\n"/" "/g;
$str =~ s/","/";"/g;
This will do it. You don't need to use quotations around them.
$str =~ s/\n/ /g;
$str =~ s/,/;/g;
Explanation of modifier options for the Substitution Operator (s///)
e Forces Perl to evaluate the replacement pattern as an expression.
g Replaces all occurrences of the pattern in the string.
i Ignores the case of characters in the string.
m Treats the string as multiple lines.
o Compiles the pattern only once.
s Treats the string as a single line.
x Lets you use extended regular expressions.
You don't need to quote in your search and replace, only to represent a space in your first example (or you could just do / / too).
$str =~ s/\n/" "/g;
$str =~ s/,/;/g;
I'd use tr:
$str =~ tr/\n,/ ;/;
I have a two strings, XXXXXXnumber and XXXXXXdate and I want to strip all the XXXXXX from each string. The actual number of character represented by XXXXXX can vary. The suffixes 'number' and 'date' are constant. XXXXXXnumber and XXXXXXXdate should become XXXXXX.
my ($prefix) = ($string =~ /\A (.+?) (?:date|number) \z/x);
Alternatively:
$string =~ s/ (?:date|number) \z//x;
I would use a regular expression like $line =~ s/(number|date)$// for that task, where $line can be either line.
If your line has additional characters after number or date, they must be filtered out, too. An alternative approach would be using an expression like ($num) = ($line =~ /^(.*)(number|date).*$/);
use regexes:
($newvar = $oldvar) =~ s/^(.*)(number|date)$/$1/;
if you have no mor euse for $oldvar's original value (including the Xes) this simplifies to
$oldvar =~ s/^(.*)(number|date)$/$1/;
A simple substitution takes care of it:
$str =~ s/(?:number|date)\z/;