Perl regex digit - regex

Consider my regex in this code section:
use strict;
my #list = ("1", "2", "123");
&chk(#list);
sub chk {
my #num = split (" ", "#_");
foreach my $chk (#num) {
chomp $chk;
if ($chk =~ m/\d{1,2}?/) {
print "$chk\n";
}
}
}
The \d{4} will print nothing. The \d{3} will print only 123. But if I change to \d{1,2}? it will print all. I thought, according to all the sources I read so far, that {1,2} mean: one digit but no more than two. So it should have printed only 1 and 2, correct?
What do I need to extract items that contains only one to two digits?

\d{1,2} succeeds if it finds 1 or 2 digits anywhere in the string provided. Additional string content is does not cause the match to fail. If you want to match only when the string contains exactly 1 or 2 digits, do this: ^\d{1,2}$

You should anchor your regular expression for the desired effect. The built-in function grep suits better here since it is a selection from an array that is to be done:
#!/usr/bin/env perl
use strict;
use warnings;
my #list = ( 1, 2, 123 );
print join "\n", grep /^\d{1,2}$/, #list;

It appears to be working perfectly!
Here's a hint: Use the Perl variables $`, $&, and $'. These variables are special regular expression variables that show the part of the string before the match, what was matched, and the post matched string.
Here's a sample program:
#! /usr/bin/env perl
use strict;
use warnings;
use feature qw(say);
use Scalar::Util;
my #list = ("1", "2", "123");
foreach my $string (#list) {
if ($string =~ /\d{1,2}?/) {
say qq(We have a match for "string"!);
say qq("$`" "$&" "$'");
}
else {
say "No match makes David Sad";
}
}
The output will be:
We have a match for "1"!
"" "1" ""
We have a match for "2"!
"" "2" ""
We have a match for "123"!
"" "1" "23"
What this does is divide up the string into three sections: The section of the string before the regular expression match, the section of the string that matches the regular expression, and the section of the string after the regular expression match.
In each case, there was no pre-match because the regular expression matches from the start of the string. We also see that \d{1,2}? matches a single digit in each case even through 123 could have matched two digits. Why? Because the question mark on the end of the match specifier tells the regular expression not to be greedy. In this case, we tell the regular expression to match either one or two characters. Fine, it matches on one. Remove the question mark, and the last line would have looked like this:
We have a match for "123"!
"" "12" "3"
If you want to match on one or two digits, but not three or more digits, you'll have to specify the part of your string before and after the one or two digits. Something like this:
/\D\d{1,2}\D/
This would match your string foo12bar, but not foo123bar. But what if the string is 12? In that case, we want to say that either we have the beginning of the string, or a non-digit before our one or two character match, and we either have a non-digit or the end of the string at the end of our one or two character match:
/(\D|^)\d{1,2}(/D|$)/
A quick explanation:
(\D|^): A non-digit or the beginning of the string (The ^ anchor)
d{1,2}: One or two digits
(\D|$): A non-digit or the end of the string (The $ anchor)
Now, this will match 12, but not 123, and it will match foo12 and foo12bar, but not foo123 or foo123bar.
Just looking for a one or two digit number, we can simply specify the anchors:
/^\d{1,2}$/;
Now, that will match 1, 12, but not foo12 or 123.
The main thing is to use the $`, $&, and $' variables in order to help see exactly what your regular expression is matching on and what's before and after your match.

No, because while the regex only matches two digits, $chk still contains 123. If you want to only print the part that is matched, use
if ($chk =~ m/(\d{1,2})/) {
print "$1\n";
}
Note the parentheses and the $1. This causes it to print only that which is in the parentheses.
Also, this code doesn't make much sense:
sub chk {
my #num = split (" ", "#_");
Because #_ already is an array it makes no sense to make it into a string and then split it. Simply do:
sub chk {
foreach my $chk (#_) {
You also do not need to use chomp for data that is not coming from user input, as it is intended to remove the trailing newline. There is no newline in any of this data.

#!/usr/bin/perl
use strict;
my #list = ("1", "2", "123");
&chk(\#list);
sub chk {
foreach my $chk (#{$_[0]}) {
print "$chk\n" if $chk =~ m/^\d{1,2}$/ ;
}
}

#!/usr/bin/perl
use strict;
use warnings;
my #list = ("1", "2", "123");
&chk(#list);
sub chk {
my #num = split (" ", "#_");
foreach my $chk (#num) {
chomp $chk;
if ($chk =~ m/\d{1,2}/ && length($chk) <= 2) {
print "$chk\n";
}
}
}

Related

How to verify if a variable value contains a character and ends with a number using Perl

I am trying to check if a variable contains a character "C" and ends with a number, in minor version. I have :
my $str1 = "1.0.99.10C9";
my $str2 = "1.0.99.10C10";
my $str3 = "1.0.999.101C9";
my $str4 = "1.0.995.511";
my $str5 = "1.0.995.AC";
I would like to put a regex to print some message if the variable has C in 4th place and ends with number. so, for str1,str2,str3 -> it should print "matches". I am trying below regexes, but none of them working, can you help correcting it.
my $str1 = "1.0.99.10C9";
if ( $str1 =~ /\D+\d+$/ ) {
print "Candy match1\n";
}
if ( $str1 =~ /\D+C\d+$/ ) {
print "Candy match2\n";
}
if ($str1 =~ /\D+"C"+\d+$/) {
print "candy match3";
}
if ($str1 =~ /\D+[Cc]+\d+$/) {
print "candy match4";
}
if ($str1 =~ /\D+\\C\d+$/) {
print "candy match5";
}
if ($str1 =~ /C[^.]*\d$/)
C matches the letter C.
[^.]* matches any number of characters that aren't .. This ensures that the match won't go across multiple fields of the version number, it will only match the last field.
\d matches a digit.
$ matches the end of the string. So the digit has to be at the end.
I found it really helpful to use https://www.regextester.com/109925 to test and analyse my regex strings.
Let me know if this regex works for you:
((.*\.){3}(.*C\d{1}))
Following your format, this regex assums 3 . with characters between, and then after the third . it checks if the rest of the string contains a C.
EDIT:
If you want to make sure the string ends in a digit, and don't want to use it to check longer strings containing the formula, use:
^((.*\.){3}(.*C\d{1}))$
Lets look what regex should look like:
start{digit}.{digit}.{2-3 digits}.{2-3 digits}C{1-2 digits}end
very very strict qr/^1\.0\.9{2,3}\.101?C\d+\z/ - must start with 1.0.99[9]?.
very strict qr/^1\.\0.\d{2,3}\.\d{2,3}C\d{1,2}\z/ - must start with 1.0.
strict qr/^\d\.\d\.\d{2,3}\.\d{2,3}C\d{1,2}\z/
relaxed qr/^\d\.\d\.\d+\.\d+C\d+\z/
very relaxed qr/\.\d+C\d+\z/
use strict;
use warnings;
use feature 'say';
my #data = qw/1.0.99.10C9 1.0.99.10C10 1.0.999.101C9 1.0.995.511 1.0.995.AC/;
#my $re = qr/^\d\.\d\.\d+\.\d+C\d+\z/;
my $re = qr/^\d\.\d\.\d{2,3}\.\d{2,3}C\d+\z/;
say '--- Input Data ---';
say for #data;
say '--- Matching -----';
for( #data ) {
say 'match ' . $_ if /$re/;
}
Output
--- Input Data ---
1.0.99.10C9
1.0.99.10C10
1.0.999.101C9
1.0.995.511
1.0.995.AC
--- Matching -----
match 1.0.99.10C9
match 1.0.99.10C10
match 1.0.999.101C9

Matching a variable in a string in Perl from the end

I want to match a variable character in a given string, but from the end.
Ideas on how to do this action?
for example:
sub removeCharFromEnd {
my $string = shift;
my $char = shift;
if($string =~ m/$char/){ // I want to match the char, searching from the end, $doesn't work
print "success";
}
}
Thank you for your assistance.
There is no regex modifier that would force Perl regex engine to parse the string from right to left. Thus, the most convenient way to achieve that is via a negative lookahead:
m/$char(?!.*$char)/
The (?!.*$char) negative lookahead will require the absence (=will fail the match if found) of a $char after any 0+ chars other than linebreak chars (use s modifier if you are running the regex against a multiline string input).
The regex engine works from left to right.
You can use the natural greediness of quantifiers to reach the end of the string and find the last char with the backtracking mechanism:
if($string =~ m/.*\K$char/s) { ...
\K marks the position of the match result beginning.
Other ways:
you can also reverse the string and use your previous pattern.
you can search all occurrences and take the last item in the list
I'm having trouble understanding what you want. Your subroutine is called removeCharFromEnd, so perhaps you want to remove $char from $string if it appears at the end of the string
You can do that like this
sub removeCharFromEnd {
my ( $string, $char ) = #_;
if ( $string =~ s/$char\z// ) {
print "success";
}
$string;
}
Or perhaps you want to remove the last occurrence of $char wherever it is. You can do that with
s/.*\K$char//
The subroutine I have written returns the modified string, so you would have to assign the result to a variable to save it. You can write
my $s = 'abc';
$s = removeCharFromEnd($s, 'c');
say $s;
output
ab
If you just want to modify the string in place then you should write
$ARGV[0] =~ s/$char\z//
using whichever substitution you choose. Then you can do this
my $s = 'abc';
removeCharFromEnd($s, 'c');
say $s;
This produces the same output
To get Perl to search from the end of a string, reverse the string.
sub removeCharFromEnd {
my $string = reverse shift #_;
my $char = quotemeta reverse shift #_;
$string =~ s/$char//;
$string = reverse $string;
return $string;
}
print removeCharFromEnd(qw( abcabc b )), "\n";
print removeCharFromEnd(qw( abcdefabcdef c )), "\n";
print removeCharFromEnd(qw( !"/$%?&*!"/$%?&* $ )), "\n";

Regex to read the id

I have the log file with the following content:
(2947:_dRW00T3WEeSkhZ9pqkt5dQ) ---$ ABC XY "Share" 16-Sep-2014 03:22 PM
(2948:_3nFSwz3TEeSkhZ9pqkt5dQ) ---$ ABC XY "Share" 16-Sep-2014 03:05 PM
(2949:_voeYED3AEeSkhZ9pqkt5dQ) ---$ ABC XY "Initial for Re,oved" 16-Sep-2014 12:44 PM
I want to read the unique id say _dRW00T3WEeSkhZ9pqkt5dQ from each line and store it in a array.
My current code is:
while(<$fh>) {
if ($_ =~ /\((.*?)\)/) {
push #cs_ids , $1;
}
}
Try this:
while(<$fh>) {
if ($_ =~ /\(\d+:(.+?)\)/) {
push #cs_ids , $1;
}
}
The regexp checks all string which starts with ( then one or more digits a double point and than one or more characters ( Which will be stored in $1). THe end of the string is a ).
You were almost there:
perl -e '$string = "(2947:_dRW00T3WEeSkhZ9pqkt5dQ)"; if ($string =~ /^\((\d+:)(.*?)\)$/) { die $2; }'
_dRW00T3WEeSkhZ9pqkt5dQ at -e line 1.
Change your regular expression condition to:
/^\((\d+:)(.*?)\)$/
What that does is match and group the 4 digits and colon into special var $1 and the id you want into special var $2.
If every line of the log file is guaranteed to have an ID string, then you can write just
while (<$fh>) {
/:(\w+)/ and push #cs_ids , $1;
}
The \w ("word") character class matches alphanumeric characters or underscore, and this regex just snags the first sequence of word characters that follow a colon. It is best to avoid the non-greedy modifier if possible as it is a sloppy specification and can be much slower than a simple multiple character match.

regular expression for matching a string

I'm trying to remove a part of a given string using the either of the two rules:
Eliminate all the consonant(s) at the beginning of a string
Eliminate all but the consonants at the beginning of a string.
Suppose my string is str. Is ${str%%[aeoui]{1}*} correct for the second rule? I'm not sure what to do for the first rule.
I'm not sure what language you are trying to implement this in, so I'll just use some generic syntax.
1. s/^[^aeiouAEIOU]*(.*)/\1/
2. s/^[aeiouAEIOU]*(.*)/\1/
There are ways to make it case insensitive, but I like being specific like this just for clarity.
The only difference between the two is ^ inside the [] in #1 which just negates it.
* means zero or more. If you use +, for instance, there would have to be at least one consonant in #1 and at least one vowel in #2 or the test would fail.
In my generic syntax here \1 returns what was found by (.*).
Here's some very crude Perl to demonstrate (where $1 in the print statements behaves as \1 in my example above):
#!/usr/bin/perl
$string1="abcdef";
$string2="fedcba";
if ($string1 =~ /^[aeiouAEIOU]*(.*)/) {
print "Test 1 on $string1: $1\n";
}
if ($string2 =~ /^[aeiouAEIOU]*(.*)/) {
print "Test 1 on $string2: $1\n";
}
if ($string1 =~ /^[^aeiouAEIOU]*(.*)/) {
print "Test 2 on $string1: $1\n";
}
if ($string2 =~ /^[^aeiouAEIOU]*(.*)/) {
print "Test 2 on $string2: $1\n";
}
And here's the output:
Test 1 on abcdef: bcdef
Test 1 on fedcba: fedcba
Test 2 on abcdef: abcdef
Test 2 on fedcba: edcba

Regex to match only innermost delimited sequence

I have a string that contains sequences delimited by multiple characters: << and >>. I need a regular expression to only give me the innermost sequences. I have tried lookaheads but they don't seem to work in the way I expect them to.
Here is a test string:
'do not match this <<but match this>> not this <<BUT NOT THIS <<this too>> IT HAS CHILDREN>> <<and <also> this>>'
It should return:
but match this
this too
and <also> this
As you can see with the third result, I can't just use /<<[^>]+>>/ because the string may have one character of the delimiters, but not two in a row.
I'm fresh out of trial-and-error. Seems to me this shouldn't be this complicated.
#matches = $string =~ /(<<(?:(?!<<|>>).)*>>)/g;
(?:(?!PAT).)* is to patterns as [^CHAR]* is to characters.
$string = 'do not match this <<but match this>> not this <<BUT NOT THIS <<this too>> IT HAS CHILDREN>> <<and <also> this>>';
#matches = $string =~ /(<<(?:[^<>]+|<(?!<)|>(?!>))*>>)/g;
Here's a way to use split for the job:
my $str = 'do not match this <<but match this>> not this <<BUT NOT THIS <<this too>> IT HAS CHILDREN>> <<and <also> this>>';
my #a = split /(?=<<)/, $str;
#a = map { split /(?<=>>)/, $_ } #a;
my #match = grep { /^<<.*?>>$/ } #a;
Keeps the tags in there, if you want them removed, just do:
#match = map { s/^<<//; s/>>$//; $_ } #match;