I'm trying to get a regex that matches: (It should not match any other string)
Word1 or Word2 or Word3 or Wordn
Capturing the words between before or after an "or"
1: Word1
2: Word2
3: Word3
n: Wordn
I've tried modifying a csv regex:
(?:^|,)(\"(?:[^\"]+|\"\")*\"|[^,]*)
to
(?:^|(?:or)((?:[^(?:or)]+)*|[^(?:or)]*)
But that does not give me what I want.
I'm sure I'm missing something, but I've been banging my head for hours.
How about:
my $string = " foo or bar or foobar ";
if ( $string =~ m|^\s*[^\s]+(\s+or\s+[^\s]+)+\s*$| ) {
my $tmp = "$string";
$tmp =~ s|^\s+||;
$tmp =~ s|\s+$||;
my #words = split( /\s+or\s+/, $tmp );
printf( "Found %d words:\n", scalar( #words ) );
foreach my $word ( #words ) {
print( "\t'$word'\n" );
}
} else {
print( "No match\n" );
}
The above will output:
Found 3 words:
'foo'
'bar'
'foobar'
Try splitting the string on ' or '.
You know, this isn't something for which I'd naturally reach for regex. I'd try a split first.
my #words = split / or /, $string;
This regex will match any string that has at least word1 or word2, and any number more or's after that. It must have no whitespace at the beginning or end of the string as well, but you can remove the ^ and $ if you want to search for a string of this form within a larger string
(?:^(\w+)(?=\s+or))|(?:\s+or\s+(\w+))+
RegexPal
The real solution is to split on ' or '. A regex solution is not so straight forward.
$sm =~ / or / and #between_or = $sm =~ /(?:^\s*|(?<= or ))(.+?)(?= or |\s*$)/sg;
Related
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";
I'm just starting to learn Perl. I need to parse JavaScript file. I came up with the following subroutine, to do it:
sub __settings {
my ($_s) = #_;
my $f = $config_directory . "/authentic-theme/settings.js";
if ( -r $f ) {
for (
split(
'\n',
$s = do {
local $/ = undef;
open my $fh, "<", $f;
<$fh>;
}
)
)
{
if ( index( $_, '//' ) == -1
&& ( my #m = $_ =~ /(?:$_s\s*=\s*(.*))/g ) )
{
my $m = join( '\n', #m );
$m =~ s/[\'\;]//g;
return $m;
}
}
}
}
I have the following regex, that removes ' and ; from the string:
s/[\'\;]//g;
It works alright but if there is a mentioned chars (' and ;) in string - then they are also removed. This is undesirable and that's where I stuck as it gets a bit more complicated for me and I'm not sure how to change the regex above correctly to only:
Remove only first ' in string
Remove only last ' in string
Remove ont last ; in string if exists
Any help, please?
You can use the following to match:
^'|';?$|;$
And replace with '' (empty string)
See DEMO
Remove only first ' in string
Remove only last ' in string
^[^']*\K'|'(?=[^']*$)
Try this .See demo.
https://regex101.com/r/oF9hR9/8
Remove ont last ; in string if exists
;(?=[^;]*$)
Try this.See demo.
https://regex101.com/r/oF9hR9/9
All three in one
^[^']*\K'|'(?=[^']*$)|;(?=[^;]*$)
See Here
You can use this code:
#!/usr/bin/perl
$str = "'string; 'inside' another;";
$str =~ s/^'|'?;?$//g;
print $str;
IDEONE demo
The main idea is to use anchors: ^ beginning of string, $ end of string and ;? matches the ";" symbol at the end only if it is present (? quantifier is making the pattern preceding it optional).EDIT: Also, ; will get removed even if there is no preceding '.
I suggest that your original code should look more like this. It is much more idiomatic Perl and I think more straightforward to follow
sub __settings {
my ($_s) = #_;
my $file = "$config_directory/authentic-theme/settings.js";
return unless -r $file;
open my $fh, '<', $file or die qq{Unable to open "$file" for input: $!};
my #file = <$fh>;
chomp #file;
for ( #file ) {
next if m{//};
if ( my #matches = $_ =~ /(?:$_s\s*=\s*(.*))/g ) {
my $matches = join "\n", #matches;
$matches =~ tr/';//d;
return $matches;
}
}
}
I want to match only the strings that have exactly one vowel.
I tried this code, and it works but it also matches those strings that haven't any vowels (for example hshs, ksks, lslsl) and I need only the strings that have just one vowel
if ( $string !~ /\*w[aeiou]\w*[aeiou]\W*/ ) {
print $string;
}
You can use tr/// to count the occurrences of letters in a string.
Something like this perhaps
use strict;
use warnings;
for my $string ( qw/ a fare is paid for every cab /) {
if ( $string =~ tr/aeiuoAEIOU// == 1 ) {
print $string, "\n";
}
}
output
a
is
for
cab
Make it simple, at least one vowel:
if ($string =~ /[aeiou]/i) {
print $string;
}
exactly one vowel:
if ($string =~ /^[^aeiou]*[aeiou][^aeiou]*$/i) {
print $string;
}
Make a pattern that will match three consecutive copies of whatever is currently contained in $what. That is, if $what is fred, your pattern should match fredfredfred. If $what is fred|barney, your pattern should match fredfredbarney, barneyfredfred, barneybarneybarney, or many other variations. (Hint: You should set $what at the top of the pattern test program with a statement like my $what = 'fred|barney';)
But my solution to this is just too easy so I'm assuming its wrong. My solution is:
#! usr/bin/perl
use warnings;
use strict;
while (<>){
chomp;
if (/fred|barney/ig) {
print "pattern found! \n";
}
}
It display what I want. And I didn't even have to save the pattern in a variable. Can someone help me through this? Or enlighten me if I'm doing/understanding the problem wrong?
This example should clear up what was wrong with your solution:
my #tests = qw(xxxfooxx oofoobar bar bax rrrbarrrrr);
my $str = 'foo|bar';
for my $test (#tests) {
my $match = $test =~ /$str/ig ? 'match' : 'not match';
print "$test did $match\n";
}
OUTPUT
xxxfooxx did match
oofoobar did match
bar did match
bax did not match
rrrbarrrrr did match
SOLUTION
#!/usr/bin/perl
use warnings;
use strict;
# notice the example has the `|`. Meaning
# match "fred" or "barney" 3 times.
my $str = 'fred|barney';
my #tests = qw(fred fredfredfred barney barneybarneybarny barneyfredbarney);
for my $test (#tests) {
if( $test =~ /^($str){3}$/ ) {
print "$test matched!\n";
} else {
print "$test did not match!\n";
}
}
OUTPUT
$ ./test.pl
fred did not match!
fredfredfred matched!
barney did not match!
barneybarneybarny did not match!
barneyfredbarney matched!
use strict;
use warnings;
my $s="barney/fred";
my #ra=split("/", $s);
my $test="barneybarneyfred"; #etc, this will work on all permutations
if ($test =~ /^(?:$ra[0]|$ra[1]){3}$/)
{
print "Valid\n";
}
else
{
print "Invalid\n";
}
Split delimited your string based off of "/". (?:$ra[0]|$ra[1]) says group, but do not extract, "barney" or "fred", {3} says exactly three copies. Add an i after the closing "/" if the case doesn't matter. The ^ says "begins with," and the $ says "ends with."
EDIT:
If you need the format to be barney\fred, use this:
my $s="barney\\fred";
my #ra=split(/\\/, $s);
If you know that the matching will always be on fred and barney, then you can just replace $ra[0], $ra[1] with fred and barney.
How can I count the amount of spaces at the start of a string in Perl?
I now have:
$temp = rtrim($line[0]);
$count = ($temp =~ tr/^ //);
But that gives me the count of all spaces.
$str =~ /^(\s*)/;
my $count = length( $1 );
If you just want actual spaces (instead of whitespace), then that would be:
$str =~ /^( *)/;
Edit: The reason why tr doesn't work is it's not a regular expression operator. What you're doing with $count = ( $temp =~ tr/^ // ); is replacing all instances of ^ and with itself (see comment below by cjm), then counting up how many replacements you've done. tr doesn't see ^ as "hey this is the beginning of the string pseudo-character" it sees it as "hey this is a ^".
You can get the offset of a match using #-. If you search for a non-whitespace character, this will be the number of whitespace characters at the start of the string:
#!/usr/bin/perl
use strict;
use warnings;
for my $s ("foo bar", " foo bar", " foo bar", " ") {
my $count = $s =~ /\S/ ? $-[0] : length $s;
print "'$s' has $count whitespace characters at its start\n";
}
Or, even better, use #+ to find the end of the whitespace:
#!/usr/bin/perl
use strict;
use warnings;
for my $s ("foo bar", " foo bar", " foo bar", " ") {
$s =~ /^\s*/;
print "$+[0] '$s'\n";
}
Here's a script that does this for every line of stdin. The relevant snippet of code is the first in the body of the loop.
#!/usr/bin/perl
while ($x = <>) {
$s = length(($x =~ m/^( +)/)[0]);
print $s, ":", $x, "\n";
}
tr/// is not a regex operator. However, you can use s///:
use strict; use warnings;
my $t = (my $s = " \t\n sdklsdjfkl");
my $n = 0;
++$n while $s =~ s{^\s}{};
print "$n \\s characters were removed from \$s\n";
$n = ( $t =~ s{^(\s*)}{} ) && length $1;
print "$n \\s characters were removed from \$t\n";
Since the regexp matcher returns the parenthesed matches when called in a list context, CanSpice's answer can be written in a single statement:
$count = length( ($line[0] =~ /^( *)/)[0] );
This prints amount of white space
echo " hello" |perl -lane 's/^(\s+)(.*)+$/length($1)/e; print'
3