I have a string:
set a "ODUCTP-1-1-1-2P1"
regexp {.*?\-(.*)} $a match sub
I expect the value of sub to be 1-1-1-2P1
But I'm getting empty string. Can any one tell me how to properly use the regex?
The problem is that the non-greediness of the .*? is leaking over to the .* later on, which is a feature of the RE engine being used (automata-theoretic instead of stack-based).
The simplest fix is to write the regular expression differently.
Because Tcl has unanchored regular expressions (by default) and starts matches as soon as it can, a greedy match from the first - to the end of the string is perfect (with sub being assigned everything after the -). That's a very simple RE: -(.*). To use that, you do this:
regexp -- {-(.*)} $a match sub
Note the --; it's needed here because the regular expression starts with a - symbol and is otherwise confused as weird (and unsupported) option. Apart from that one niggle, it's all entirely straight-forward.
$str = "ODUCTP-1-1-1-2P1";
$str =~ s/^.*?-//;
print $str;
or:
$str =~ /^.*?-(.*)$/;
print $1;
Related
I'm trying to match if a word such as *FOO (* as a normal character) is in a line. My input is a C++ source code. I need to use a pre-compiled regex for this due to program flow requirements, so I tried the following:
$pattern = qr/[^a-zA-Z](\*FOO)[^a-zA-Z]|^\s*(\*FOO)[^a-zA-Z]/;
And I use it like this:
if ($line =~ m/$pattern/) { ... }
It works and catches lines containing *FOO such as hey *FOO.BAR but also matches lines such as:
//FOO programming using stuff and things
which I want to ignore. What am I missing? Is \* not the right way to escape * in a pre-compiled regex in perl? If *FOO is stored in $word and the pattern looks like this:
$pattern = qr/[^a-zA-Z](\\$word)[^a-zA-Z]|^\s*(\\$word)[^a-zA-Z]/;
Is that different from the previous pattern? Because I tried both and the result seems to be the same.
I found a way to bypass this problem by removing the first char of $word and escaping * in the pattern, but if $word = "**.?FOO" for example, how do I create a qr// with $word so that all the meta-characters are escaped?
You do need to escape the *. One way to do it is by the quotemeta \Q operator:
use warnings;
use strict;
my $qr = qr/\Q*FOO/;
while (<DATA>) { print if /$qr/ }
__DATA__
//FOO programming using stuff and things
hey *FOO.BAR
Note that this escapes all ASCII non-"word" characters through the rest of the pattern. If you need to limit its action to only a part of the pattern then stop it using \E. Please see linked docs.
The above determines whether *FOO is in the line, regardless of whether it is a word or a part of one. It is not clear to me which is needed. Once that is specified the pattern can be adjusted.
Note that /\*FOO/ works, too. What you tried failed probably because of all the rest that you are trying to match, which purpose I do not understand. If you only need to detect whether the pattern is present the above does it. if there is a more specific requirement please clarify.
As for the examples: for me that string //FOO... is not matched by the main (first) $pattern you show. The second one won't interpolate $word -- but is firstly much too convoluted. The regex can really tie one in nasty knots when pushed; I suggest to keep it simple as much as possible.
Question 1:
my $word = '*FOO';
my $pattern = qr/\\$word/;
is equivalent to
my $pattern = qr/\\*FOO/; # zero or more '\' followed by 'FOO'
The $word is simply interpolated as is.
To get something equivalent to
my $pattern = qr/\*FOO/;
you should use
my $word = '*FOO';
my $pattern = qr/\Q$word\E/;
By default, an interpolated variable is considered a mini-regular expression, meta characters in the variable such as *, +, ? are still interpreted as meta character. \Q...\E will add a backslash before any character not matching /[A-Za-z_0-9]/, thus any meta characters in the interpolated variable is interpreted as literal ones. Refer to perldoc.
Question 2
I tried
my $pattern = qr/[^a-zA-Z](\*FOO)[^a-zA-Z]|^\s*(\*FOO)[^a-zA-Z]/;
my $line = '//FOO programming using stuff and things';
if($line =~ m/$pattern/){
print "$&\n";
}
else{
print "No match!";
}
and it printed "No match!". I can't explain how you get it matched.
I am trying to get the a sub string from a string using regular expression but it getting error as my regular expression is not working. Can any one help me out in writing correct one :
Here is the Pattern on which i am trying to write the regular expression :
MSM8_BD_V4.3_1-1_idle-Kr_Run3.xlsx
MSM8_BD_V4.3_2-6_mp3-Kr_Run2.xlsx
MSM8_BD_V4.3_Camera_snap-7.xlsx
MSM8_BD_V4.3_Camera_snap-8.xlsx
MSM8_BD_V4.3_Radio_202.16-0.xlsx
I am trying to get the bold part of the substring .
below is the Regular expression i tried:
my $line = "MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx";
my ($captured) = $line =~ /MSM8939_BD_V4\.\3\_[d]*(.+?)\w/gx;
print "$captured\n";
[d] matches nothing but the literal letter d. You want \d, without the brackets, to match a digit. However, it looks like you also want to include underscores. That would be [\d_].
Try this:
/^MSM8_BD_V4\.3_[\d_]*-?([^-]+)/
If I run this on your input (with e.g. perl -nE 'say $1 if /^MSM8_BD_V4\.3_[\d_]*-?([^-]+)/'), I get this output:
1_idle
6_mp3
Camera_snap
Camera_snap
Radio_202.16
my $line = "MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx";
for (qw(
MSM8939_BD_V4.3_1-1_idle-Kratos_Run3.xlsx
MSM8939_BD_V4.3_2-6_mp3-Kratos_Run2.xlsx
MSM8939_BD_V4.3_Camera_snap-7.xlsx
MSM8939_BD_V4.3_Camera_snap-8.xlsx
MSM8939_BD_V4.3_Radio_202.16-0.xlsx
)) {
my ($captured) = ($_ =~ /.*[-_]([^\W_]+_[\w.]+)-/gx);
print "$captured\n";
}
Use a greedy pattern to go as far as possible, then grab the last two strings that look like what you want which are still followed by a hyphen.
As does the other answer which was just edited while I was typing, this produces:
1_idle
6_mp3
Camera_snap
Camera_snap
Radio_202.16
This one may be more general in that the beginning of the substring is not hard-coded, i.e., you could use it in other cases which did not necessarily start with MSM8_BD_V4.3.
I want to find a regular expression in Perl which matches a pattern such as this:
my $sumthing = "people say
for -->";
Over here after say there is a single newline character. So I need to find a regular expression which could match such a pattern which includes a newline within a pattern. Please help me to find this as I'm new to Perl & regular expression.
The possible methods I tried were these:
if (($sumthing !~ (/\n+$/)) && ($sumthing !~ (/^\n+/m)))
They kindly help me to find out an expression to match this kind of a pattern, but not getting the output as desired.
It's not clear what you want. Do you want match that string exactly? If so, you could use
$sumthing =~ /^people say\nfor -->\z/
or
$sumthing eq "people say\nfor -->"
Or maybe what you need to know is that . matches any character including newline when /s is used?
/people .* -->/s
The following will check for anything then new line then anything. Not sure if I totally understood your question.
if($sumthing =~ m/.*\n.*/)
Have a look at the /s modifier which causes .to match anything, including a newline.
my $str = "people say for\nsomething...";
$str =~ m{say(.*)}s and print "'$1'\n";
This would print:
' for
something...'
I think this question is repeated, but searching wasn't helpful for me.
my $pattern = "javascript:window.open\('([^']+)'\);";
$mech->content =~ m/($pattern)/;
print $1;
I want to have an external $pattern in the regular expression. How can I do this? The current one returns:
Use of uninitialized value $1 in print at main.pm line 20.
$1 was empty, so the match did not succeed. I'll make up a constant string in my example of which I know that it will match the pattern.
Declare your regular expression with qr, not as a simple string. Also, you're capturing twice, once in $pattern for the open call's parentheses, once in the m operator for the whole thing, therefore you get two results. Instead of $1, $2 etc. I prefer to assign the results to an array.
my $pattern = qr"javascript:window.open\('([^']+)'\);";
my $content = "javascript:window.open('something');";
my #results = $content =~ m/($pattern)/;
# expression return array
# (
# q{javascript:window.open('something');'},
# 'something'
# )
When I compile that string into a regex, like so:
my $pattern = "javascript:window.open\('([^']+)'\);";
my $regex = qr/$pattern/;
I get just what I think I should get, following regex:
(?-xism:javascript:window.open('([^']+)');)/
Notice that it it is looking for a capture group and not an open paren at the end of 'open'. And in that capture group, the first thing it expects is a single quote. So it will match
javascript:window.open'fum';
but not
javascript:window.open('fum');
One thing you have to learn, is that in Perl, "\(" is the same thing as "(" you're just telling Perl that you want a literal '(' in the string. In order to get lasting escapes, you need to double them.
my $pattern = "javascript:window.open\\('([^']+)'\\);";
my $regex = qr/$pattern/;
Actually preserves the literal ( and yields:
(?-xism:javascript:window.open\('([^']+)'\);)
Which is what I think you want.
As for your question, you should always test the results of a match before using it.
if ( $mech->content =~ m/($pattern)/ ) {
print $1;
}
makes much more sense. And if you want to see it regardless, then it's already implicit in that idea that it might not have a value. i.e., you might not have matched anything. In that case it's best to put alternatives
$mech->content =~ m/($pattern)/;
print $1 || 'UNDEF!';
However, I prefer to grab my captures in the same statement, like so:
my ( $open_arg ) = $mech->content =~ m/($pattern)/;
print $open_arg || 'UNDEF!';
The parens around $open_arg puts the match into a "list context" and returns the captures in a list. Here I'm only expecting one value, so that's all I'm providing for.
Finally, one of the root causes of your problems is that you do not need to specify your expression in a string in order for your regex to be "portable". You can get perl to pre-compile your expression. That way, you only care what instructions the characters are to a regex and not whether or not you'll save your escapes until it is compiled into an expression.
A compiled regex will interpolate itself into other regexes properly. Thus, you get a portable expression that interpolates just as well as a string--and specifically correctly handles instructions that could be lost in a string.
my $pattern = qr/javascript:window.open\('([^']+)'\);/;
Is all that you need. Then you can use it, just as you did. Although, putting parens around the whole thing, would return the whole matched expression (and not just what's between the quotes).
You do not need the parentheses in the match pattern. It will match the whole pattern and return that as $1, which I am guess is not matching, but I am only guessing.
$mech->content =~ m/$pattern/;
or
$mech->content =~ m/(?:$pattern)/;
These are the clustering, non-capturing parentheses.
The way you are doing it is correct.
The solutions have been already given, I'd like to point out that the window.open call might have multiple parameters included in "" and grouped by comma like:
javascript:window.open("http://www.javascript-coder.com","mywindow","status=1,toolbar=1");
There might be spaces between the function name and parentheses, so I'd use a slighty different regex for that:
my $pattern = qr{
javascript:window.open\s*
\(
([^)]+)
\)
}x;
print $1 if $text =~ /$pattern/;
Now you have all parameters in $1 and can process them afterwards with split /,/, $stuff and so on.
It reports an uninitialized value because $1 is undefined. $1 is undefined because you have created a nested matching group by wrapping a second set of parentheses around the pattern. It will also be undefined if nothing matches your pattern.
How do I return the first 5 digits of a string of characters in Regular Expressions?
For example, if I have the following text as input:
15203 Main Street
Apartment 3 63110
How can I return just "15203".
I am using C#.
This isn't really the kind of problem that's ideally solved by a single-regex approach -- the regex language just isn't especially meant for it. Assuming you're writing code in a real language (and not some ill-conceived embedded use of regex), you could do perhaps (examples in perl)
# Capture all the digits into an array
my #digits = $str =~ /(\d)/g;
# Then take the first five and put them back into a string
my $first_five_digits = join "", #digits[0..4];
or
# Copy the string, removing all non-digits
(my $digits = $str) =~ tr/0-9//cd;
# And cut off all but the first five
$first_five_digits = substr $digits, 0, 5;
If for some reason you really are stuck doing a single match, and you have access to the capture buffers and a way to put them back together, then wdebeaum's suggestion works just fine, but I have a hard time imagining a situation where you can do all that, but don't have access to other language facilities :)
it would depend on your flavor of Regex and coding language (C#, PERL, etc.) but in C# you'd do something like
string rX = #"\D+";
Regex.replace(input, rX, "");
return input.SubString(0, 5);
Note: I'm not sure about that Regex match (others here may have a better one), but basically since Regex itself doesn't "replace" anything, only match patterns, you'd have to look for any non-digit characters; once you'd matched that, you'd need to replace it with your languages version of the empty string (string.Empty or "" in C#), and then grab the first 5 characters of the resulting string.
You could capture each digit separately and put them together afterwards, e.g. in Perl:
$str =~ /(\d)\D*(\d)\D*(\d)\D*(\d)\D*(\d)/;
$digits = $1 . $2 . $3 . $4 . $5;
I don't think a regular expression is the best tool for what you want.
Regular expressions are to match patterns... the pattern you are looking for is "a(ny) digit"
Your logic external to the pattern is "five matches".
Thus, you either want to loop over the first five digit matches, or capture five digits and merge them together.
But look at that Perl example -- that's not one pattern -- it's one pattern repeated five times.
Can you do this via a regular expression? Just like parsing XML -- you probably could, but it's not the right tool.
Not sure this is best solved by regular expressions since they are used for string matching and usually not for string manipulation (in my experience).
However, you could make a call to:
strInput = Regex.Replace(strInput, "\D+", "");
to remove all non number characters and then just return the first 5 characters.
If you are wanting just a straight regex expression which does all this for you I am not sure it exists without using the regex class in a similar way as above.
A different approach -
#copy over
$temp = $str;
#Remove non-numbers
$temp =~ s/\D//;
#Get the first 5 numbers, exactly.
$temp =~ /\d{5}/;
#Grab the match- ASSUMES that there will be a match.
$first_digits = $1
result =~ s/^(\d{5}).*/$1/
Replace any text starting with a digit 0-9 (\d) exactly 5 of them {5} with any number of anything after it '.*' with $1, which is the what is contained within the (), that is the first five digits.
if you want any first 5 characters.
result =~ s/^(.{5}).*/$1/
Use whatever programming language you are using to evaluate this.
ie.
regex.replace(text, "^(.{5}).*", "$1");