I have to extract string with particular format from a file. i.e string format is 1 followed by hyphen and 7 digits.
for ex.
#CARES# AR_NUMBER=1-4742637
here I have to extract only 1-4742637.
Help me, how to extract?
The following will capture that: /\b(1-\d{7})\b/
As demonstrated:
use strict;
use warnings;
my $text = <<'END_TEXT';
for ex.
#CARES# AR_NUMBER=1-4742637
END_TEXT
if ($text =~ /\b(1-\d{7})\b/) {
print "$1";
}
Outputs:
1-4742637
if ($subject =~ m/(1-[\d]+)/) {
# Successful match
} else {
# Match attempt failed
}
(1-[\d]+)
Match the regular expression below and capture its match into backreference number 1 «(1-[\d]+)»
Match the characters “1-” literally «1-»
Match a single digit 0..9 «[\d]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
1-\d{7}
will select the required part.
Please try
$var =~ /1-\d{7}/
{} for number of matches
You can do that as:
([0-9-]+)
or specifically for your case:
(1-\d{7})
and the first captured group \1 or $1 will contain what you want.
Demo: http://regex101.com/r/pY4gB6
Related
i searching to find some Perl Regular Expression Syntax about some requirements i have in a project.
First i want to exclude strings from a txt file (dictionary).
For example if my file have this strings:
path.../Document.txt |
tree
car
ship
i using Regular Expression
a1testtre -- match
orangesh1 -- match
apleship3 -- not match [contains word from file ]
Also i have one more requirement that i couldnt solve. I have to create a Regex that not allow a String to have over 3 times a char repeat (two chars).
For example :
adminnisstrator21 -- match (have 2 times a repetition of chars)
kkeeykloakk -- not match have over 3 times repetition
stack22ooverflow -- match (have 2 times a repetition of chars)
for this i have try
\b(?:([a-z])(?!\1))+\b
but it works only for the first char-reppeat
Any idea how to solve these two?
To not match a word from a file you might check whether a string contains a substring or use a negative lookahead and an alternation:
^(?!.*(?:tree|car|ship)).*$
^ Assert start of string
(?! negative lookahead, assert what is on the right is not
.*(?:tree|car|ship) Match 0+ times any char except a newline and match either tree car or ship
) Close negative lookahead
.* Match any char except a newline
$ Assert end of string
Regex demo
To not allow a string to have over 3 times a char repeat you could use:
\b(?!(?:\w*(\w)\1){3})\w+\b
\b Word boundary
(?! Negative lookahead, assert what is on the right is not
(?: NOn capturing group
\w*(\w)\1 Match 0+ times a word character followed by capturing a word char in a group followed by a backreference using \1 to that group
){3} Close non capturing group and repeat 3 times
) close negative lookahead
\w+ Match 1+ word characters
\b word boundary
Regex demo
Update
According to this posted answer (which you might add to the question instead) you have 2 patterns that you want to combine but it does not work:
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\1){4}))*$)
In those 2 patterns you use 2 capturing groups, so the second pattern has to point to the second capturing group \2.
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\2){4}))*$)
^
Pattern demo
One way to exclude strings that contain words from a given list is to form a pattern with an alternation of the words and use that in a regex, and exclude strings for which it matches.
use warnings;
use strict;
use feature qw(say);
use Path::Tiny;
my $file = shift // die "Usage: $0 file\n"; #/
my #words = split ' ', path($file)->slurp;
my $exclude = join '|', map { quotemeta } #words;
foreach my $string (qw(a1testtre orangesh1 apleship3))
{
if ($string !~ /$exclude/) {
say "OK: $string";
}
}
I use Path::Tiny to read the file into a a string ("slurp"), which is then split by whitespace into words to use for exclusion. The quotemeta escapes non-"word" characters, should any happen in your words, which are then joined by | to form a string with a regex pattern. (With complex patterns use qr.)
This may be possible to tweak and improve, depending on your use cases, for one in regards to the order of of patterns with common parts in alternation.†
The check that successive duplicate characters do not occur more than three times
foreach my $string (qw(adminnisstrator21 kkeeykloakk stack22ooverflow))
{
my #chars_that_repeat = $string =~ /(.)\1+/g;
if (#chars_that_repeat < 3) {
say "OK: $string";
}
}
A long string of repeated chars (aaaa) counts as one instance, due to the + quantifier in regex; if you'd rather count all pairs remove the + and four as will count as two pairs. The same char repeated at various places in the string counts every time, so aaXaa counts as two pairs.
This snippet can be just added to the above program, which is invoked with the name of the file with words to use for exclusion. They both print what is expected from provided samples.
† Consider an example with exclusion-words: so, sole, and solely. If you only need to check whether any one of these matches then you'd want shorter ones first in the alternation
my $exclude = join '|', map { quotemeta } sort { length $a <=> length $b } #words;
#==> so|sole|solely
for a quicker match (so matches all three). This, by all means, appears to be the case here.
But, if you wanted to correctly identify which word matched then you must have longer words first,
solely|sole|so
so that a string solely is correctly matched by its word before it can be "stolen" by so. Then in this case you'd want it the other way round,
sort { length $b <=> length $a }
I hope someone else will come with a better solution, but this seems to do what you want:
\b Match word boundary
(?: Start capture group
(?:([a-z0-9])(?!\1))* Match all characters until it encounters a double
(?:([a-z0-9])\2)+ Match all repeated characters until a different one is reached
){0,2} Match capture group 0 or 2 times
(?:([a-z0-9])(?!\3))+ Match all characters until it encounters a double
\b Match end of word
I changed the [a-z] to also match numbers, since the examples you gave seem to also include numbers. Perl regex also has the \w shorthand, which is equivalent to [A-Za-z0-9_], which could be handy if you want to match any character in a word.
My problem is that i have 2 regex that working:
Not allow over 3 pairs of chars:
(?=^(?!(?:\w*(.)\1){3}).+$)
Not allow over 4 times a char to repeat:
(?=^(?:(.)(?!(?:.*?\1){4}))*$)
Now i want to combine them into one row like:
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\1){4}))*$)
but its working only the regex that is first and not both of them
As mentioned in comment to #zdim's answer, take it a bit further by making sure that the order in which your words are assembled into the match pattern doesn't trip you. If the words in the file are not very carefully ordered to start, I use a subroutine like this when building the match string:
# Returns a list of alternative match patterns in tight matching order.
# E.g., TRUSTEES before TRUSTEE before TRUST
# TRUSTEES|TRUSTEE|TRUST
sub tight_match_order {
return #_ unless #_ > 1;
my (#alts, #ordered_alts, %alts_seen);
#alts = map { $alts_seen{$_}++ ? () : $_ } #_;
TEST: {
my $alt = shift #alts;
if (grep m#$alt#, #alts) {
push #alts => $alt;
} else {
push #ordered_alts => $alt;
}
redo TEST if #alts;
}
#ordered_alts
}
So following #zdim's answer:
...
my #words = split ' ', path($file)->slurp;
#words = tight_match_order(#words); # add this line
my $exclude = join '|', map { quotemeta } #words;
...
HTH
I am trying to match strings of digits that contain non-digits within them. Using the default text in http://regexr.com/, the following should match:
v2.1
-98.7
3.141
.6180
9,000
+42
555.123.4567
+1-(800)-555-2468
The following should not match:
0123456789
12345
I tried:
/[^\n\ ]{1,}\d+\S+\d/g
But it would not match +42 and it incorrectly matched 0123456789 and 12345, and it treated "555.123.4567 +1-(800)-555-2468" as one string.
I tried to alleviate it by putting a $ at the end but that matched nothing. Not sure what I am doing wrong.
You can use this regex to match any text with at least one non-digit:
/^\d*[^\d\n]+\d.*$/mg
RegEx Demo
RegEx Breakup:
^ - Start
\d* - Match 0 or more digits
[^\d\n]+ - Match 1 or more of any character that is not a digit and not a newline
\d - Match a digit
.* - Match 0 or more of any character
$ - End
Try this:
^(?=.*\d)(?=.*[^\d\s])\S+$
This means "at least one digit and one non-digit and no whitespace".
See live demo.
If no newlines were in your input, you could use slightly simpler:
^(?=.*\d)(?=.*\D)\S+$
Aren't you over-thinking this massively? What's wrong with using /\D/ to match a string that contains a non-digit?
I'm not sure what your exact requirements are, but if you're looking for a string that contains at least one digit and at least one non-digit, then the easiest approach is to use to regex matches - /\d/ && /\D/.
#!/usr/bin/perl
use strict;
use warnings;
use 5.010;
while (<DATA>) {
chomp;
say "$_: " . (/\d/ && /\D/ ? 'matches' : 'doesn\'t match');
}
__DATA__
v2.1
-98.7
3.141
.6180
9,000
+42
555.123.4567
+1-(800)-555-2468
0123456789
12345
Looks like you want to dodge strings made up entirely of digits, or entirely of letters. So you can exclude those. That will also let in strings without any numbers, so also require a number.
my $exclude = qr/(?: [0-9]+ | [A-Za-z]+ )/x;
my #res = grep { not /^$exclude$/ and /\d/ } #strings;
If any other characters need be excluded (underscore?), add it to the list.
It is not clear how your input is coming, this takes a list of ready strings. Add word boundaries and/or /s, depending on the input. Or parse the input into a list of strings for this.
If input comes as as a multi-line string, my #strings = split '\n|\s+', $text;.
I have a string that can read something like this (although not always, numbers can vary).
Board Length,45,inches,color,board height,8,inches,black,store,wal-mart,Board weight,20,dollars
I am trying to match the 45 that follows the Board Length this regex expression.
if ($string =~/Board Length,(\d+\.\d+)/){
print $string;
}
Is the formatting wrong? I thought d+ would match as many numbers as needed, . would match a literal '.', and d+ would match any numbers after the decimal (if there are any).
As you have put it, decimal . and following digits are mandatory. Thus (\.\d+)? to make it optional,
if ($string =~/Board Length,(\d+(?:\.\d+)?)/)
You are absolutely right about what that should match. However, without the '?' character, you are specifying that all of those pieces must be present.
\d+\.\d+
This means "1 or more numbers, period, 1 or more numbers"
1.5, 253333.7, 0.0 would all be matched. However, your example uses 45, which has no "." in it, nor numbers afterward. There are a few solutions to your problem, the most full proof of which was stated above by mpapac. Allow the decimal and following digits to be optional.
(\.\d+)?
The problem with this as such is that putting a () around it makes it another capture group. You may or may not want this. Putting the ?: inside it means "Use this as a group, but don't capture it." Hence:
(?:\.\d+)?
The other option is not to do the grouping, and instead make both the decimal itself optional and the digits after the decimal ZERO or more instead of ONE or more. That would look something like this:
\d+\.?\d*
You are not printing what you capture. You're printing $_ which we don't know what it is.
if ($string =~/Board Length,(\d+\.\d+)/){
print $_;
}
What I think you want is:
if ($string =~/Board Length,(\d+\.\d+)/){
print $1;
}
You have the following expression:
$string =~/Board Length,(\d+\.\d+) /
Your string is this:
Board Length,45,inches
The string Board Length will match the pattern Board Length,. However, the rest of our pattern is matching one or more digits followed by a period follows by one or more digits. This doesn't match the string 45. There's no decimal there.
The question is what are you trying to match. For example, if the number is surrounded by commas, you could do this:
$string =~ /Board Length,([^,]+),/;
my $number = $1;
The [^,] means Not a comma. You're capturing everything after a comma to the next comma. This will allow you to capture 45, 45.32, and even 4.5e+10. Just anything between the two commas.
Note that you use $1 for your first capture group and not $_.
Another way is to use non-greedy matching:
$string =~ /Board Length,(.+?),/;
my $number = $1;
What happens if what is captured isn't a number? You can test for that using the looks_like_number function from Scalar::Util (which has been included in Perl distributions for a long time).:
use Scalar::Util qw(looks_like_number);
my $string = "Board Length,Extra long,feet,...";
...
$string =~ /Board Length,(.+?),/;
my $number = $1;
if ( looks_like_number( $number ) ) {
print "$number is a number\n";
}
else {
print "Nope. $number isn't a number\n";
}
I have a body of text I'm looking to pull repeat sets of 4-digit numbers out from.
For Example:
The first is 1234 2) The Second is 2098 3) The Third is 3213
Now I know i'm able to get the first set of digits out by simply using:
/\d{4}/
...returning 1234
But how do I match the second set of digits, or the third, and so on...?
edit: How do i return 2098, or 3213
You don't appear to have a proper answer to your question yet.
The solution is to use the /g modifier on your regex. In list context it will find all of the numbers in your string at once, like this
my $str = 'The first is 1234 2) The Second is 2098 3) The Third is 3213';
my #numbers = $str =~ /\b \d{4} \b/gx;
print "#numbers\n";
output
1234 2098 3213
Or you can iterate through them, using scalar context in a while loop, like this
while ($str =~ /\b (\d{4}) \b/gx) {
my $number = $1;
print $number, "\n";
}
output
1234
2098
3213
I have added the \b patterns to the regex so that it only matches whole four-digit numbers and doesn't, for example, find 1234 in 1234567. The /x modifier just allows me to add spaces so that the pattern is more intelligible.
See http://perldoc.perl.org/perlre.html for discussion on the use of the 'g' modifier which will cause your regular expression to match ALL occurrances of its pattern, not just the first.
If you want a pattern that finds the $n'th 4-digit group, this seems to work:
$pat = "^(?:.*?\\b(\\d{4})\\b){$n}";
if ($s =~ /$pat/) {
print "Found $1\n";
} else {
print "Not found\n";
}
I did this by building a string pattern because I couldn't get a variable interpolated into a quantifier {$n}.
This pattern finds 4-digit groups that are on word boundaries (the \b tests); I don't know if that meets your requirements. The pattern uses .*? to ensure that as few characters as possible are matched between each four-digit group. The pattern is matched $n times, and the capture group $1 is set to whatever it was in the last iteration, i.e. the $n'th one.
EDIT: When I just tried it again, it seemed to interpolate $n in a quantifier just fine. I don't know what I did differently that it didn't work last time. So maybe this will work:
if ($s =~ /^(?:.*?\b(\d{4}\b){$n}/) { ...
If not, see amon's comment about qr//.
If the regex is only matched once, then match all three in one regex and extract them using matched groups:
^.*\b(\d{4})\b.*\b(\d{4})\b.*\b(\d{4})\b.*$
The three 4-digit numbers will be captured in group 1. 2 and 3.
Ajb's answer with "gx" is the best. If you know you will have three numbers, this straighforward line does the trick:
my $str = 'The first is 1234 2) The Second is 2098 3) The Third is 3213';
my ($num1, $num2, $num3) = $str =~ /\b \d{4} \b/gx;
print "$num1, $num2, $num3\n";
I have a String variable containing something like ABCD.asd.qwe.com:/dir1.
I want to extract the ABCD portion i.e. the portion from beginning till the first appearance of .. The problem is that there can be almost any characters (only alphanumeric) of any length before the .. So I created this regexp.
if($arg =~ /(.*?\.?)/)
{
my $temp_name = $1;
}
However it is giving me blank string. The logic is that :
.*? - any character non-greedily
\.? - till first or none appearance of .
What could be wrong?
You can instead use negative character class like this
^[^.]+
[^.] would match any character except .
[^.]+ would match 1 to many characters(except .)
^ depicts the start of string
OR
^.+?(?=\.|$)
(?=) is a lookahead which checks for a particular pattern after the current position..So for text abcdad with regex a(?=b) only a would match
$ depicts the end of line(if used with multiline option) or end of string(if used with singleline option)
\.? doesn't mean "till first or none appearance of .". It means "a . here or not".
If the first character of the string is .:
.*? matches 0 chars at position 0.
\.? matches 1 char at position 0.
$1 contains ..
If the first character of the string isn't .:
.*? matches 0 chars at position 0.
\.? matches 0 chars at position 0.
$1 is empty.
To match ABCD, the following would do:
/^(.*?)\./
However, I hate the non-greedy modifier. It's fragile, in the sense that it stops doing what you want if you use two in the same pattern. I'd use the following instead ("match non-periods"):
/^([^.]*)\./
or even just
/^([^.]*)/
use strict;
my $string = "ABCD.asd.qwe.com:/dir1";
$string =~ /([^.]+)/;
my $capture = $1;
print"$capture\n";
OR you can also use Split function like,
my $sub_string = ( split /\./, $string )[0];
print"$sub_string\n";
Note in general: For the explaination of Regex (understanding the complex Regex), take a look at YAPE::Regex::Explain module.
This should work:
if($arg =~ /(.*?)\..+/)
{
my $temp_name = $1;
}
That would match anything before the first . .
You could change the .+ to .* if your input may end after the first ..
You could change the first .*? to .+? if you are sure that there is always at least one character before the first ..