I am trying to grab any digits in front of a known line number of a phone, if they exist (in Perl). There will be no dashes, only digits.
For example, say I know the line number will always be 8675309. 8675309 may or may not have leading digits, if it does I want to capture them. There is not really a limit on the number of leading digits.
$input $digits $number
'8675309' '' '8675309'
'8008675309' '800' '8675309'
'18888675309' '1888' '8675309'
'18675309' '1' '8675309'
'86753091' not a match
/8675309$/ this will match how to capture the pre-digits in one regex?
Some regexes work better backwards than forwards. So sometimes it is useful to use sexeger, rather than regexes.
my $pn = '18008675309';
reverse($pn) =~ /^9035768(\d*)/;
my $got = reverse $1;
The regex is cleaner and avoids a lot of back tracking at the cost of some fummery with reversing the input and captured values.
The backtracking gain is smaller in this case than it would be if you had a general phone number extraction regex:
Regex: /^(\d*)\d{7}$/
Sexeger: /^\d{7}(\d*)/
There is a whole class of problems where this technique is useful. For more info see the sexeger post on Perlmonks.
my($digits,$number);
if ($input =~ /^(\d*)(8675309)$/) {
($digits,$number) = ($1,$2);
}
The * quantifier is greedy, but that means it matches as much as possible while still allowing a match. So initially, yes, \d* tries to gobble up all the digits in $number, but it reluctantly gives up character-by-character what it's matched until the whole pattern matches successfully.
Another approach is to chop off the tail:
(my $digits = $input) =~ s/8675309$//;
You could do the same without using a regular expression:
my $digits = $input;
substr($digits, -7) = "";
The above, at least with perl-5.10-1, could even be condensed to
substr(my $digits = $input, -7) = "";
The regex special variables $` and $& are another way of grabbing those pieces of information. They hold the contents of the data preceding the match and the match itself respectively.
if ( /8675309$/ )
{
printf( "%s,%s,%s\n", $_, $`, $& );
}
else
{
printf( "%s,Not a match\n", $_ );
}
There's a Perl package that deals with at least UK and US phone numbers.
It's called Number::Phone and the code is somewhere on the cpan.org site.
How about /(\d)?(8675309)/?
UPDATE:
whoops that should haev been /(\d*)(8675309)/
I might not understand the problem. Why is there a difference between the first and fourth examples:
'8675309' '' '8675309'
...
'8675309' '1' '8675309'
If all you want is to separate the last seven digits from everything else, you could have said it that way rather than provide confusing examples. A regex for that would be:
/(\d*)(\d{7,7})$/
If you weren't just providing a hypothetical number, and really are only looking for lines with '8675309' (seems strange), replace the '\d{7,7}' with '8675309'.
Related
I am trying to split texts into "steps"
Lets say my text is
my $steps = "1.Do this. 2.Then do that. 3.And then maybe that. 4.Complete!"
I'd like the output to be:
"1.Do this."
"2.Then do that."
"3.And then maybe that."
"4.Complete!"
I'm not really that good with regex so help would be great!
I've tried many combination like:
split /(\s\d.)/
But it splits the numbering away from text
I would indeed use split. But you need to exclude the digit from the match by using a lookahead.
my #steps = split /\s+(?=\d+\.)/, $steps;
All step-descriptions start with a number followed by a period and then have non-numbers, until the next number. So capture all such patterns
my #s = $steps =~ / [0-9]+\. [^0-9]+ /xg;
say for #s;
This works only if there are surely no numbers in the steps' description, like any approach relying on matching a number (even if followed by a period, for decimal numbers)†
If there may be numbers in there, we'd need to know more about the structure of the text.
Another delimiting pattern to consider is punctuation that ends a sentence (. and ! in these examples), if there are no such characters in steps' description and there are no multiple sentences
my #s = $steps =~ / [0-9]+\. .*? [.!] /xg;
Augment the list of patterns that end an item's description as needed, say with a ?, and/or ." sequence as punctuation often goes inside quotes.‡
If an item can have multiple sentences, or use end-of-sentence punctuation mid-sentence (as a part of a quotation perhaps) then tighten the condition for an item's end by combining footnotes -- end-of-sentence punctuation and followed by number+period
my #s = $steps =~ /[0-9]+\. .*? (?: \."|\!"|[.\!]) (?=\s+[0-9]+\. | \z)/xg;
If this isn't good enough either then we'd really need a more precise description of that text.
† An approach using a "numbers-period" pattern to delimit item's description, like
/ [0-9]+\. .*? (?=\s+[0-9]+\. | \z) /xg;
(or in a lookahead in split) fails with text like
1. Only $2.50 or 1. Version 2.4.1 ...
‡ To include text like 1. Do "this." and 2. Or "that!" we'd want
/ [0-9]+\. .*? (?: \." | !" | [.!?]) /xg;
Following sample code demonstrates power of regex to fill up %steps hash in one line of code.
Once the data obtained you can dice and slice it anyway your heart desires.
Inspect the sample for compliance with your problem.
use strict;
use warnings;
use feature 'say';
use Data::Dumper;
my($str,%steps,$re);
$str = '1.Do this. 2.Then do that. 3.And then maybe that. 4.Complete!';
$re = qr/(\d+)\.(\D+)\./;
%steps = $str =~ /$re/g;
say Dumper(\%steps);
say "$_. $steps{$_}" for sort keys %steps;
Output
$VAR1 = {
'1' => 'Do this',
'2' => 'Then do that',
'3' => 'And then maybe that'
};
1. Do this
2. Then do that
3. And then maybe that
My string is $tables="newdb1.table1:100,db2.table2:90,db1.table1:90". My search string is db1.table1 and my aim is to extract the value after : (i.e 90 in this case).
I am using:
if ($tables =~ /db1.table1:(\d+)/) { print $1; }
but the problem is it is matching newdb1.table1:100 and printing 100.
Can you please give my a regular expression to match a string which either starts with a newline or has comma before it.
Use word boundaries:
if ($tables =~ /\bdb1.table1:(\d+)/) { print $1; }
here __^^
if ($tables =~ /(^|,)db1.table1:(\d+)/) { print $2; }
To answer your exact question, that is to match just after the start of the string or a comma, you want a positive look-behind assertion. You may be tempted to write a pattern of
/(?<=^|,)db1\.table1:(\d+)/
but that may fail with an error of
Variable length lookbehind not implemented in regex m/(?<=^|,)db1\.table1:(\d+)/ ...
So hold the regex engine’s hand a bit by making the alternatives equal in length—tricky to do in the general case but workable here.
/(?<=^d|,)db1\.table1:(\d+)/
While we are locking it down, let’s be sure to bracket the end with a look-ahead assertion.
while ($tables =~ /(?<=^d|,)db1\.table1:(\d+)(?=,|$)/g) {
print "[$1]\n";
}
Output:
[90]
You could also use \b for a regex word boundary, which has the same output.
while ($tables =~ /\bdb1\.table1:(\d+)(?=,|$)/g) {
print "[$1]\n";
}
For the most natural solution, follow the rule of thumb proposed by Randal Schwartz, author of Learning Perl. Use capturing when you know what you want to keep and split when you know what you want to throw away. In your case you have a mixture: you want to discard the comma separators, and you want to keep the digits after the colon for a certain table. Write that as
for (split /\s*,\s*/, $tables) { # / to fix Stack Overflow highlighting
if (my($value) = /^db1\.table1:(\d+)$/) {
print "[$value]\n";
}
}
Output:
[90]
I want to use a regular expression to check a string to make sure 4 and 5 are in order. I thought I could do this by doing
'$string =~ m/.45./'
I think I am going wrong somewhere. I am very new to Perl. I would honestly like to put it in an array and search through it and find out that way, but I'm assuming there is a much easier way to do it with regex.
print "input please:\n";
$input = <STDIN>;
chop($input);
if ($input =~ m/45/ and $input =~ m/5./) {
print "works";
}
else {
print "nata";
}
EDIT: Added Info
I just want 4 and 5 in order, but if 5 comes before at all say 322195458900023 is the number then where 545 is a problem 5 always have to come right after 4.
Assuming you want to match any string that contains two digits where the first digit is smaller than the second:
There is an obscure feature called "postponed regular expressions". We can include code inside a regular expression with
(??{CODE})
and the value of that code is interpolated into the regex.
The special verb (*FAIL) makes sure that the match fails (in fact only the current branch). We can combine this into following one-liner:
perl -ne'print /(\d)(\d)(??{$1<$2 ? "" : "(*FAIL)"})/ ? "yes\n" :"no\n"'
It prints yes when the current line contains two digits where the first digit is smaller than the second digit, and no when this is not the case.
The regex explained:
m{
(\d) # match a number, save it in $1
(\d) # match another number, save it in $2
(??{ # start postponed regex
$1 < $2 # if $1 is smaller than $2
? "" # then return the empty string (i.e. succeed)
: "(*FAIL)" # else return the *FAIL verb
}) # close postponed regex
}x; # /x modifier so I could use spaces and comments
However, this is a bit advanced and masochistic; using an array is (1) far easier to understand, and (2) probably better anyway. But it is still possible using only regexes.
Edit
Here is a way to make sure that no 5 is followed by a 4:
/^(?:[^5]+|5(?=[^4]|$))*$/
This reads as: The string is composed from any number (zero or more) characters that are not a five, or a five that is followed by either a character that is not a four or the five is the end of the string.
This regex is also a possibility:
/^(?:[^45]+|45)*$/
it allows any characters in the string that are not 4 or 5, or the sequence 45. I.e., there are no single 4s or 5s allowed.
You just need to match all 5 and search fails, where preceded is not 4:
if( $str =~ /(?<!4)5/ ) {
#Fail
}
Say I have a text file to parse, which contains some fixed length content:
123jackysee 45678887
456charliewong 32145644
<3><------16------><--8---> # Not part of the data.
The first three characters is ID, then 16 characters user name, then 8 digit phone number.
I would like to write a regular expression to match and verify the input for each line, the one I come up with:
(\d{3})([A-Za-z ]{16})(\d{8})
The user name should contains 8-16 characters. But ([A-Za-z ]{16}) would also match null value or space. I think of ([A-Za-z]{8,16} {0,8}) but it would detect more than 16 characters. Any suggestions?
No, no, no, no! :-)
Why do people insist on trying to pack so much functionality into a single RE or SQL statement?
My suggestion, do something like:
Ensure the length is 27.
Extract the three components into separate strings (0-2, 3-18, 19-26).
Check that the first matches "\d{3}".
Check that the second matches "[A-Za-z]{8,} *".
Check that the third matches "\d{8}".
If you want the entire check to fit on one line of source code, put it in a function, isValidLine(), and call it.
Even something like this would do the trick:
def isValidLine(s):
if s.len() != 27 return false
return s.match("^\d{3}[A-za-z]{8,} *\d{8}$"):
Don't be fooled into thinking that's clean Python code, it's actually PaxLang, my own proprietary pseudo-code. Hopefully, it's clear enough, the first line checks to see that the length is 27, the second that it matches the given RE.
The middle field is automatically 16 characters total due to the first line and the fact that the other two fields are fixed-length in the RE. The RE also ensures that it's eight or more alphas followed by the right number of spaces.
To do this sort of thing with a single RE would be some monstrosity like:
^\d{3}(([A-za-z]{8} {8})
|([A-za-z]{9} {7})
|([A-za-z]{10} {6})
|([A-za-z]{11} {5})
|([A-za-z]{12} )
|([A-za-z]{13} )
|([A-za-z]{14} )
|([A-za-z]{15} )
|([A-za-z]{16}))
\d{8}$
You could do it by ensuring it passes two separate REs:
^\d{3}[A-za-z]{8,} *\d{8}$
^.{27}$
but, since that last one is simply a length check, it's no different to the isValidLine() above.
I would use the regex you suggested with a small addition:
(\d{3})([A-Za-z]{3,16} {0,13})(\d{8})
which will match things that have a non-whitespace username but still allow space padding. The only addition is that you would then have to check the length of each input to verify the correct number of characters.
Hmm... Depending on the exact version of Regex you're running, consider:
(?P<id>\d{3})(?=[A-Za-z\s]{16}\d)(?P<username>[A-Za-z]{8,16})\s*(?P<phone>\d{8})
Note 100% sure this will work, and I've used the whitespace escape char instead of an actual space - I get nervous with just the space character myself, but you may want to be more restrictive.
See if it works. I'm only intermediate with RegEx myself, so I might be in error.
Check out the named groups syntax for your version of RegEx a) exists and b) matches the standard I've used above.
EDIT:
Just to expand what I'm trying to do (sorry to make your eyes bleed, Pax!) for those without a lot of RegEx experience:
(?P<id>\d{3})
This will try to match a named capture group - 'id' - that is three digits in length. Most versions of RegEx let you use named capture groups to extract the values you matched against. This lets you do validation and data capture at the same time. Different versions of RegEx have slightly different syntaxes for this - check out http://www.regular-expressions.info/named.html for more detail regarding your particular implementation.
(?=[A-Za-z\s]{16}\d)
The ?= is a lookahead operator. This looks ahead for the next sixteen characters, and will return true if they are all letters or whitespace characters AND are followed by a digit. The lookahead operator is zero length, so it doesn't actually return anything. Your RegEx string keeps going from the point the Lookahead started. Check out http://www.regular-expressions.info/lookaround.html for more detail on lookahead.
(?P<username>[A-Za-z]{8,16})\s*
If the lookahead passes, then we keep counting from the fourth character in. We want to find eight-to-sixteen characters, followed by zero or more whitespaces. The 'or more' is actually safe, as we've already made sure in the lookahead that there can't be more than sixteen characters in total before the next digit.
Finally,
(?P<phone>\d{8})
This should check the eight-digit phone number.
I'm a bit nervous that this won't exactly work - your version of RegEx may not support the named group syntax or the lookahead syntax that I'm used to.
I'm also a bit nervous that this Regex will successfully match an empty string. Different versions of Regex handle empty strings differently.
You may also want to consider anchoring this Regex between a ^ and $ to ensure you're matching against the whole line, and not just part of a bigger line.
Assuming you mean perl regex and if you allow '_' in the username:
perl -ne 'exit 1 unless /(\d{3})(\w{8,16})\s+(\d{8})/ && length == 28'
#OP,not every problem needs a regex. your problem is pretty simple to check. depending on what language you are using, they would have some sort of built in string functions. use them.
the following minimal example is done in Python.
import sys
for line in open("file"):
line=line.strip()
# check first 3 char for digit
if not line[0:3].isdigit(): sys.exit()
# check length of username.
if len(line[3:18]) <8 or len(line[3:18]) > 16: sys.exit()
# check phone number length and whether they are digits.
if len(line[19:26]) == 8 and not line[19:26].isdigit(): sys.exit()
print line
I also don't think you should try to pack all the functionality into a single regex. Here is one way to do it:
#!/usr/bin/perl
use strict;
use warnings;
while ( <DATA> ) {
chomp;
last unless /\S/;
my #fields = split;
if (
( my ($id, $name) = $fields[0] =~ /^([0-9]{3})([A-Za-z]{8,16})$/ )
and ( my ($phone) = $fields[1] =~ /^([0-9]{8})$/ )
) {
print "ID=$id\nNAME=$name\nPHONE=$phone\n";
}
else {
warn "Invalid line: $_\n";
}
}
__DATA__
123jackysee 45678887
456charliewong 32145644
678sdjkfhsdjhksadkjfhsdjjh 12345678
And here is another way:
#!/usr/bin/perl
use strict;
use warnings;
while ( <DATA> ) {
chomp;
last unless /\S/;
my ($id, $name, $phone) = unpack 'A3A16A8';
if ( is_valid_id($id)
and is_valid_name($name)
and is_valid_phone($phone)
) {
print "ID=$id\nNAME=$name\nPHONE=$phone\n";
}
else {
warn "Invalid line: $_\n";
}
}
sub is_valid_id { ($_[0]) = ($_[0] =~ /^([0-9]{3})$/) }
sub is_valid_name { ($_[0]) = ($_[0] =~ /^([A-Za-z]{8,16})\s*$/) }
sub is_valid_phone { ($_[0]) = ($_[0] =~ /^([0-9]{8})$/) }
__DATA__
123jackysee 45678887
456charliewong 32145644
678sdjkfhsdjhksadkjfhsdjjh 12345678
Generalizing:
#!/usr/bin/perl
use strict;
use warnings;
my %validators = (
id => make_validator( qr/^([0-9]{3})$/ ),
name => make_validator( qr/^([A-Za-z]{8,16})\s*$/ ),
phone => make_validator( qr/^([0-9]{8})$/ ),
);
INPUT:
while ( <DATA> ) {
chomp;
last unless /\S/;
my %fields;
#fields{qw(id name phone)} = unpack 'A3A16A8';
for my $field ( keys %fields ) {
unless ( $validators{$field}->($fields{$field}) ) {
warn "Invalid line: $_\n";
next INPUT;
}
}
print "$_ : $fields{$_}\n" for qw(id name phone);
}
sub make_validator {
my ($re) = #_;
return sub { ($_[0]) = ($_[0] =~ $re) };
}
__DATA__
123jackysee 45678887
456charliewong 32145644
678sdjkfhsdjhksadkjfhsdjjh 12345678
You can use lookahead: ^(\d{3})((?=[a-zA-Z]{8,})([a-zA-Z ]{16}))(\d{8})$
Testing:
123jackysee 45678887 Match
456charliewong 32145644 Match
789jop 12345678 No Match - username too short
999abcdefghijabcde12345678 No Match - username 'column' is less that 16 characters
999abcdefghijabcdef12345678 Match
999abcdefghijabcdefg12345678 No Match - username column more that 16 characters
I am looking for a regex that will find repeating letters. So any letter twice or more, for example:
booooooot or abbott
I won't know the letter I am looking for ahead of time.
This is a question I was asked in interviews and then asked in interviews. Not so many people get it correct.
You can find any letter, then use \1 to find that same letter a second time (or more). If you only need to know the letter, then $1 will contain it. Otherwise you can concatenate the second match onto the first.
my $str = "Foooooobar";
$str =~ /(\w)(\1+)/;
print $1;
# prints 'o'
print $1 . $2;
# prints 'oooooo'
I think you actually want this rather than the "\w" as that includes numbers and the underscore.
([a-zA-Z])\1+
Ok, ok, I can take a hint Leon. Use this for the unicode-world or for posix stuff.
([[:alpha:]])\1+
I Think using a backreference would work:
(\w)\1+
\w is basically [a-zA-Z_0-9] so if you only want to match letters between A and Z (case insensitively), use [a-zA-Z] instead.
(EDIT: or, like Tanktalus mentioned in his comment (and as others have answered as well), [[:alpha:]], which is locale-sensitive)
Use \N to refer to previous groups:
/(\w)\1+/g
You might want to take care as to what is considered to be a letter, and this depends on your locale. Using ISO Latin-1 will allow accented Western language characters to be matched as letters. In the following program, the default locale doesn't recognise é, and thus créé fails to match. Uncomment the locale setting code, and then it begins to match.
Also note that \w includes digits and the underscore character along with all the letters. To get just the letters, you need to take the complement of the non-alphanum, digits and underscore characters. This leaves only letters.
That might be easier to understand by framing it as the question:
"What regular expression matches any digit except 3?"
The answer is:
/[^\D3]/
#! /usr/local/bin/perl
use strict;
use warnings;
# uncomment the following three lines:
# use locale;
# use POSIX;
# setlocale(LC_CTYPE, 'fr_FR.ISO8859-1');
while (<DATA>) {
chomp;
if (/([^\W_0-9])\1+/) {
print "$_: dup [$1]\n";
}
else {
print "$_: nope\n";
}
}
__DATA__
100
food
créé
a::b
The following code will return all the characters, that repeat two or more times:
my $str = "SSSannnkaaarsss";
print $str =~ /(\w)\1+/g;
Just for kicks, a completely different approach:
if ( ($str ^ substr($str,1) ) =~ /\0+/ ) {
print "found ", substr($str, $-[0], $+[0]-$-[0]+1), " at offset ", $-[0];
}
FYI, aside from RegExBuddy, a real handy free site for testing regular expressions is RegExr at gskinner.com. Handles ([[:alpha:]])(\1+) nicely.
How about:
(\w)\1+
The first part makes an unnamed group around a character, then the back-reference looks for that same character.
I think this should also work:
((\w)(?=\2))+\2
/(.)\\1{2,}+/u
'u' modifier matching with unicode