Simple, I am trying to see if a field has 9 digits and nothing else.
my $var = 123456789
( my $nine ) = ( $var =~ /\d{9}/ );
from my understanding this says, "match a digit 9 times and nothing else"
this outputs 1 but not the 123456789 that i was expecting. Why?
Your pattern matches a sequence of nine (international) digit character anywhere in the string. The 1 you are seeing is a true value that the pattern match returns to say that the match was successful.
If you just want to verify that the contents of a variable are exactly nine ASCII digits, then you should write
if ( $var =~ /\A[0-9]{9}\z/ ) { ... }
or, if you have the ASCII /a modifier available (any version of Perl 5 since 14) then you can say
if ( $var =~ /\A\d{9}\z/a ) { ... }
There is nothing coming back in list context within the regex,
so the default is that it sets your variable $nine to the scalar result
of the regex function call. And that function call returns in scalar context
the number of matches.
If you were to change that to (my $nine) = ($var =~ /\d/g);
$nine would contain 9.
Add a capture buffer around the digits.
ie:
( my $nine ) = ( $var =~ /(\d{9})/ );
However, even though your syntax for your assignment will work,
its conventional to write it as
my ($nine) = $var =~ /(\d{9})/;
You really want this:
( my $nine ) = ( $var =~ /(\d{9})/ );
The problem is that =~ is a comparison operator, and the result of the expression ( $var =~ /\d{9}/ ) is true (1). Adding the parentheses in my example makes the regex capture its matches, so it returns your 123456789.
Related
I have a string of the following format:
word1.word2.word3
What are the ways to extract word2 from that string in perl?
I tried the following expression but it assigns 1 to sub:
#perleval $vars{sub} = $vars{string} =~ /.(.*)./; 0#
EDIT:
I have tried several suggestions, but still get the value of 1. I suspect that the entire expression above has a problem in addition to parsing. However, when I do simple assignment, I get the correct result:
#perleval $vars{sub} = $vars{string} ; 0#
assigns word1.word2.word3 to variable sub
. has a special meaning in regular expressions, so it needs to be escaped.
.* could match more than intended. [^.]* is safer.
The match operator (//) simply returns true/false in scalar context.
You can use any of the following:
$vars{sub} = $vars{string} =~ /\.([^.]*)\./ ? $1 : undef;
$vars{sub} = ( $vars{string} =~ /\.([^.]*)\./ )[0];
( $vars{sub} ) = $vars{string} =~ /\.([^.]*)\./;
The first one allows you to provide a default if there's no match.
Try:
/\.([^\.]+)\./
. has a special meaning and would need to be escaped. Then you would want to capture the values between the dots, so use a negative character class like ([^\.]+) meaning at least one non-dot. if you use (.*) you will get:
word1.stuff1.stuff2.stuff3.word2 to result in:
stuff1.stuff2.stuff3
But maybe you want that?
Here is my little example, I do find the perl one liners a little harder to read at times so I break it out:
use strict;
use warnings;
if ("stuff1.stuff2.stuff3" =~ m/\.([^.]+)\./) {
my $value = $1;
print $value;
}
else {
print "no match";
}
result
stuff2
. has a special meaning: any character (see the expression between your parentheses)
Therefore you have to escape it (\.) if you search a literal dot:
/\.(.*)\./
You've got to make sure you're asking for a list when you do the search.
my $x= $string =~ /look for (pattern)/ ;
sets $x to 1
my ($x)= $string =~ /look for (pattern)/ ;
sets $x to pattern.
I have this text $line = "config.txt.1", and I want to match it with regex and extract the number
part of it. I am using two versions:
$line = "config.txt.1";
(my $result) = $line =~ /(\d*).*/; #ver 1, matched, but returns nothing
(my $result) = $line =~ /(\d).*/; #ver 2, matched, returns 1
(my $result) = $line =~ /(\d+).*/; #ver 3, matched, returns 1
I think the * was sort of messing things around, I have been looking at this, but still
don't the greedy mechanism in the regex engine. If I start from left of the regex, and potentially there might be no digits in the text, so for ver 1, it will match too. But for
ver 3, it won't match. Can someone give me an explanation for why it is that and how
I should write for what I want? (potentially with a number, not necessarily single digit)
Edit
Requirement: potentially with a number, not necessarily single digit, and match can not capture anything, but should not fail
The output must be as follows (for the above example):
config.txt 1
The regex /(\d*).*/ always matches immediately, because it can match zero characters. It translates to match as many digits at this position as possible (zero or more). Then, match as many non-newline characters as possible. Well, the match starts looking at the c of config. Ok, it matches zero digits.
You probably want to use a regex like /\.(\d+)$/ -- this matches an integer number between a period . and the end of string.
Use the literal '.' as a reference to match before the number:
#!/usr/bin/perl
use strict;
use warnings;
my #line = qw(config.txt file.txt config.txt.1 config.foo.2 config.txt.23 differentname.fsdfsdsdfasd.2444);
my (#capture1, #capture2);
foreach (#line){
my (#filematch) = ($_ =~ /(\w+\.\w+)/);
my (#numbermatch) = ($_ =~ /\w+\.\w+\.?(\d*)/);
my $numbermatch = $numbermatch[0] // $numbermatch[1];
push #capture1, #filematch;
push #capture2, #numbermatch;
}
print "$capture1[$_]\t$capture2[$_]\n" for 0 .. $#capture1;
Output:
config.txt
file.txt
config.txt 1
config.foo 2
config.txt 23
differentname.fsdfsdsdfasd 2444
Thanks guys, I think I figured out myself what I want:
my ($match) = $line =~ /\.(\d+)?/; #this will match and capture any digit
#number if there was one, and not fail
#if there wasn't one
To capture all digits following a final . and not fail the match if the string doesn't end with digits, use /(?:\.(\d+))?$/
perl -E 'if ("abc.123" =~ /(?:\.(\d+))?$/) { say "matched $1" } else { say "match failed" }'
matched 123
perl -E 'if ("abc" =~ /(?:\.(\d+))?$/) { say "matched $1" } else { say "match failed" }'
matched
You do not need .* at all. These two statements assign the exact same number:
my ($match1) = $str =~ /(\d+).*/;
my ($match1) = $str =~ /(\d+)/;
A regex by default matches partially, you do not need to add wildcards.
The reason your first match does not capture a number is because * can match zero times as well. And since it does not have to match your number, it does not. Which is why .* is actually detrimental in that regex. Unless something is truly optional, you should use + instead.
Say I have $foo = "bar.baz"
I want to use the scalar $foo to find strings that contain "bar.baz" (anywhere in the string), but not the regex-evaluted version of $foo.
So the line: if( $other =~ m/$foo/ ) ...
isn't working, because $foo is being evaluated such that the '.' is evaluated to any character. How do I stop that?
Pick one:
$foo = quotemeta("bar.baz");
if ($other =~ m/\Q$foo/)
(Both are actually the same thing, just done at different times.)
My program read other programs source code and colect information about used SQL queries. I have problem with getting substring.
...
$line = <FILE_IN>;
until( ($line =~m/$values_string/i && $line !~m/$rem_string/i) || eof )
{
if($line =~m/ \S{2}DT\S{3}/i)
{
# here I wish to get (only) substring that match to pattern \S{2}DT\S{3}
# (7 letter table name) and display it.
$line =~/\S{2}DT\S{3}/i;
print $line."\n";
...
In result print prints whole line and not a substring I expect. I tried different approach, but I use Perl seldom and probably make basic concept error. ( position of tablename in line is not fixed. Another problem is multiple occurrence i.e.[... SELECT * FROM AADTTAB, BBDTTAB, ...] ). How can I obtain that substring?
Use grouping with parenthesis and store the first group.
if( $line =~ /(\S{2}DT\S{3})/i )
{
my $substring = $1;
}
The code above fixes the immediate problem of pulling out the first table name. However, the question also asked how to pull out all the table names. So:
# FROM\s+ match FROM followed by one or more spaces
# (.+?) match (non-greedy) and capture any character until...
# (?:x|y) match x OR y - next 2 matches
# [^,]\s+[^,] match non-comma, 1 or more spaces, and non-comma
# \s*; match 0 or more spaces followed by a semi colon
if( $line =~ /FROM\s+(.+?)(?:[^,]\s+[^,]|\s*;)/i )
{
# $1 will be table1, table2, table3
my #tables = split(/\s*,\s*/, $1);
# delim is a space/comma
foreach(#tables)
{
# $_ = table name
print $_ . "\n";
}
}
Result:
If $line = "SELECT * FROM AADTTAB, BBDTTAB;"
Output:
AADTTAB
BBDTTAB
If $line = "SELECT * FROM AADTTAB;"
Output:
AADTTAB
Perl Version: v5.10.0 built for MSWin32-x86-multi-thread
I prefer this:
my ( $table_name ) = $line =~ m/(\S{2}DT\S{3})/i;
This
scans $line and captures the text corresponding to the pattern
returns "all" the captures (1) to the "list" on the other side.
This psuedo-list context is how we catch the first item in a list. It's done the same way as parameters passed to a subroutine.
my ( $first, $second, #rest ) = #_;
my ( $first_capture, $second_capture, #others ) = $feldman =~ /$some_pattern/;
NOTE:: That said, your regex assumes too much about the text to be useful in more than a handful of situations. Not capturing any table name that doesn't have dt as in positions 3 and 4 out of 7? It's good enough for 1) quick-and-dirty, 2) if you're okay with limited applicability.
It would be better to match the pattern if it follows FROM. I assume table names consist solely of ASCII letters. In that case, it is best to say what you want. With those two remarks out of the way, note that a successful capturing regex match in list context returns the matched substring(s).
#!/usr/bin/perl
use strict;
use warnings;
my $s = 'select * from aadttab, bbdttab';
if ( my ($table) = $s =~ /FROM ([A-Z]{2}DT[A-Z]{3})/i ) {
print $table, "\n";
}
__END__
Output:
C:\Temp> s
aadttab
Depending on the version of perl on your system, you may be able to use a named capturing group which might make the whole thing easier to read:
if ( $s =~ /FROM (?<table>[A-Z]{2}DT[A-Z]{3})/i ) {
print $+{table}, "\n";
}
See perldoc perlre.
Parens will let you grab part of the regex into special variables: $1, $2, $3...
So:
$line = ' abc andtabl 1234';
if($line =~m/ (\S{2}DT\S{3})/i) {
# here I wish to get (only) substring that match to pattern \S{2}DT\S{3}
# (7 letter table name) and display it.
print $1."\n";
}
Use a capturing group:
$line =~ /(\S{2}DT\S{3})/i;
my $substr = $1;
$& contains the string matched by the last pattern match.
Example:
$str = "abcdefghijkl";
$str =~ m/cdefg/;
print $&;
# Output: "cdefg"
So you could do something like
if($line =~m/ \S{2}DT\S{3}/i) {
print $&."\n";
}
WARNING:
If you use $& in your code it will slow down all pattern matches.
I thought this would have done it...
$rowfetch = $DBS->{Row}->GetCharValue("meetdays");
$rowfetch = /[-]/gi;
printline($rowfetch);
But it seems that I'm missing a small yet critical piece of the regex syntax.
$rowfetch is always something along the lines of:
------S
-M-W---
--T-TF-
etc... to represent the days of the week a meeting happens
$rowfetch =~ s/-//gi
That's what you need for your second line there. You're just finding stuff, not actually changing it without the "s" prefix.
You also need to use the regex operator "=~" for this.
Here is what your code presently does:
# Assign 'rowfetch' to the value fetched from:
# The function 'GetCharValue' which is a method of:
# An Value in A Hash Identified by the key "Row" in:
# Either a Hash-Ref or a Blessed Hash-Ref
# Where 'GetCharValue' is given the parameter "meetdays"
$rowfetch = $DBS->{Row}->GetCharValue("meetdays");
# Assign $rowfetch to the number of times
# the default variable ( $_ ) matched the expression /[-]/
$rowfetch = /[-]/gi;
# Print the number of times.
printline($rowfetch);
Which is equivalent to having written the following code:
$rowfetch = ( $_ =~ /[-]/ )
printline( $rowfetch );
The magic you are looking for is the
=~
Token instead of
=
The former is a Regex operator, and the latter is an assignment operator.
There are many different regex operators too:
if( $subject =~ m/expression/ ){
}
Will make the given codeblock execute only if $subject matches the given expression, and
$subject =~ s/foo/bar/gi
Replaces ( s/) all instances of "foo" with "bar", case-insentitively (/i), and repeating the replacement more than once(/g), on the variable $subject.
Using the tr operator is faster than using a s/// regex substitution.
$rowfetch =~ tr/-//d;
Benchmark:
use Benchmark qw(cmpthese);
my $s = 'foo-bar-baz-blee-goo-glab-blech';
cmpthese(-5, {
trd => sub { (my $a = $s) =~ tr/-//d },
sub => sub { (my $a = $s) =~ s/-//g },
});
Results on my system:
Rate sub trd
sub 300754/s -- -79%
trd 1429005/s 375% --
Off-topic, but without the hyphens, how will you know whether a "T" is Tuesday or Thursday?