CODEWITHSUNDEEP
regexperl
user595234
user595234

Reputation: 6259

perl regex expression extraction

how to extract this string using regular expression

||03/15/2012||10:17:11|FOR TEST

I want to extract "03/15/2012" . I have tried this command 

my $firstLine =~ m/^\|\|\d{2}\/\d{2}\/\d{4}\|\|/i ;
trace("first line extraction : $firstLine first $1");

It doesn't work. could you please help ?

thanks

Upvotes: 1

Views: 141

Answers (4)

stema
stema

Reputation: 93056

You can assign the value of the a capturing group directly like this:

my $in = "||03/15/2012||10:17:11|FOR TEST";
(my $Date) = $in =~ /^\|\|(\d{2}\/\d{2}\/\d{4})\|\|/;

You will find the result of the capturing group 1 in $Date.

Btw. you don't need the modifier i, because this makes letters matching case insensitive, since you have no letters in your regex, this option is useless.

Upvotes: 1

Ωmega
Ωmega

Reputation: 43703

Script:

my $firstLine = '||03/15/2012||10:17:11|FOR TEST';
print $1 if $firstLine =~ /(\d{2}\/\d{2}\/\d{4})/;

Output:

03/15/2012

Test this code here: http://ideone.com/99iUg

Upvotes: 0

Derek Risling
Derek Risling

Reputation: 364

Try: my $firstLine =~m/^\|\|(.+)\|\|.*$/i

You only really care about what's between the first two sets of double pipes

Upvotes: 0

maerics
maerics

Reputation: 156652

You need to use a capturing group () around the part you want to extract:

my $line = '||03/15/2012||10:17:11|FOR TEST';
if ($line =~ m/^\|\|(\d{2}\/\d{2}\/\d{4})\|\|/i) {
  # Capturing group ┴───────────────────┘
  print("Group 1: $1\n"); # 03/15/2012
}

Upvotes: 4

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