How can I use a regular expression to validate month input?
I am setting up this example Perl snippet to validate for months in a date:
Some scenarios I want to accept are:
MM M
#!/usr/bin/perl
use strict;
use warnings;
my $pattern;
my $month = "(0[1-9]|1[012])";
my $day = "(0[1-9]|[12]\d|3[01])";
system("cls");
do {
print "Enter in a month: ";
chomp($pattern = <STDIN>);
# We only want to print if the pattern matches
print "Pattern matches\n" if ($pattern =~ /$month/);
} while ($pattern ne "Q");
When I run this, it correctly filters fro开发者_开发知识库m 01-12 but when I change the regex to:
$month = "(0?[1-9]|1[012])";
then the regex allows 13, 14, etc... what gives?
If you really like to use regex, you need to put ^ and $, like
"^(0?[1-9]|1[012])$"
it will not match 13, 14 ....
You shouldn't use a regular expression to do numeric range validation. The regular expression you want is:
/^(\d+)$/
Then,
if ($1 >= 1 && $1 <= 12) {
# valid month
}
This is much easier to read than any regular expression to validate a numeric range.
As an aside, Perl evaluates regular expressions by searching within the target for a matching expression. So:
/(0[1-9]|1[012])/
searches for a 0 followed by 1 to 9, or a 1 followed by 0, 1, or 2. This would match "202" for example, and many other numbers. On the other hand:
/(0?[1-9]|1[012])/
searches for an optional 0 1 to 9, or a 1 followed by 0, 1, or 2. So "13" matches here because it contains a 1, matching the first half of the regex. To make your regular expressions work as you expect,
/^(0?[1-9]|1[012])$/
The ^
and $
anchor the search to the start and end of the string, respectively.
To give you hint - month number "120" also matches in your version :-)
Change:
my $month = "(0[1-9]|1[012])";
to
my $month = /^(0[1-9]|1[012])$/;
and then play more with it
(0[1-9]|1[012])
the parens are so you can use this inside another block, for example if matching for an entire yyyy-MM-dd date format
source: http://www.regular-expressions.info/dates.html
Don't use regular expressions.
Perl has the ability to automatically evaluate as a number or a string based on context. 01-09 will evaluate to 1-9 in the numeric context. So, you can simply check for a value:
print "Enter in a month: ";
chomp($pattern = <STDIN>);
# We only want to print if the pattern matches
print "Pattern matches\n" if ($pattern < 13 && $pattern > 0);
"^(1[012]|0?[1-9])$"
would be better because regular expression is assessed first one first. Let's say you want to match '12' and you write "^(0?[1-9]|1[012])$"
, then '1' will be picked because 0?[1-9]
is taken first.
here's one way
while(1){
print "Enter in a month: ";
$pattern = <STDIN>;
chomp($pattern);
if ($pattern =~ /^(Q|q)$/ ){last;}
if ($pattern =~ /^[0-9]$/ || $pattern =~ /^[0-9][12]$/ ) {
print "Pattern matches\n";
}else{
print "try again\n";
}
}
output
$ perl perl.pl
Enter in a month: 01
Pattern matches
Enter in a month: 000
try again
Enter in a month: 12
Pattern matches
Enter in a month: 00
try again
Enter in a month: 02
Pattern matches
Enter in a month: 13
try again
Enter in a month:
To test month
/ year
:
^(0?[1-9]|1[012])\/([2-9][0-9)]{3})$
[^d>12|d<0] OR ^[d>12|d<0]
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