Number Validation 3

[wood]

I am new to perl, and I need to validate a number that is entered. It needs to be in this format, ###.#. It can have up to 3 digits before the decimal and one after.

I am trying: /\d\d\d\.\d/ but that doesn't seem to work. It allows 1234, and 123.45 through.

Could someone please assist me. Thanks!

 

[azzazzello]

your pattern matching is testing for ###.# ANYWHERE in the number - 123.45 matches that. You need to say that ###.# is the, basically, "from start to finish", so you do /^\d\d\d\.\d$/

^ means basically "begins here" and $ means "ends here"

 

[wood]

Ok, that worked for 123.45, but it still allows 1234 or 12.23. Any suggestions?

I am trying to validate that the number can have 3 digits before the decimal point and one after.

Thanks!

 

[PaulTEG]

I'd suggest range checking

if (($number >= -1000) and ($number<=1000)) {
  print "Winner alright";
} else {
  print "outta bounds!";
}

HTH
--Paul

cigless ...

 

[wood]

Ok, that works for the 1234. Thank you for your suggestion.

How do I stop the person from entering more than one decimal place?

 

[azzazzello]

err...Wood, I think you may have mistyped something. I have just tried the code I gave you, and it works

-------
my $number = <STDIN>;
chomp $number;
print "Number entered is: *$number*\n";
($number =~ /^\d\d\d\.\d$/) ? print "Validated!\n" : print "NOT Validated\n";
-------

This will validate ONLY if the format is as you suggested (digit,digit,digit,perriod,digit)

 

[wood]

Thank you both for your help. I figured out what I needed:


if (! (($ARGS{mileage} =~ /^\d\.\d$/ ) | ($ARGS{mileage} =~ /^\d\d\.\d$/ ) | ($ARGS{mileage} =~ /^\d\d\d\.\d$/ ) | ($ARGS{mileage} =~ /^\d\d\d$/ )))

because the person can enter 123 or 1.2 or 12.3 or 123.4 but not 123.45.

Thanks again!

 

[azzazzello]

a cleaner way to do that would be

if (! #ARGS{$mileage} =~ /^\d{1,3}\.\d$/)

 

[ishnid]

I suspect azzazzello has typoed. Should be:

if ( ! $ARGS{$mileage} =~ /^\d{1,3}\.\d$/ )

wood - in the expression you posted, you should be using the logical OR operator `||', rather than its bitwise equivalent `|'. They'll work similarly in some situations but are not the same.

 

[tchatzi]

wood
because the person can enter 123 or 1.2 or 12.3 or 123.4 but not 123.45.

you need something like this

=~ /^\d{1,3}\.?\d$/;

the '?' goes because you want to accept numbers like 123 , so we match an optional '.'
without '?' if he enters 123 it will say that it is not valid.

TIMTOWTDI

 

[ishnid]

@pengo
I'd change that slightly again. Yours won't allow single-digit integers, so the last digit should be optional too:

=~ /^\d{1,3}(?:\.\d)?$/;

 

[tchatzi]

What ishnid said!

TIMTOWTDI

 

[wood]

Thanks ishnid. That worked the way I wanted.

A BIG thank you to all of those who helped me. I appreciate you assistance.