Regex Pattern to validate Numerical Value

Question

I need to validate the following string to match only if the Number is greater than 0.

[D:\ifs\Pearl_Sync_Test\Client\Pub\Checkout\crm\source\crm\client\Ifs.SM.Common.MSOffice\Ifs.SM.Common.MSOffice.csproj]
47 Warning(s)
5 Error(s)

My solution so far is to match the above string

((\d)\sError\(s\))

Resulting in extracting this string, 5 Error(s)

So, is it possible to check if the Number is greater than 0 ?

Thanks

No it is not, this is just the part I wanted. I've updated the content — wishman
– wishman, Commented Jan 27, 2015 at 15:33
A bit unclear: Do you want to know a second regexp, in addition to yours, in order to find whether the number you extracted is greater than zero? Or do you want a regexp that improves yours, and will not match it at all if it is zero? And do you want to match the "Warnings" as well? — RealSkeptic
– RealSkeptic, Commented Jan 27, 2015 at 15:37
I just want to match 5 Error(s), and yes I need to improve my current Pattern. Not a second regex pattern. Thanks — wishman
– wishman, Commented Jan 27, 2015 at 15:42

m0skit0 · Accepted Answer · 2015-01-27 15:36:06Z

4

Regular expressions are not really fit for this problem but assuming you will not get negative errors, you can just check if it's not 0.

"[1-9]\\d*\\sError\\(s\\)"

Demo

answered Jan 27, 2015 at 15:36

m0skit0

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1 Comment

m0skit0 Over a year ago

The answer you accepted was posted after mine, but anyway, glad to have helped :)

Mena · Accepted Answer · 2015-01-27 15:37:07Z

3

You can use the following regular expression idiom for error numbers that start with a digit > 1:

String[] errors = {"5 Error(s)", "50 Error(s)", "0 Error(s)"};
//                           | starts with digit > 0
//                           |    | optionally ends with 0 or more digits
//                           |    |    | rest of the pattern
Pattern p = Pattern.compile("[1-9]\\d* Error\\(s\\)");
for (String s: errors) {
    System.out.println(s.matches(p.pattern()));
}

Output

true
true
false

answered Jan 27, 2015 at 15:37

Mena

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5 Comments

m0skit0 Over a year ago

+ not really needed

Aran-Fey Over a year ago

Not that it really matters here, but [1-9]\\d* does the job too, and it's less catastrophic than [1-9]+\\d*.

Mena Over a year ago

@Rawing true, thanks for noticing. Although I fail to understand why "catastrophic"!

m0skit0 Over a year ago

Because it can match strings like ` Error(s)`

Aran-Fey Over a year ago

@Mena: Reference to catastrophic backtracking.

Skinner · Accepted Answer · 2015-01-27 15:53:43Z

0

[1-9][0-9]*\sError(s)?

thanks rawling.

edited Jan 27, 2015 at 15:53

answered Jan 27, 2015 at 15:43

Skinner

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1 Comment

Aran-Fey Over a year ago

The question mark does not make the second digit optional. It makes the + quantifier non-greedy. This won't match single digit numbers.

Steven Stip · Accepted Answer · 2015-01-27 15:42:37Z

-1

in this case wouldn't you want to strip the errors bit off and check the value after you cast the first part to an int? It seems like you're trying to do string validation while you actually want to know what a certain number is.

answered Jan 27, 2015 at 15:42

Steven Stip

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