Regex help

[John Landon]

Hi guys,
I need regex for:
String must fail if:
1) Non numeric
2) If any leading zeroes are entered without a decimal point
3) If there are any zeroes trailing a decimal point
4) If there is a decimal point with no numbers before it
5) If there is decimal point more then 10 digits after the zero

I came up with this but its not quite right it still matches if there are zeroes
^\d+(\.\d{1,10})?$

Thanks!

[Campbell Ritchie]

Go through the Scanner class documentation and the Pattern class. You might be lucky and find regexes for numbers ready made there. Is no 5 written correctly? Do you mean not more than ten digits after the decimal point?
Of course \d will match 0. You would have to try [1-9] or something like that.
You do realise that your specification will fail for this valid double literal: 123.0.

[fred rosenberger]

Why do you want a single regex to do all these things? Wouldn't it be easier to write several, smaller ones that each do a picece of the checks you need?

[John Landon]

fred rosenberger wrote:Why do you want a single regex to do all these things? Wouldn't it be easier to write several, smaller ones that each do a picece of the checks you need?

No because I need to do it in validation annotation

[Ron McLeod]

John Landon wrote:No because I need to do it in validation annotation

You could make your annotation take multiple expressions.

@Target(value = ElementType.FIELD)
@Retention(value = RetentionPolicy.RUNTIME)
public @interface MyValidationAnnotation {
	String[] regexPatterns() default {};
}

[John Landon]

Ron McLeod wrote:

John Landon wrote:No because I need to do it in validation annotation

You could make your annotation take multiple expressions.

@Target(value = ElementType.FIELD)
@Retention(value = RetentionPolicy.RUNTIME)
public @interface MyValidationAnnotation {
	String[] regexPatterns() default {};
}

Thanks so how would you put this inside? (its regex that does what I want except trailing zeroes):^\d+(\.\d{1,10})?$

[Junilu Lacar]

So...

by #2,
this (leading zeroes without a decimal) would fail: 0053
and this (leading zeroes with a decimal) would pass: 0053.

But by #3, this would fail: 0053.00

I don't get what those rules with leading and trailing zeroes are going for.

[Ron McLeod]

John Landon wrote:Thanks so how would you put this inside?

Are you asking how to put multiple expression in the annotation?

@MyValidationAnnotation(regexPatterns = { "\\d+(\\.\\d{1,10})?", "other expression", " another expression" })
private String magicNumber;

[John Landon]

05.5 will fail
5.50 will fail
0.5 will succeed

[Carey Brown]

"^(([1-9][0-9]*)|"  // integer, no decimal
+ "([1-9][0-9]*\\.[\\d{0,9}[1-9])|"  // leading integer with decimal
+ "(0\\.\\d{0,9}[1-9]))$"  // zero with decimal

Haven't tested it but I think this should work.

[John Landon]

Carey Brown wrote:

"^(([1-9][0-9]*)|"  // integer, no decimal
+ "([1-9][0-9]*\\.[\\d{0,9}[1-9])|"  // leading integer with decimal
+ "(0\\.\\d{0,9}[1-9]))$"  // zero with decimal

Haven't tested it but I think this should work.

How about leading zeros? Pleas how do I apply them all at once?

[Carey Brown]

How about leading zeros? Pleas how do I apply them all at once?

If you look closely you'll see that it is one long regex String put together with String (+) concatenation.

I just broke it out so that you can see the separate rules separated by an OR (|).

[Henry Wong]

First, I guess I can take a shot at it (and without actually compiling or testing it, so take with a grain of salt)...

String regexPattern = "(0|[1-9]\\d*)(\\.(0|\\d{0,9}[1-9]))?";

Second, assuming that this even works, I highly recommend that you don't use it without actually understanding it !!

Henry

[John Landon]

Carey Brown wrote:

How about leading zeros? Pleas how do I apply them all at once?

If you look closely you'll see that it is one long regex String put together with String (+) concatenation.

I just broke it out so that you can see the separate rules separated by an OR (|).

thanks.
Yeas I saw but this website denies it
http://www.freeformatter.com/java-regex-tester.html

[John Landon]

Henry Wong wrote:
First, I guess I can take a shot at it (and without actually compiling or testing it, so take with a grain of salt)...

String regexPattern = "(0|[1-9]\\d*)(\\.(0|\\d{0,9}[1-9]))?";

Second, assuming that this even works, I highly recommend that you don't use it without actually understanding it !!

Henry

It doesnt. Does the same thing as mine: ^\d+(\.\d{1,10})?$

It passes even if you do 75.550 (the extra zero should fail it)

[Henry Wong]

John Landon wrote:
It passes even if you do 75.550 (the extra zero should fail it)

The test failed, as expected, when I tried it...

public static void main(String[] args) {
		String regexPattern = "(0|[1-9]\\d*)(\\.(0|\\d{0,9}[1-9]))?";
		
		boolean ret = "75.550".matches(regexPattern);
		System.out.println(ret);
}

It printed "false", as expected.

Henry

[Carey Brown]

When working with complex regular expressions I find it useful to create a 20 line regex tester program that takes as parameters an array of "good" test cases that are expected to match, and an array of "bad" test cases that are expected to fail. If any test doesn't result in the expected outcome it should report the failed string.

[John Landon]

Carey Brown wrote:When working with complex regular expressions I find it useful to create a 20 line regex tester program that takes as parameters an array of "good" test cases that are expected to match, and an array of "bad" test cases that are expected to fail. If any test doesn't result in the expected outcome it should report the failed string.

It matches the 0
http://www.freeformatter.com/java-regex-tester.html#ad-output

[John Landon]

John Landon wrote:

Carey Brown wrote:When working with complex regular expressions I find it useful to create a 20 line regex tester program that takes as parameters an array of "good" test cases that are expected to match, and an array of "bad" test cases that are expected to fail. If any test doesn't result in the expected outcome it should report the failed string.

It matches the 0
http://www.freeformatter.com/java-regex-tester.html#ad-output

Also it doesnt match number 5

[Knute Snortum]

I like to break long regexes into chunks and comment them.  I think it makes working with them easier and it certainly helps anyone maintaining the code, if only to state my intent.

	private static String REGEX = 
			  "("            // Start group 1
			+   "0"          // Match zero
			+   "|"          // or
			+   "[1-9]"      // match one digit, one through nine
			+   "\\d*"       // followed by any number of digits
			+ ")"            // End group 0
			+ "("            // Start group 2
			+   "\\."        // Match a literal dot
			+   "("          // Start group 3
			+     "0"        // Match zero
			+     "|"        // or
			+     "\\d{0,9}" // match from zero up to nine digits
			+     "[1-9]"    // followed by one digit, one through nine 
			+   ")"          // End group 3
			+ ")"            // End group 2
			+ "?";           // Make group 2 optional 
	
	public static void main(String[] args) {
        boolean ret = "75.550".matches(REGEX);
        System.out.println(ret);
	}

I can see from this that the regex will fail to match "75.50", which I think it should.  Edit: I guess not, looking at the specs.

[Knute Snortum]

John Landon wrote:Also it doesnt match number 5

Henry Wong's regex?  It does for me, running the little program.

[John Landon]

Knute Snortum wrote:

John Landon wrote:Also it doesnt match number 5

Henry Wong's regex?  It does for me, running the little program.

This is awesome thanks (the closest I've seen) but it does match 5.0.......

[Henry Wong]

John Landon wrote:
This is awesome thanks (the closest I've seen) but it does match 5.0.......

It matches "5.0" because that is how the regex is coded. I took your original post, as disallowing trailing zeros, but allowing one digit after the decimal point. So, "5.00" is not allowed, but "5.0" is allowed.

The fix to this, is actually very simple. Just get rid of the alternation after the decimal, that allows it.


And .... Remember when I said that you should never use a regex that you don't understand? Well, this fix is ridiculously easy to do, and this is a chance for you to understand the regex. So, give it a try at fixing it yourself.

Henry

[John Landon]

Henry Wong wrote:

John Landon wrote:
This is awesome thanks (the closest I've seen) but it does match 5.0.......

It matches "5.0" because that is how the regex is coded. I took your original post, as disallowing trailing zeros, but allowing one digit after the decimal point. So, "5.00" is not allowed, but "5.0" is allowed.

The fix to this, is actually very simple. Just get rid of the alternation after the decimal, that allows it.


And .... Remember when I said that you should never use a regex that you don't understand? Well, this fix is ridiculously easy to do, and this is a chance for you to understand the regex. So, give it a try at fixing it yourself.

Henry

LOL Thanks Henry. I removed //+     "0"        // Match zero and it works. Is there any chance to limit the number of digits total (except the dot) say to 11?

[Stephan van Hulst]

No, not without a regex that's much more complex.

[Henry Wong]

John Landon wrote:Is there any chance to limit the number of digits total (except the dot) say to 11?

Sure. You can do something like using a negative look-ahead, at the beginning of the regex -- to match, 12 or more digits til the end-of-input, which will, of course, cause the regex to fail, when it encounters more than 11 digits.

Now, if what I said sounds like gibberish ... ... it is likely because this is an advanced feature. At this point, I am not even convinced that you understand the regex that you currently have (that we did for you)....

Anyway, I highly recommend doing a little studying of regular expressions, to fully understand the regex that you are currently using. And once you feel comfortable with that, then do a little research on look-aheads. This should lead you to the solution that I described.

Hope this helps,
Henry

[Campbell Ritchie]

What is wrong with 5.0? That is a normal format for a number. Similarly 0.5.

[Henry Wong]

Campbell Ritchie wrote:What is wrong with 5.0? That is a normal format for a number. Similarly 0.5.

Agreed. And my regular expression allowed for both!

Henry