Java Regular Expression help

[royalibrahim]

Hi,

I need to write a java regular expression that has the following rules:

a) should not start with space or '.' (dot) character(s)
b) should not contain spaces in between of the characters
c) total characters length must be less than 80
d) should not be empty string


so far I have written this:

"(^[^\\.\\s])(.*[^\\s])"

[r035198x]

A simple approach is use negative look behind for (a) and (b) and normal string methods for (c) and (d).

[iceagecoming]

It's just the second part of your expression that doesn't work (for the spaces). To match on a string with no spaces, you should use \\S*, so here, this should work:

(^[^\\.\\s])(\\S*)

[JosAH]

Hi,

I need to write a java regular expression that has the following rules:

a) should not start with space or '.' (dot) character(s)
b) should not contain spaces in between of the characters
c) total characters length must be less than 80
d) should not be empty string


so far I have written this:

"(^[^\\.\\s])(.*[^\\s])"

I've answered this question before but I can't remember on what forum; my answer was (and is) that your regular expression should match:

1) a non-space-and-not-a-dot character.
2) up to 78 non-space characters.

1) can be solved with the range structure [ ... ] and 2) is solved by \S{0,78}

kind regards,

Jos

ps. I just found your crosspost here: http://forums.devshed.com/java-help-...lp-651883.html, you should've mentioned that fact.

[royalibrahim]

Thanks a lot for all your inputs.

I have recoded my regular expression like this:
"(^[^\\.\\s])(\\S{0,79})"

still I need to add another one condition, I should not allow consecutive "." characters, (i.e) 'aaa.bbb' is right, but 'aaa..bbb' is wrong and 'aaa.bbb.cc' is right.

How to make the regular expression now?

[iceagecoming]

I think this will work, but I don't have time to test all of the cases:

(^[^\\.\\s])(\\.?[^\\.\\s]){0,79}

Basically, the second part looks for a '.' and if it finds one, it makes sure that the next character is not a '.'.

Here are my test cases:
Input ( blah) - No Match
Input (.blah) - No Match
Input () - No Match
Input (b l a h) - No Match
Input (b..lah) - No Match
Input (b.la h) - No Match
Input (blah) - Match
Input (b) - Match
Input (b.l.a.h) - Match

[CodesAway]

Using that regex, the max number of characters is no longer 80.

Instead, use a negative forward assertion, since a '.' should NOT follow (i.e. negative, forward assertion).

The second part has two branches:
1) A period not followed by a period
2) Not a period, nor a space

The two branches are grouped in a non-capture group, which repeats between 0 and 79 times.

String regex = "(^[^\\.\\s])((?:\\.(?!\\.)|[^.\\s]){0,79})";

[iceagecoming]

Using that regex, the max number of characters is no longer 80.

I don't know why you say that. What is the max number of characters according to you? Here I've done these tests with my regex and it looks like 80 to me:

Input (0123456789.123456789.123456789.123456789.123456789.123456789.123456789.123456789) - Match
Input (01234567891123456789212345678931234567894123456789512345678961234567897123456789) - Match
Input (0123456789.123456789.123456789.123456789.123456789.123456789.123456789.123456789.) - No Match
Input (012345678911234567892123456789312345678941234567895123456789612345678971234567898) - No Match

And in addition, I think my regex is much less complicated.

[CodesAway]

I don't know why you say that. What is the max number of characters according to you?

And in addition, I think my regex is much less complicated.

Sorry about that, you're right, I should have provided a counter-example - I really wasn't thinking. BTW, I agree, you're regex is much simpler.

This is the regex you gave

(^[^\\.\\s])[COLOR="Red"](\\.?[^\\.\\s])[/COLOR]{0,79}

Notice how the second part can be two characters (at most) - a '.' followed by another character. You can do this at most 79 times. So, the maximum number of characters is 1 + 79 * 2 = 159.

To create a "worst-case scenario", the first character can be any valid character. Then, have 79 occurrences of a '.' followed by (for example) the letter 'a', which matches the [^\\.\\s] part. A 159 character VALID match.

StringBuilder input = new StringBuilder();

// matches [^\\.\\s]
input.append('a');

for (int i = 0; i < 79; i++) {
	// matches (\\.?[^\\.\\s])
	// (which can occur, at most, 79 times)
	input.append(".a");
}

String regex = "(^[^\\.\\s])(\\.?[^\\.\\s]){0,79}";
Pattern pattern = Pattern.compile(regex);

Matcher matcher = pattern.matcher(input);

// outputs "159 true"
System.out.println(input.length() + " " + matcher.matches());

[iceagecoming]

ok, you're right. I also noticed that my regex rejects any string ending in '.'.

[Mekonom]

I prefer to use Automation. I worked on project that it checks that email addres sysntax valid or not.

[Fubarable]

I prefer to use Automation. I worked on project that it checks that email addres sysntax valid or not.

What do you mean by this, and how does it answer the original poster's question?