Re: Regex in java

From:
"programmer.sajjad@gmail.com" <programmer.sajjad@gmail.com>
Newsgroups:
comp.lang.java.programmer
Date:
Sun, 20 Apr 2008 22:32:00 -0700 (PDT)
Message-ID:
<066df984-1637-4331-9d7e-69fe2835a076@a22g2000hsc.googlegroups.com>
On Apr 18, 12:51 pm, Mark Space <marksp...@sbc.global.net> wrote:

programmer.saj...@gmail.com wrote:

Hi computer scientists/developers/engineers....
Can some one solve my problem, actually i want to check validity for
string data that is actually a BCD number.
Valid data is in following format:
String value=
"5.5" or "-5.5" or "54.598" or "-.5" or "5" or "128" etc
Regex that i have used is some thing like this

void checkValid(String value){

if((value.trim().matches("-??[0-9]{1,}[^A-Za-z].??[0-9]{1,}")) ||

(value.trim().matches("[^A-Za-z]-??.??[0-9]{1,}")) ){
      this.value=value.trim();
    }
    else{
      throw new IllegalArgumentException("Invalid BCD character("+value
+") ");
    }

but when i give only '5' its not working


I'm not a regex expert (say that three times fast), but I'll give it a go.

First, is trim() really a good idea? Seems to me regex can skip over
white space.

I don't really understand why you are using quantifiers. Those "??"
everywhere don't seem to be needed. I not saying it's wrong, just
pointing out I didn't parse that part very well.

Also, shouldn't the . in ??.?? be escaped? I assume that's a literal
decimal point there, not a regex reserved word.

Ok, my first attempt is:

"\\s*-?[0-9]*\\.?[0-9]*\\s*"

That's the Java string. The regex would be "\s*-?[0-9]*\.?[0-9]*\s"
after Java is done with the \ escaping. This seems to work ok. It
doesn't require a trim() either.

But the above regex does allow just one "-" or "." to match. So I think
it needs either one digit before the . or one after to be ok.

Second attempt:

"\\s*-?([0-9]+\\.?[0-9]*|[0-9]*\\.?[0-9]+)\\s*"

This is better. It does match "5.", and it is a little verbose. Maybe
you don't want to match patterns like "5.". Some thinking on my part
yeilds:

"\\s*-?[0-9]*\\.?[0-9]+\\s*"

So I think that's it. (Now I'm not sure what you were trying to do with
{1,}.) Clearly, having no leading digits like "-.5, or trailing digits
like just "5", is ok. My regex uses "one or more" ("+") on the second
part, relying on the greedy "*" to accept digits until a "." or a white
space is found. If a "." is found, it must be followed by at least one
digit. The "+" also says that at least one digit must be present --
white space alone won't match.

Here's my test harness:

public class Main {

     static String [] tests = {
         "5",
         "-.5",
         "5.5",
         " ",
         " 5 ",
         " -5.5",
         " - 3.1419",
         "5.",
         "5.5.5.5",
         "-.",
         ".",
         "-",
         ""
     };

     static String regex ="\\s*-?[0-9]*\\.?[0-9]*\\s*";

     public static void main(String[] args) {
         for( String s : tests )
         {
             if( s.matches( regex ) )
             {
                 System.out.println( s + " matches." );
             }
             else
             {
                 System.out.println( "No match on "+s );
             }
         }
     }

}


Thnx, for your kind reply. I have tried the following regex and its
working fine as well, but you code is pretty one...thnx

String value=.......
if((value.trim().matches("-??[0-9]++"))||(value.trim().matches("-??
[0-9]*+[.][0-9]++")) ){
      //this.value=value.trim();
    }
    else{
      throw new IllegalArgumentException("Invalid BCD character("+value
+") ");
    }

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