Extension of "Detecting a Character" question


Basically the question is answered but i need more...

What i need is to hav a statement that detects if the string has items other than:

any integer, (, ), +, -, / and *.

if any of the above r detected thats ok... anything else is bad

so basically,

if ( not any integer, (, ), +, -, / and *) {
    System.err.println("INVALID");
    System.exit(1);
}
AnthonyCosenzaAsked:
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zzynxConnect With a Mentor Software engineerCommented:
Anthony,

I think you want to allow
1) all digits
2) (
3) )
4) +
5) -
6) /
7) *

No commas, right?

Then this is the reg expression:

     if ( !str.matches("[0-9\\(\\)\\+\\-\\*/]+") ) {
        System.err.println("INVALID");
        System.exit(1);
    }

If you want to allow spaces too:  (because now "5 + 6" is invalid)

     if ( !str.matches("[0-9 \\(\\)\\+\\-\\*/]+") ) {
        System.err.println("INVALID");
        System.exit(1);
    }

Remarks:
[1]
This doesn't recognize weird strings like
        String str = "6+*/-5-15*-6"
as invalid.

[2]
All six comments you accepted already are graded with a B.
Is that the maximum grade you'll ever give?
Here at EE, were used to get an A for correct answers/solutions,
a B for answers that helped to find the correct answer/solution and
a C for answers that help a little bit in finding the solution.

Did you know that giving an A-grade doesn't cost *you* any more points than the # originally set, while *we* get more points?
(A=x4, B=x3 and C=x2)
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sudhakar_koundinyaCommented:
String str="Abcd";

if(str.matches("A-Za-z"))
{
//ok
}
else
{
//not ok
}
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sudhakar_koundinyaCommented:
oops sorry wrong answer
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AnthonyCosenzaAuthor Commented:
ok, i'm not just worried about characters... i'm worried about things like "#" or "\"

basically any small little piece of text that could cause mischief

Thanks
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sudhakar_koundinyaCommented:
booleab b=str.matches.(", \\(\\ ) \\+\\ - / \\ *");
if(b)
for(int i=0;i<str.length);i++)
{
  if(!Character.isDigit(str.charAt(i)
  {
      b=false;
        break;  
  }

}

if(b)
//ok
else
//not ok
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AnthonyCosenzaAuthor Commented:
May i ask wat this line is doin??

if(!Character.isDigit(str.charAt(i)

Thanks again
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sudhakar_koundinyaCommented:
ignore previous one
try this

    if(str.matches("[0-9\\,\\(\\)\\+\\-\\*]+"))
    {
      System.err.println("ok");
    }
    else
    {
              System.err.println("not ok");
    }
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sudhakar_koundinyaConnect With a Mentor Commented:
missed / character

This works

if(str.matches("[0-9\\,\\(\\)\\+\\-\\*/]+"))
    {
      System.err.println("ok");
    }
    else
    {
              System.err.println("not ok");
    }
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zzynxSoftware engineerCommented:
>>May i ask wat this line is doin??
>>if(!Character.isDigit(str.charAt(i)

It checks if the character at position i in the String str is a digit
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AnthonyCosenzaAuthor Commented:
I've only been givin B grades because they are helpful solutions yet they r only parts of a bigger problem...
I had no idea that they affected the points system.

Anyway.. i hav another problem..

I hav something like this

while (st.hasMoreTokens( )) {
    if ( !token.matches("[0-9\\(\\)\\+\\-\\*/]+") ) {
        System.err.println("INVALID EXPRESSION");
        System.exit(1);
    } else {
        System.out.println(token);
        token = st.nextToken( );
    }
}
System.exit(1);

Where i hav a StringTokenizer goin through the string which is in a file...
What happens here is if i hav a expression in the file, for example ( 4 + 3 )..
it prints:

(
4
+
3

.. It doesnt print watever the last item is...

Puzzled..
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zzynxSoftware engineerCommented:
Can you show us what StringTokenizer you use?

Remark:
StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead.

Thanks for accepting
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sudhakar_koundinyaCommented:
Thanks for points :)

zzynx,
The last question was already answered here http://www.experts-exchange.com/Programming/Programming_Languages/Java/Q_21165896.html by CEHJ  

;-)
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zzynxSoftware engineerCommented:
>> The last question was already answered
Always handy to know. ;°)
Thanks sudhakar.
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