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Oct 21st, 2009

Count characters in a code

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I am trying to get this program to count the number of characters in a string. By number of characters l mean not punctuation or a space.What is wrong with my code because l don't understand why the program won't work?

Java Syntax (Toggle Plain Text)
import javax.swing.*;
 
publiemc class ExamPractise
{
	public static void main (String [] geek)
 
	{
		String sentence = JOptionPane.showInputDialog("Enter a any sentence");
 
		int countCharacters=0;
 
       for (int i = 0; i < sentence.length(); i++)
       {
		   char c = sentence.charAt(i);		  
 
			    	if (c !=' '||c!='?'||c!='.'||c!=' ')
			    	{
						countCharacters++;
					}
	   }
 
		  System.out.println("There are" + " " + countCharacters + " " + "word");
 
	}
}
import javax.swing.*;

publiemc class ExamPractise
{
	public static void main (String [] geek)

	{
		String sentence = JOptionPane.showInputDialog("Enter a any sentence");

		int countCharacters=0;

       for (int i = 0; i < sentence.length(); i++)
       {
		   char c = sentence.charAt(i);		  
			
			    	if (c !=' '||c!='?'||c!='.'||c!=' ')
			    	{
						countCharacters++;
					}
	   }

		  System.out.println("There are" + " " + countCharacters + " " + "word");

	}
}

Last edited by chengeto; Oct 21st, 2009 at 1:25 am. Reason: Typo

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chengeto

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Oct 21st, 2009

Re: Count characters in a code

(Toggle Plain Text)

publiemc class ExamPractise

Make it public

And:

(Toggle Plain Text)

if (c !=' '||c!='?'||c!='.'||c!=')
{
       countCharacters++;
}

^What do you wanna c not to be there?

Should be find after that.
-----

And I checked out the program, it still counts for spaces and punctuation.

EDIT AGAIN

Since your counting the characters you should make you if statement to check the ASCII value for an example:

Java Syntax (Toggle Plain Text)
if ((c >= 65 && c > 91))
{
	countCharacters++;
}
if ((c >= 65 && c > 91))
{
	countCharacters++;
}

That will check for uppercase letters only. Heres an ascii table.
http://web.cs.mun.ca/~michael/c/ascii-table.html

Just OR it with the other conditionals you make i.e.

Java Syntax (Toggle Plain Text)
if ((c >= 65 && c > 91) || (lower case condition) || (numbers condition))
{
	countCharacters++;
}
if ((c >= 65 && c > 91) || (lower case condition) || (numbers condition))
{
	countCharacters++;
}

Last edited by .11; Oct 21st, 2009 at 1:45 am.

.11

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Oct 21st, 2009

Re: Count characters in a code

Click to Expand / Collapse Quote originally posted by .11 ...

(Toggle Plain Text)
publiemc class ExamPractise
publiemc class ExamPractise
Make it public

And:
(Toggle Plain Text)
if (c !=' '||c!='?'||c!='.'||c!=')
{
       countCharacters++;
}
if (c !=' '||c!='?'||c!='.'||c!=') { countCharacters++; }
^What do you wanna c not to be there?

Should be find after that.
-----

And I checked out the program, it still counts for spaces and punctuation.

Right now if l enter the word Obama Obama ? it counts the question mark and space between the two words as a character and l don't want it to do that.I don't know how do l get around that ?

chengeto

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Oct 21st, 2009

Re: Count characters in a code

Click to Expand / Collapse Quote originally posted by chengeto ...

Right now if l enter the word Obama Obama ? it counts the question mark and space between the two words as a character and l don't want it to do that.I don't know how do l get around that ?

Check out the ASCII table above I edit my post, the upper case condition is checked for, just check for the other two. And it will eliminate the characters that aren't a letter and number.

.11

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Oct 21st, 2009

Re: Count characters in a code

try using this method to check each character:

if (Character.isLetter(c)) {
countCharacters++;
}

don't forget, you'll need to import java.lang.Character;

nola_Coder

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Oct 21st, 2009

Re: Count characters in a code

> By number of characters l mean not punctuation or a space

Look into the isLetter method of the Character class; modify your logic to increment the count only if the above method returns true .

Last edited by ~s.o.s~; Oct 21st, 2009 at 1:56 am.

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