Checking if a character is a special character in Java [duplicate]

Question

Possible Duplicate:
JAVA: check a string if there is a special character in it

I am a novice programmer and am looking for help determining if a character is a special character. My program asks the user to input the name of a file, and the program reads the text in the file and determines how many blanks spaces, digits, letters, and special characters are in the text. I have the code completed to determine the blanks, digits, and letters, however am unsure of how to check if a character is a special character. Any help you can offer is appreciated and if something was not clear enough I can try to elaborate. My code so far is:

import java.util.Scanner;
import java.io.*;

public class TextFile{

public static void main(String[] args){

  Scanner input = new Scanner (System.in);
  String fileName;
  boolean goodName = false;
  int blankCount = 0;
  int letterCount = 0;
  int digitCount = 0;
  int specialcharCount = 0;
  String currentLine;
  char c;
  Scanner lineFile= null;
  FileReader infile;

  System.out.println("Please enter the name of the file: ");
  fileName = input.nextLine();

  while (!goodName) {
    try{
      infile = new FileReader(fileName);
      lineFile = new Scanner(infile);

      goodName= true;
    }
    catch(IOException e) {
      System.out.println("Invalid file name, please enter correct file name: ");
      fileName=input.nextLine();
    }
  }

  while (lineFile.hasNextLine()){
    currentLine = lineFile.nextLine();
    for(int j=0; j<currentLine.length();j++){
      c=currentLine.charAt(j);
      if(c== ' ') blankCount++;
      if(Character.isDigit(c)) digitCount++;
      if(Character.isLetter(c)) letterCount++;
      if() specialcharCount++;
    }
  }
}
}

I need something to put in the if statement at the end to increment specialcharCount.

Answer 1

This method checks if a String contains a special character (based on your definition).

/**
 *  Returns true if s contains any character other than 
 *  letters, numbers, or spaces.  Returns false otherwise.
 */

public boolean containsSpecialCharacter(String s) {
    return (s == null) ? false : s.matches("[^A-Za-z0-9 ]");
}

You can use the same logic to count special characters in a string like this:

/**
 *  Counts the number of special characters in s.
 */

 public int getSpecialCharacterCount(String s) {
     if (s == null || s.trim().isEmpty()) {
         return 0;
     }
     int theCount = 0;
     for (int i = 0; i < s.length(); i++) {
         if (s.substring(i, 1).matches("[^A-Za-z0-9 ]")) {
             theCount++;
         }
     }
     return theCount;
 }

Another approach is to put all the special chars in a String and use String.contains:

/**
 *  Counts the number of special characters in s.
 */

 public int getSpecialCharacterCount(String s) {
     if (s == null || s.trim().isEmpty()) {
         return 0;
     }
     int theCount = 0;
     String specialChars = "/*!@#$%^&*()\"{}_[]|\\?/<>,.";
     for (int i = 0; i < s.length(); i++) {
         if (specialChars.contains(s.substring(i, 1))) {
             theCount++;
         }
     }
     return theCount;
 }

NOTE: You must escape the backslash and " character with a backslashes.

---

The above are examples of how to approach this problem in general.

For your exact problem as stated in the question, the answer by @LanguagesNamedAfterCoffee is the most efficient approach.

Answer 2

You can use regular expressions.

String input = ...
if (input.matches("[^a-zA-Z0-9 ]"))

If your definition of a 'special character' is simply anything that doesn't apply to your other filters that you already have, then you can simply add an else. Also note that you have to use else if in this case:

if(c == ' ') {
    blankCount++;
} else if (Character.isDigit(c)) {
    digitCount++;
} else if (Character.isLetter(c)) {
    letterCount++;
} else { 
  specialcharCount++;
}

Answer 3

Take a look at class java.lang.Character static member methods (isDigit, isLetter, isLowerCase, ...)

Example:

String str = "Hello World 123 !!";
int specials = 0, digits = 0, letters = 0, spaces = 0;
for (int i = 0; i < str.length(); ++i) {
   char ch = str.charAt(i);
   if (!Character.isDigit(ch) && !Character.isLetter(ch) && !Character.isSpace(ch)) {
      ++specials;
   } else if (Character.isDigit(ch)) {
      ++digits;
   } else if (Character.isSpace(ch)) {
      ++spaces;
   } else {
      ++letters;
   }
}

Answer 4

What I would do:

char c;
int cint;
for(int n = 0; n < str.length(); n ++;)
{
    c = str.charAt(n);
    cint = (int)c;
    if(cint <48 || (cint > 57 && cint < 65) || (cint > 90 && cint < 97) || cint > 122)
    {
        specialCharacterCount++
    }
}

That is a simple way to do things, without having to import any special classes. Stick it in a method, or put it straight into the main code.

ASCII chart: http://www.gophoto.it/view.php?i=http://i.msdn.microsoft.com/dynimg/IC102418.gif#.UHsqxFEmG08