Java Programming Finding a character

Gman_Ireland

hi i am trying to find how i can print the number of times the character 'e' appears in your name. can anybody help me with this.

lets say my name is gerry byrne erin

hope this make sense to you,

Joneser

I would say add the string to a charArray and iterate through it, increasing the count by 1 every time a char = e, or the relevant ascii value, "lower case e has an ascii value of 101)

I can give some help with pseudo code if u like, but is prob better if u give it a go yourself first

Gman_Ireland

@Joneser if you could provide the code that would be great, have tried and will look at it again. thanks

dlofnep

There are a few ways to do this.

One is as mentioned, iterating through the array.

Let's say you have a string called name and it's value is "teddy".

You could do the following

int characterCount = 0;
String search = userinput(); // use whatever scanner / i/o class you want
for(int i = 0; i < name.length(); i++){
 if(name.charAt(i)).equals(search)) {
  characterCount++;
  }
}

System.out.println("The character: " + search + " has been found: " + characterCount + " times.");

That's it more or less I'd imagine.

Sparks

Gman_Ireland wrote: »

@Joneser if you could provide the code that would be great, have tried and will look at it again. thanks

http://www.boards.ie/vbulletin/showpost.php?p=52731901&postcount=2

Gman_Ireland

@Sparks thanks
here is what i have done so far,

: 

import java.util.Scanner;

class  week5_scanner {

		public static void main(String args[]){
		
			String name;
			Scanner in = new Scanner(System.in);
			
			System.out.println("What is your name" );
			name = in.nextLine();
			char c1 = name.charAt(0);
			int pos = name.indexOf(" ");
			char c2 = name.charAt(pos + 1);
			in.close();
			
			String lowername= name.toLowerCase();
			System.out.println("Hello " + lowername );
			
			String uppername= name.toUpperCase();
			System.out.println("Hello " + uppername );
			
			int len = name.length();
			System.out.println("STR LEN = " + name.length() );
			
			System.out.println("My initials are :- " + c1 + c2);
			
			char at = name.charAt(0); 
			System.out.println("first char is = " + name.charAt(0));
			
		}
		
		}

dlofnep

Sorry, forgot to convert the character to a string. Here's full working code for you.

public class countChar{
	public static void main(String[] args){
		int count = 0;
		String name = "Teddy"; 
		String search = "d"; // replace this with user input
		String currentChar;

		for(int i=0; i<name.length(); i++){
			currentChar = Character.toString(name.charAt(i));
			if(currentChar.equals(search)){
				count++;
			}
		}

		System.out.println("The character: " + search + " has been found: " + count + " times.");


	}
}

Gman_Ireland

dlofnep wrote: »

Sorry, forgot to convert the character to a string. Here's full working code for you.

public class countChar{
	public static void main(String[] args){
		int count = 0;
		String name = "Teddy"; 
		String search = "d"; // replace this with user input
		String currentChar;

		for(int i=0; i<name.length(); i++){
			currentChar = Character.toString(name.charAt(i));
			if(currentChar.equals(search)){
				count++;
			}
		}

		System.out.println("The character: " + search + " has been found: " + count + " times.");


	}
}

@dlofnep Thanks that helps alot.

Gman_Ireland

@dlofnep Thanks got it and have it in, really appreciate the help.

how do you put the code in the way you did on boards.

import java.util.Scanner;

class  week5_scanner {

		public static void main(String args[]){
		
			String name;
			Scanner in = new Scanner(System.in);
				


			System.out.println("What is your name" );
			name = in.nextLine();
			char c1 = name.charAt(0);
			int pos = name.indexOf(" ");
			char c2 = name.charAt(pos + 1);
			int count = 0;
			String search = "e";
			String currentChar; 
			in.close();
			
			String lowername= name.toLowerCase();
			System.out.println("Hello " + lowername );
			
			int len = name.length();
			System.out.println("STR LEN = " + name.length() );
			
			System.out.println("My initials are :- " + c1 + c2);
			
			char at = name.charAt(0); 
			System.out.println("first char is = " + name.charAt(0));
			
			for(int i=0; i<name.length(); i++){
			currentChar = Character.toString(name.charAt(i));
			if(currentChar.equals(search)){
				count++;
			}
		}
			System.out.println("The character: " + search + " has been found: " + count +  " times.");
			
			 
					
	} 
}

dlofnep

No probs

You place it within the CODE tags. If you have any questions about the code, just ask and I'll explain it all to you.

Reku

dlofnep wrote: »

Sorry, forgot to convert the character to a string. Here's full working code for you.

public class countChar{
	public static void main(String[] args){
		int count = 0;
		String name = "Teddy"; 
		String search = "d"; // replace this with user input
		String currentChar;

		for(int i=0; i<name.length(); i++){
			currentChar = Character.toString(name.charAt(i));
			if(currentChar.equals(search)){
				count++;
			}
		}

		System.out.println("The character: " + search + " has been found: " + count + " times.");


	}
}

Why convert the character you wish to search for and the character at index i to strings? Are you just trying to make it easy to extend it to searching for actual strings (e.g. just switch the charAt function for a subString one)?

char currentChar = 'd'; //whatever character you wish to search for.

for(int i=0; i<name.length(); i++)
{
	if(currentChar == name.charAt(i))
        {
		count++;
	}
}

Gman_Ireland

@dlofnep

thanks for that, 


                               think i got it now. 


defo learning something everyday this week. 

dlofnep

Reku wrote: »

Why convert the character you wish to search for and the character at index i to strings? Are you just trying to make it easy to extend it to searching for actual strings (e.g. just switch the charAt function for a subString one)?

.equals requires the character to be a string. I just figured that charAt() would be more intuitive than substring() for the OP to read. No other reason tbh. But yes, either way would work.