check if the input is integer

[sang]

Hi

I am writing a application program that checks if the input is integer, if it is integer print the integer value only. i have no idea about that please any one help me.

Thanks
Sang

[r035198x]

Hi

I am writing a application program that checks if the input is integer, if it is integer print the integer value only. i have no idea about that please any one help me.

Thanks
Sang

Use Exception handling
as in

String input = ....
try {
    int x = Integer.parseInt(input);
    System.out.println(x);
}
catch(NumberFormatException nFE) {
    System.out.println("Not an Integer");
}

[sang]

Thanks for your reply.

This coding is print if the input is integer otherwise not printed. I am trying to get only integer values i am not able to do this by using the isNumber() but i am got the error.

so,please give me the code for

if the input is like this "There is 3 apples in 1 tree" the output is only 1 & 3

that is only print the integer.

Thanks
Sang

[D_C]

If you are going to search for a number within a string, then print the number, you don't even need to convert it to an integer. If you are using negatives, be sure to modify the if statement to accept minus sign as well as digit.

for each character in the string
{
  if(character is a digit)
  {
    while(next character is a digit); 
    print the substring of all digits
  }
}

[sang]

Thanks for your advice but i am not able to do that because i am new to java for that reason i am strugle in this.

The following code is copmlied but it is not correct please correct it and then sent to me

import java.io.*;
class str {
public boolean isNumeric(String input){
try {
char[] ch=charArray(input);
for( int i=0 ; i<input.length() ; i++)
{
if(Character.isNumber)
{
System.out.println(i);
}
else { return false;}
}
}catch(NumberFormatException nfe) {
System.out.println(nfe);
}
}
public static void main(String arg[]) {
int c = str.isNumeric("java 123");
System.out.println(c);
}
}

Thanks
Sang

[r035198x]

Thanks for your advice but i am not able to do that because i am new to java for that reason i am strugle in this.

The following code is copmlied but it is not correct please correct it and then sent to me

import java.io.*;
class str {
public boolean isNumeric(String input){
try {
char[] ch=charArray(input);
for( int i=0 ; i<input.length() ; i++)
{
if(Character.isNumber)
{
System.out.println(i);
}
else { return false;}
}
}catch(NumberFormatException nfe) {
System.out.println(nfe);
}
}
public static void main(String arg[]) {
int c = str.isNumeric("java 123");
System.out.println(c);
}
}

Thanks
Sang

An initial attempt would be

public class Numbers {
    public static void main(String[] args) {
        String str = "java5.02ds77dfsff";
        char[] all = str.toCharArray();
        String numbers = "";
        for(int i = 0; i < all.length;i++) {
            if(Character.isDigit(all[i])) {
                numbers = numbers + all[i];
            }
        }
        System.out.println(numbers);
 
    }
}

[sang]

Thanks a lot I find out my mistakes i will change it

Thank you for your guidens.

[kadsoft]

public static boolean isNumeric(String aStringValue) {
Pattern pattern = Pattern.compile( "\\d+" );

Matcher matcher = pattern.matcher(aStringValue);
return matcher.matches();
}

[abhishek kush]

import java.io.*;
class abhi
{
public static void main(String args[])throws IOException
{
DataInputStream d=new DataInputStream(System.in);
int i,as,l;
String na;
na=d.readLine();
l=na.length();
char ch;
for(i=0;i<l;i++)
{
ch=na.charAt(i);
as=(int)ch;
if(as>=48&&as<=57)
System.out.println(ch);
}
}
}

[venkat4u]

hi,
you can do like this way also

String st=input
 
Integer i=new Integer(st)
 
if(i instance of Integer){
int j=i.intValue();
System.out.println(j);}

[r035198x]

That won't even compile. Please only suggest things when you are sure you know what you are talking about.

[Karidev]

The best solution is to use a Pattern like kadsoft suggested before.