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check if the input is integer

83
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
Sep 28 '06 #1
11 188025
r035198x
13,262 8TB
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
Expand|Select|Wrap|Line Numbers
  1. String input = ....
  2. try {
  3.     int x = Integer.parseInt(input);
  4.     System.out.println(x);
  5. }
  6. catch(NumberFormatException nFE) {
  7.     System.out.println("Not an Integer");
  8. }
Sep 28 '06 #2
sang
83
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
Sep 28 '06 #3
D_C
293 100+
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.
Expand|Select|Wrap|Line Numbers
  1. for each character in the string
  2. {
  3.   if(character is a digit)
  4.   {
  5.     while(next character is a digit); 
  6.     print the substring of all digits
  7.   }
  8. }
Sep 28 '06 #4
sang
83
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
Sep 29 '06 #5
r035198x
13,262 8TB
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

Expand|Select|Wrap|Line Numbers
  1. public class Numbers {
  2.     public static void main(String[] args) {
  3.         String str = "java5.02ds77dfsff";
  4.         char[] all = str.toCharArray();
  5.         String numbers = "";
  6.         for(int i = 0; i < all.length;i++) {
  7.             if(Character.isDigit(all[i])) {
  8.                 numbers = numbers + all[i];
  9.             }
  10.         }
  11.         System.out.println(numbers);
  12.  
  13.     }
  14. }
Sep 29 '06 #6
sang
83
Thanks a lot I find out my mistakes i will change it

Thank you for your guidens.
Sep 29 '06 #7
public static boolean isNumeric(String aStringValue) {
Pattern pattern = Pattern.compile( "\\d+" );

Matcher matcher = pattern.matcher(aStringValue);
return matcher.matches();
}
Sep 28 '10 #8
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);
}
}
}
Nov 5 '11 #9
hi,
you can do like this way also

Expand|Select|Wrap|Line Numbers
  1. String st=input
  2.  
  3. Integer i=new Integer(st)
  4.  
  5. if(i instance of Integer){
  6. int j=i.intValue();
  7. System.out.println(j);}
Mar 13 '13 #10
r035198x
13,262 8TB
That won't even compile. Please only suggest things when you are sure you know what you are talking about.
Mar 14 '13 #11
The best solution is to use a Pattern like kadsoft suggested before.
Oct 12 '16 #12

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