-
/*Check if string contains valid number example.This example shows how to check if string contains valid numberor not using parseDouble and parseInteger methods ofDouble and Integer wrapper classes.*/
public class CheckValidNumberExample {
public static void main(String[] args) {
String[] str = new String[]{“10.20”, “123456”, “12.invalid”};
for(int i=0 ; i < str.length ; i ++){
if( str[i].indexOf(“.”) > 0 ){
try{/** To check if the number is valid decimal number, use* double parseDouble(String str) method of* Double wrapper class.** This method throws NumberFormatException if the* argument string is not a valid decimal number.*/Double.parseDouble(str[i]);System.out.println(str[i] + ” is a valid decimal number”);}catch(NumberFormatException nme){System.out.println(str[i] + ” is not a valid decimal number”);}
}else{try{/** To check if the number is valid integer number, use* int parseInt(String str) method of* Integer wrapper class.** This method throws NumberFormatException if the* argument string is not a valid integer number.*/
Integer.parseInt(str[i]);System.out.println(str[i] + ” is valid integer number”);}catch(NumberFormatException nme){System.out.println(str[i] + ” is not a valid integer number”);}}}
}}
/*Output would be10.20 is a valid decimal number123456 is valid integer number12.invalid is not a valid decimal number*/
Friday 13 April 2018
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