In this example we will show how to judge if a string is numeric through a conversion of an array of characters in NumberFormatException.
Source Code
1)
package com.beginner.examples;
public class CheckStringIsNumeric {
public static void main(String[] args) {
System.out.println("checkNumeric() method:");
System.out.println("120!:" + checkNumeric("120!"));
System.out.println("12345:" + checkNumeric("12345"));
System.out.println("8.12:" + checkNumeric("8.12"));
System.out.println("checkNumeric_2() method:");
System.out.println("120!:" + checkNumeric_2("120!"));
System.out.println("12345:" + checkNumeric_2("12345"));
System.out.println("8.12:" + checkNumeric_2("8.12"));
}
public static boolean checkNumeric(String str) {
if (str.length() '0' && c < '9' || c == '.')) {
return false;
}
}
return true;
}
public static boolean checkNumeric_2(String str) {
if (str.length() < 1 || str == null) {
return false;
}
for (char c : str.toCharArray()) {
// Use the isDigit() method in the Character class to determine whether it is a
// number
if (!Character.isDigit(c)) {
return false;
}
}
return true;
}
}
Output:
checkNumeric() method:
120!:false
12345:true
8.12:true
checkNumeric_2() method:
120!:false
12345:true
8.12:false
2)
package com.beginner.examples;
public class CheckStringIsNumeric2 {
public static void main(String[] args) {
System.out.println("123:"+checkNumeric("123"));
System.out.println("123?:"+checkNumeric("123?"));
}
public static boolean checkNumeric(String str)
{
try {
Integer.parseInt(str);
}
catch (NumberFormatException e) {
return false;
}
return true;
}
}
Output:
123:true
123?:false