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Java Cookbook
book

Java Cookbook

by Ian F. Darwin
June 2001
Intermediate to advanced content levelIntermediate to advanced
888 pages
21h 1m
English
O'Reilly Media, Inc.
Content preview from Java Cookbook

Checking Whether a String Is a Valid Number

Problem

You need to check if a given string contains a valid number, and if so, convert it to binary (internal) form.

Solution

Use the appropriate wrapper class’s conversion routine and catch the NumberFormat Exception. This code converts a string to a double :

// StringToDouble.java 
public static void main(String argv[]) { 
    String aNumber = argv[0];    // not argv[1] 
    double result; 
    try { 
        result = Double.parseDouble(aNumber);  
    } catch(NumberFormatException exc) { 
        System.out.println("Invalid number " + aNumber); 
        return; 
    } 
    System.out.println("Number is " + result); 
}

Discussion

Of course, that lets you validate only numbers in the format that the designers of the wrapper classes expected. If you need to accept a different definition of numbers, you could use regular expressions (see Chapter 4) to make the determination.

There may also be times when you want to tell if a given number is an integer number or a floating-point number. One way is to check for the characters ., d, or e in the input; if it is present, convert the number as a double, otherwise, convert it as an int:

// GetNumber.java System.out.println("Input is " + s); if (s.indexOf('.') >0 || s.indexOf('d') >0 || s.indexOf('e') >0) try { dvalue = Double.parseDouble(s); System.out.println("It's a double: " + dvalue); return; } catch (NumberFormatException e) { System.out.println("Invalid a double: " + s); return; } else // did not contain . or d or e, so try as int. try { ivalue ...
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Publisher Resources

ISBN: 0596001703Supplemental ContentCatalog PageErrata