Getting an information about type of input ( main args )

This question already has an answer here:

How do I convert a String to an int in Java? 30 answers

---

最简单的方法是使用parseInt进行检查。

public static void main(String[] args) throws InterruptedException
{
    try
    {
        Integer.parseInt(args[0]);
    }
    catch(NumberFormatException e)
    {
        System.out.println("It is not a number.");
    }
}

---

args[] takes in all the arguments supplied to the main method, so args[0] although as a String, if you expected the user to enter an Integer you could cast this with...

Integer.parseInt(args[0]);

Consider the fact though it may through a java.lang.NumberFormatException so you'd have to check that in case the user entered an invalid input. If you wanted to do something based on if this is truly a number, look at StringUtils.isNumeric which is in Apache Commons Lang, why reinvent the wheel?

The reason I would go for this rather than try{} catch() is because exceptions should be used for exceptional circumstances.

---

The type of your argument is always just String . If that string is "", "Foo" or "182" is at this point totally arbitrary.

These strings may, or may not be valid representations of a number. You need to convert the strings to your number type (eg Integer.parseInt ) for int .

链接地址: http://www.djcxy.com/p/20930.html

上一篇: 如何将String []或Object []的组件转换为其他类型

下一篇: 获取关于输入类型的信息（主要参数）