如何确定一个字符串只有任何长度的两倍？

Question

我需要一个正则表达式，我想看看一个字符串是否只是一个双精度数，但它可以是任意长度，有什么建议吗？

Answer 1

Could you try and convert the string to a double and catch the exception if it fails?

try{
  Double aDouble = Double.parseDouble(aString);
}catch (NumberFormatException nfe){
// handle it not being a Double here
}

Answer 2

A double would match the following regexp:

^[-+]?\d*\.?\d+([eE][-+]?\d+)?$

Note that for a Java String you would need to escape the backslashes, and that \\d is a shorthand for [0-9] .

Answer 3

You might also like to look at the NumberUtils from Apache Commons

http://commons.apache.org/lang/api-2.4/org/apache/commons/lang/math/NumberUtils.html

Answer 4

尝试^(\\d+|\\d+\\.|\\d*\\.\\d+)[dD]?$

Answer 5

If I take your meaning correctly, this regular expression might work

"/^[+-]?\d+\.\d+$/"

In English -> A + or - sign (maybe) followed by one or more digits followed by a . (decimal point) followed by one or more digits. This regular expression assumes that numbers are not formatted with commas and also assumes that the decimal separator is a dot which is not true in all locales.

Answer 6

I stole this expression from perldoc faq4 -- "is a decimal number".

import java.util.regex.Pattern;

    public class Test {

        public static void main(String[] args) {
            Pattern pattern = 
                Pattern.compile("^-?(?:\\d+(?:\\.\\d*)?|\\.\\d+)$");

            System.out.println(((Boolean) pattern.matcher("1").find()).toString());
            System.out.println(((Boolean) pattern.matcher("1.1").find()).toString());
            System.out.println(((Boolean) pattern.matcher("abc").find()).toString());


        }

    }

Prints:

true
true
false