问题描述

是否有一种本地方式(最好不实现自己的方法)来检查字符串是否可以用 Double.parseDouble() 解析?

Is there a native way (preferably without implementing your own method) to check that a string is parseable with Double.parseDouble()?

推荐答案

常见的方法是使用正则表达式检查它，就像 Double.valueOf(String) 文档.

The common approach would be to check it with a regular expression like it's also suggested inside the Double.valueOf(String) documentation.

那里提供(或包含在下面)的正则表达式应该涵盖所有有效的浮点情况，所以你不需要摆弄它，因为你最终会错过一些更好的点.

The regexp provided there (or included below) should cover all valid floating point cases, so you don't need to fiddle with it, since you will eventually miss out on some of the finer points.

如果您不想这样做，try catch 仍然是一种选择.

If you don't want to do that, try catch is still an option.

JavaDoc 建议的正则表达式如下:

The regexp suggested by the JavaDoc is included below:

final String Digits     = "(\p{Digit}+)";
final String HexDigits  = "(\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally 
// signed decimal integer.
final String Exp        = "[eE][+-]?"+Digits;
final String fpRegex    =
    ("[\x00-\x20]*"+ // Optional leading "whitespace"
    "[+-]?(" +         // Optional sign character
    "NaN|" +           // "NaN" string
    "Infinity|" +      // "Infinity" string

    // A decimal floating-point string representing a finite positive
    // number without a leading sign has at most five basic pieces:
    // Digits . Digits ExponentPart FloatTypeSuffix
    // 
    // Since this method allows integer-only strings as input
    // in addition to strings of floating-point literals, the
    // two sub-patterns below are simplifications of the grammar
    // productions from the Java Language Specification, 2nd 
    // edition, section 3.10.2.

    // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
    "((("+Digits+"(\.)?("+Digits+"?)("+Exp+")?)|"+

    // . Digits ExponentPart_opt FloatTypeSuffix_opt
    "(\.("+Digits+")("+Exp+")?)|"+

    // Hexadecimal strings
    "((" +
    // 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "(\.)?)|" +

    // 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "?(\.)" + HexDigits + ")" +

    ")[pP][+-]?" + Digits + "))" +
    "[fFdD]?))" +
    "[\x00-\x20]*");// Optional trailing "whitespace"

if (Pattern.matches(fpRegex, myString)){
    Double.valueOf(myString); // Will not throw NumberFormatException
} else {
    // Perform suitable alternative action
}

这篇关于如何检查字符串是否可解析为双精度字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！