Java 如何检查字符串可解析为双精度？

Question

有没有一种本机的方法（最好不要实现自己的方法）来检查字符串是否可解析Double.parseDouble()？

Answer 1

常见方法是使用正则表达式进行检查，就像Double.valueOf(String)文档中也建议的那样。

此处提供的regexp（或下面包含的）应涵盖所有有效的浮点数情况，因此你无需费心处理，因为你最终会错过一些更好的点。

如果你不想这样做，try catch仍然可以选择。

JavaDoc建议的正则表达式如下：

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
}