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()?
Double.parseDouble()
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。
Double.valueOf(String)
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。
try catch
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 }