我知道有一百万种方式做到这一点,但最快的是什么?这应该包括科学记数法.
注意:我没有兴趣将值转换为Double,我只想知道是否可能.即private boolean isDouble(String value).
解决方法
您可以使用Double类使用的相同正则表达式来检查它.这里有很好的记录:
http://docs.oracle.com/javase/6/docs/api/java/lang/Double.html#valueOf%28java.lang.String%29
这是代码部分:
To avoid calling this method on an invalid string and having a NumberFormatException be thrown,the regular expression below can be used to screen the input string:
- 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
- }