如何检查字符串是否可解析为double？

Question

是否有一种本机方式(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

像往常一样,Apache以Apache Commons-Lang的形式 提供了很好的答案Apache

处理空值,不需要Apache Commons-Lang/ NumberUtils.isCreatable(String)阻止.

Answer 2

您始终可以在try catch块中包装Double.parseDouble().

try
{
  Double.parseDouble(number);
}
catch(NumberFormatException e)
{
  //not a double
}

Answer 3

常见的方法是使用正则表达式检查它,就像Double.valueOf(String)文档中也提到的那样.

在那里(或包含在下面)提供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精细的点.

如果你不想这样做,try catch仍然是一个选择.

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
}

Answer 4

像下面这样的东西就足够了: -

String decimalPattern = "([0-9]*)\\.([0-9]*)";  
String number="20.00";  
boolean match = Pattern.matches(decimalPattern, number);
System.out.println(match); //if true then decimal else not  

Answer 5

谷歌的番石榴库提供了一个很好的辅助方法来做到这一点:Doubles.tryParse(String).您可以使用它,Double.parseDouble但null如果字符串不解析为double,则返回而不是抛出异常.

Answer 6

所有答案都可以,取决于你想要的学术水平.如果您希望准确地遵循Java规范,请使用以下命令:

private static final Pattern DOUBLE_PATTERN = Pattern.compile(
    "[\\x00-\\x20]*[+-]?(NaN|Infinity|((((\\p{Digit}+)(\\.)?((\\p{Digit}+)?)" +
    "([eE][+-]?(\\p{Digit}+))?)|(\\.((\\p{Digit}+))([eE][+-]?(\\p{Digit}+))?)|" +
    "(((0[xX](\\p{XDigit}+)(\\.)?)|(0[xX](\\p{XDigit}+)?(\\.)(\\p{XDigit}+)))" +
    "[pP][+-]?(\\p{Digit}+)))[fFdD]?))[\\x00-\\x20]*");

public static boolean isFloat(String s)
{
    return DOUBLE_PATTERN.matcher(s).matches();
}

此代码基于Double的JavaDocs .