java - Reg~x 代表 N~number

Question

感谢您的帮助。

什么是优雅的正则表达式来检查一个字符串是否包含一个数字？有什么建议么？谢谢。

Answer 1

我不认为异常(即使它被触发)比正则表达式昂贵得多 - 您必须分析它并查看它是否真的有所作为。

也就是说，一个根据 Java Docs 实现 BigDecimal 语法的正则表达式是

[+-]?(?:\d+(?:\.\d*)?|\.\d+)(?:[eE][+-]?\d+)?

解释:

[+-]?       # optional sign
(?:         # required significand: either...
 \d+        # a number
 (?:\.\d*)? # optionally followed by a dot, optionally followed by more digits
 |          # or...
 \.\d+      # just a dot, followed by digits (in this case required)
)           # end of significand
(?:         # optional exponent
 [eE]       # required exponent indicator
 [+-]?      # optional sign
 \d+        # required digits
)?          # end of exponent

如果您想允许不同的数字格式(和千位分隔符)，您可以先从 DecimalFormatSymbols 中获取这些值并用它构建你的正则表达式。

像这样(我不懂 Java，所以请随时纠正我的语法错误):

// you need java.util.regex.Pattern and java.text.DecimalFormatSymbols
string ds = Pattern.quote(DecimalFormatSymbols.getDecimalSeparator())
string gs = Pattern.quote(DecimalFormatSymbols.getGroupingSeparator())
string ms = Pattern.quote(DecimalFormatSymbols.getMinusSign())
string es = Pattern.quote(DecimalFormatSymbols.getExponentSeparator())

string myre = 
    "(?xi)\n       # verbose, case-insensitive regex" +
    "[+" +ms+ "]?  # optional sign\n" +
    "(?:           # required significand: either...\n" +
    " (?:\\d{1,3}(?:" +gs+ "\\d{3}|\\d++) # a number with optional thousand separators,\n" +
    " (?:" +ds+ "\\d*)? # optionally followed by a dot, optionally followed by more digits\n" +
    " |            # or...\n" +
      ds+ "\\d+    # just a dot, followed by digits (in this case required)\n" +
    ")             # end of significand\n" +
    "(?:           # optional exponent\n" +
      es+ "        # required exponent indicator\n" +
    " [+" +ms+ "]? # optional sign\n" +
    " \\d+         # required digits\n" +
    ")?            # end of exponent"

boolean foundMatch = subjectString.matches(myregex);

然后，您可以在将数字移交给 BigDecimal 转换之前将与区域设置相关的位替换回美国对应位，如果这不是区域设置感知的话。