如何使正则表达式匹配测量单位？
我正在构建一个小型 Java 库，它必须匹配字符串中的单位。例如，如果我有“300000000 m/s^2”，我希望它与“m”和“s^2”匹配。
到目前为止，我已经尝试了最能想象到的（我自己）类似的配置（我希望这是一个好的开始）
"[[a-zA-Z]+[\\^[\\-]?[0-9]+]?]+"
为了澄清，我需要一些与 letters[^[-]numbers] 匹配的东西（其中 [ ]表示非强制性部分）。这意味着：字母，后面可能跟着一个可能为负数的指数。
我研究了一点正则表达式，但我真的不太流利，所以任何帮助将不胜感激！
非常感谢，
编辑： 我刚刚尝试了前 3 个回复
String regex1 = "([a-zA-Z]+)(?:\\^(-?\\d+))?";
String regex2 = "[a-zA-Z]+(\\^-?[0-9]+)?";
String regex3 = "[a-zA-Z]+(?:\\^-?[0-9]+)?";
，但它不起作用...我知道测试模式的代码有效，因为如果我尝试一些简单的操作，例如在“12345”中匹配“[0-9]+”，它将匹配整个字符串。所以，我不明白还有什么问题。我正在尝试在目前需要的地方更改我的方括号...
用于测试的代码：
public static void main(String[] args) {
String input = "30000 m/s^2";
// String input = "35345";
String regex1 = "([a-zA-Z]+)(?:\\^(-?\\d+))?";
String regex2 = "[a-zA-Z]+(\\^-?[0-9]+)?";
String regex3 = "[a-zA-Z]+(?:\\^-?[0-9]+)?";
String regex10 = "[0-9]+";
String regex = "([a-zA-Z]+)(?:\\^\\-?[0-9]+)?";
Pattern pattern = Pattern.compile(regex3);
Matcher matcher = pattern.matcher(input);
if (matcher.matches()) {
System.out.println("MATCHES");
do {
int start = matcher.start();
int end = matcher.end();
// System.out.println(start + " " + end);
System.out.println(input.substring(start, end));
} while (matcher.find());
}
}
I'm building a small Java library which has to match units in strings. For example, if I have "300000000 m/s^2", I want it to match against "m" and "s^2".
So far, I have tried most imaginable (by me) configurations resembling (I hope it's a good start)
"[[a-zA-Z]+[\\^[\\-]?[0-9]+]?]+"
To clarify, I need something that will match letters[^[-]numbers] (where [ ] denotes non obligatory parts). That means: letters, possibly followed by an exponent which is possibly negative.
I have studied regex a little bit, but I'm really not fluent, so any help will be greatly appreciated!
Thank you very much,
EDIT:
I have just tried the first 3 replies
String regex1 = "([a-zA-Z]+)(?:\\^(-?\\d+))?";
String regex2 = "[a-zA-Z]+(\\^-?[0-9]+)?";
String regex3 = "[a-zA-Z]+(?:\\^-?[0-9]+)?";
and it doesn't work... I know the code which tests the patterns work, because if I try something simple, like matching "[0-9]+" in "12345", it will match the whole string. So, I don't get what's still wrong. I'm trying with changing my brackets for parenthesis where needed at the moment...
CODE USED TO TEST:
public static void main(String[] args) {
String input = "30000 m/s^2";
// String input = "35345";
String regex1 = "([a-zA-Z]+)(?:\\^(-?\\d+))?";
String regex2 = "[a-zA-Z]+(\\^-?[0-9]+)?";
String regex3 = "[a-zA-Z]+(?:\\^-?[0-9]+)?";
String regex10 = "[0-9]+";
String regex = "([a-zA-Z]+)(?:\\^\\-?[0-9]+)?";
Pattern pattern = Pattern.compile(regex3);
Matcher matcher = pattern.matcher(input);
if (matcher.matches()) {
System.out.println("MATCHES");
do {
int start = matcher.start();
int end = matcher.end();
// System.out.println(start + " " + end);
System.out.println(input.substring(start, end));
} while (matcher.find());
}
}
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评论（3）
如果您匹配单个字符，则无需使用字符类
[...]。(...)这里是一个捕获括号，供您稍后提取单位和指数。(?:...)是非捕获分组。You don't need to use the character class
[...]if you're matching a single character.(...)here is a capturing bracket for you to extract the unit and exponent later.(?:...)is non-capturing grouping.您混合使用方括号来表示字符类和大括号来进行分组。请尝试这样做：
在许多正则表达式方言中，您可以使用 \d 来表示任何数字而不是 [0-9]。
You're mixing the use of square brackets to denote character classes and curly brackets to group. Try this instead:
In many regular expression dialects you can use \d to mean any digit instead of [0-9].
尝试
Try