如何在java中为匹配String和BigInteger编写正则表达式？

我试图使用这个网站。但我没有成功

我的Java正则表达式：

(.*)\\w+=[0-9][0-9]*$(.*)

根据以下内容进行测试的条目：

a = 800000000000000000000000

我的代码需要为

数据提供
true
。匹配（（.*）\\w+=[0-9][0-9]*$（.*））

的

a=80000000000000000000000

按如下操作：

public class Main {
    public static void main(String[] args) {
        String[] testStrs = { "a = 800000000000000000000000", "a = ABC800000000000000000000000",
                "a = 800000000000000000000000ABC", "a = 12.45", "a = 800000000000ABC000000000000", "a = ABC",
                "a = 900000000000000000000000", "a = -800000000000000000000000", "a = -900000000000000000000000" };
        for (String str : testStrs) {
            System.out.println(
                    str + " -> " + (str.matches("[A-Za-z]\\s+=\\s+[-]?\\d+") ? " matches." : " does not match."));
        }
    }
}

输出：

a = 800000000000000000000000 ->  matches.
a = ABC800000000000000000000000 ->  does not match.
a = 800000000000000000000000ABC ->  does not match.
a = 12.45 ->  does not match.
a = 800000000000ABC000000000000 ->  does not match.
a = ABC ->  does not match.
a = 900000000000000000000000 ->  matches.
a = -800000000000000000000000 ->  matches.
a = -900000000000000000000000 ->  matches.

说明：

a = 800000000000000000000000 ->  matches.
a = ABC800000000000000000000000 ->  does not match.
a = 800000000000000000000000ABC ->  does not match.
a = 12.45 ->  does not match.
a = 800000000000ABC000000000000 ->  does not match.
a = ABC ->  does not match.
a = 900000000000000000000000 ->  matches.
a = -800000000000000000000000 ->  matches.
a = -900000000000000000000000 ->  matches.

[A-Za-z]

用于字母表

\\s+

表示空格

[-]？

是可选的

-

\\d+

表示数字

---

在前面的答案和atmin的评论（我还不能在任何地方发表评论）的基础上展开，您还可以排除前导零，包括：

String regex = "0|(\\-?[1-9]\\d*)";
String stringToTest = "..."
boolean result = stringToTest.matches(regex);

---

你到底想实现什么？@Mureinik我希望我的程序像data.matches一样工作（（.*）\\w+=[0-9][0-9]*$（.*））==true@ArvindKumar Avinash不，它不适用于.matches（）@plaza2009-请解释您的，而不仅仅是您尝试的解决方案。不，BigInteger是任意数字，800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000。负数呢？正则表达式应该是“[-]？\\d+”，它需要匹配“a=800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000？与8000000000000不同的是，可以有任何数字，这适用于任何整数（无论多大），但它不匹配前导0（需要1-9作为第一个数字），并且没有“-0”，只需取上一个答案并替换“=\\s+”之后的部分即可；同样，这只是一个补充，因为我不能在任何地方发表评论