为什么同样的一段正则表达式， JS 和 Java 的匹配结果不一样呢？

[NeroKamin]

const regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[^a-zA-Z\d<>"'\\|])(?!.*[<>"'\\|])/;
console.log(regex.test("AaBb123!@#"));
// output: true

public static void main(String[] args) {
        Pattern compile = Pattern.compile("^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[^a-zA-Z\\d<>\"'\\\\|])(?!.*[<>\"'\\\\|])");
        Matcher matcher = compile.matcher("AaBb123!@#");
        System.out.println(matcher.matches());
        // output: false
}

丢给 AI 也看不出什么问题，AI 的结论是两段代码输出都应该一样为 true

[revlis7]

\d 多了一个斜杠

[NeroKamin]

@revlis7 #1 改成\d 会报错

[nagisaushio]

java 的 matcher.matches 是尝试匹配整个字符串吧？你在 java 的正则后面加个 .* 就对了

[javaluo]

public static void main(String[] args) {
Pattern compile = Pattern.compile("^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[^a-zA-Z\\d<>\"'\\\\|])(?!.*[<>\"'\\\\|])");
Matcher matcher = compile.matcher("AaBb123!@#");
System.out.println(matcher.matches()); }

[cheese]

Pattern compile = Pattern.compile("^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[^a-zA-Z\\d<>\"'\\\\|])(?!.*[<>\"'\\\\|])");试试这段

[revlis7]

3 楼应该是对的，或者用 matcher.find()

https://stackoverflow.com/questions/4450045/difference-between-matches-and-find-in-java-regex

[NeroKamin]

@nagisaushio #3 加上确实好使了。也就是说 js regexp.test 和 java matcher.matches 的意义是不一样的吗？