检查字符串中是否有 3 个字母，然后是 3 个数字

Question

我需要能够检查一个字符串，以便其中的前 3 个字符是字母，最后 3 个字符是数字，例如像“ABC123”这样的 rego，并确保它不是“123ABC”或“1A2B3C”之类的东西

Answer 1

使用以下内容：

string.matches("[a-zA-Z]{3}.*[0-9]{3}");

Answer 2

you can use this pattern with regex:

([a-zA-Z][a-zA-Z][a-zA-Z][0-9][0-9][0-9])

if you have anything in between you can use:

([a-zA-Z][a-zA-Z][a-zA-Z].*[0-9][0-9][0-9])

Answer 3

try this pattern ^[A-Za-z]{3}.*[0-9]{3}$

String test = "ABC123";
Matcher matcher = Pattern.compile("^[A-Za-z]{3}.*[0-9]{3}$").matcher(test);
if (matcher.find()) {
      System.out.print(matcher.group());
}

output:

ABC123

Answer 4

you can check it like this:

    if (!"AB1C23".matches("^([A-Za-z]{3}[0-9]{3})")) {
        System.out.println("Invalid!!");
    }

Answer 5

• \\d matches a digit

• [a-zA-Z] matches a letter

• {3} is the quantifier that matches exactly 3 repetitions

• () group maches

• ([a-zA-Z]{3})(\\\\d{3})

public static void main(String[] args) {

        Matcher matcher = Pattern.compile("([a-zA-Z]{3})(\\d{3})").matcher("ABC215");
//      Matcher matcher = Pattern.compile("([a-zA-Z]{3})(\\d{3})").matcher("1A2B3C"); //Macher not find :)
        if (matcher.find()) {
            System.out.println(matcher.group(0)); //ABC215
            System.out.println(matcher.group(1)); //ABC
            System.out.println(matcher.group(2)); //215
        }
    }

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