英文:
Create a Reg-ex which accept numbers and + as the first character
问题
我正在尝试在Java中创建一个正则表达式，以便我可以获得类似于 +82 11-1111-1111 的结果。第一个数字应以 +82 开头，后跟一个空格，然后它只能接受 11、12 或 13，其余数字可以接受 0-9 之间的数字。
例如：
+82 12-2234-3344+82 13-1234-5678+82 11-3456-4567
英文:
I am trying to create a regex where I get the results as +82 11-1111-1111 in java. The first number should start with +82 with a space then it should get accept only 11 or 12 or 13 and the n the rest of the numbers can accept numbers between 0-9
For eg.
+82 12-2234-3344+82 13-1234-5678+82 11-3456-4567
Please help
答案1
得分: 1
我认为你应该尝试类似这样的写法：
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class PhoneNumberRegex {
public static void main(String[] args) {
System.out.println(validate("+82 12-2234-3344"));
System.out.println(validate("+82 13-1234-5678"));
System.out.println(validate("+82 11-3456-4567"));
System.out.println(validate("+82 31-3456-4567"));
System.out.println(validate("+82 31-34564-4567"));
}
private static boolean validate(String input) {
Pattern argPattern = Pattern.compile("\\+82 (12|13|11)(?:-\\d{4}){2}");
Matcher matcher = argPattern.matcher(input);
return matcher.matches();
}
}
当我运行这段代码时，得到如下输出：
src : $ java PhoneNumberRegex
true
true
true
false
false
英文:
I think you should try something like this:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class PhoneNumberRegex {
public static void main(String[] args) {
System.out.println(validate("+82 12-2234-3344"));
System.out.println(validate("+82 13-1234-5678"));
System.out.println(validate("+82 11-3456-4567"));
System.out.println(validate("+82 31-3456-4567"));
System.out.println(validate("+82 31-34564-4567"));
}
private static boolean validate(String input) {
Pattern argPattern = Pattern.compile("\\+82 (12|13|11)(?:-\\d{4}){2}");
Matcher matcher = argPattern.matcher(input);
return matcher.matches();
}
}
When I run this, I get this output:
src : $ java PhoneNumberRegex
true
true
true
false
false
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