java - 从电话号码中删除破折号

Question

使用java的正则表达式可用于过滤掉破折号'-'并从代表电话号码的字符串中打开右圆括号......

所以 (234) 887-9999 应该给出 2348879999 并且类似地 234-887-9999 应该给出 2348879999。

谢谢，

Answer 1

phoneNumber.replaceAll("[\\s\\-()]", "");

正则表达式定义了一个字符类，它由任何空白字符 ( \s，\\s因为我们传入一个字符串而被转义)、一个破折号（因为破折号在字符类的上下文中意味着特殊的东西而被转义）和括号组成。

见String.replaceAll(String, String)。

编辑

每个gunslinger47：

phoneNumber.replaceAll("\\D", "");

用空字符串替换任何非数字。

Answer 2

    public static String getMeMyNumber(String number, String countryCode)
    {    
         String out = number.replaceAll("[^0-9\\+]", "")        //remove all the non numbers (brackets dashes spaces etc.) except the + signs
                        .replaceAll("(^[1-9].+)", countryCode+"$1")         //if the number is starting with no zero and +, its a local number. prepend cc
                        .replaceAll("(.)(\\++)(.)", "$1$3")         //if there are left out +'s in the middle by mistake, remove them
                        .replaceAll("(^0{2}|^\\+)(.+)", "$2")       //make 00XXX... numbers and +XXXXX.. numbers into XXXX...
                        .replaceAll("^0([1-9])", countryCode+"$1");         //make 0XXXXXXX numbers into CCXXXXXXXX numbers
         return out;

    }

Answer 3

Kotlin 版本的工作解决方案对我来说 - （扩展功能）

fun getMeMyNumber(number: String, countryCode: String): String? {
    return number.replace("[^0-9\\+]".toRegex(), "") //remove all the non numbers (brackets dashes spaces etc.) except the + signs
        .replace("(^[1-9].+)".toRegex(), "$countryCode$1") //if the number is starting with no zero and +, its a local number. prepend cc
        .replace("(.)(\\++)(.)".toRegex(), "$1$3") //if there are left out +'s in the middle by mistake, remove them
        .replace("(^0{2}|^\\+)(.+)".toRegex(), "$2") //make 00XXX... numbers and +XXXXX.. numbers into XXXX...
        .replace("^0([1-9])".toRegex(), "$countryCode$1") //make 0XXXXXXX numbers into CCXXXXXXXX numbers
}