Java 从电话号码中删除破折号
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Remove dash from a phone number
提问by zoom_pat277
What regular expression using java could be used to filter out dashes '-' and open close round brackets from a string representing phone numbers...
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关闭圆括号......
so that (234) 887-9999 should give 2348879999 and similarly 234-887-9999 should give 2348879999.
所以 (234) 887-9999 应该给出 2348879999，同样地 234-887-9999 应该给出 2348879999。
Thanks,
谢谢，
采纳答案by Vivin Paliath
phoneNumber.replaceAll("[\s\-()]", "");
The regular expression defines a character class consisting of any whitespace character (\s, which is escaped as \\sbecause we're passing in a String), a dash (escaped because a dash means something special in the context of character classes), and parentheses.
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一些特殊的东西而被转义）和括号组成。
See String.replaceAll(String, String).
见String.replaceAll(String, String)。
EDIT
编辑
Per gunslinger47:
phoneNumber.replaceAll("\D", "");
Replaces any non-digit with an empty string.
用空字符串替换任何非数字。
回答by Tharaka Devinda
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+"") //if the number is starting with no zero and +, its a local number. prepend cc
.replaceAll("(.)(\++)(.)", "") //if there are left out +'s in the middle by mistake, remove them
.replaceAll("(^0{2}|^\+)(.+)", "") //make 00XXX... numbers and +XXXXX.. numbers into XXXX...
.replaceAll("^0([1-9])", countryCode+""); //make 0XXXXXXX numbers into CCXXXXXXXX numbers
return out;
}
