如何使用逗号和城市代码之间的逗号将城市和州从破折号格式化为空格。 （JavaScript的）

Question

例如，我有一个字符串Atlantic-City-NJ如何使用JavaScript正则表达式（或任何其他方式）将字符串格式化为Atlantic City, NJ？

我遇到问题，因为可能会有不同的空格：Atlanta, GA vs Atlantic City, NJ vs Lake Havasu City, AZ

Answer 1

你当然可以使用U.S. state codes，但如果验证不重要，你可以像这样完成所有工作：

＆＃13;

＆＃13;

var str = "Atlantic-City-NJ";
alert(str.replace(/(.+)\-([A-Z]{2})$/, "$1, $2").replace("\-", " "));

＆＃13;

＆＃13;

＆＃13;

Answer 2

这个表达式将为您提供两个部分：

([\w-]+)([A-Z]{2})

“Atlantic-City-”和“NJ”。

你可以拿这两部分，在第一部分做一个字符串替换，将连字符转换成空格，然后从结果中形成一个字符串，一个逗号，再加上第二部分。

如果您想了解更多细节，请与我们联系。

Answer 3

您可以这样做：

var test = "Atlantic-City-NJ";
test = test.split("-");
var result = "";
for (var i = 0; i < test.length; i++) {
  if (i == test.length - 1)
    result += ",";
  result += " " + test[i];
}
//result = "Atlantic City, NJ"
alert(result);

Answer 4

以下表达式将匹配最后一个连字符（ - ）和国家/地区代码

/([\w-]+)(-(([^-])+)$)/

这样你可以做到

string.replace(/([\w-]+)(-(([^-])+)$)/, '$1, $3').replace(/-/g, ' ')

这将用逗号和国家/地区代码替换最后的连字符和国家/地区代码，并用空格替换其余的连字符。

正则表达式描述：

1st Capturing group ([\w-]+)
  [\w-]+ match a single character present in the list below
  Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
  \w match any word character [a-zA-Z0-9_]

  - the literal character -

    2nd Capturing group (-(([^-])+)$)
      - matches the character - literally

      3rd Capturing group (([^-])+)

      4th Capturing group ([^-])+
        Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
        [^-] match a single character not present in the list below
          - the literal character -

            $ assert position at end of the string

Answer 5

var cs = ["Lake-Havasu-City-AZ", "Altanta-GA", "Atlantic-City-NJ", "Atlantic-City"];
for ( var i in cs )
   console.log(cs[i].replace( new RegExp('-([A-Z]+)$'), ', $1').replace('-',' ','g'))
// prints...
Lake Havasu City, AZ
Altanta, GA
Atlantic City, NJ
Atlantic City

如果您想更简洁地了解州代码，可以将[A-Z]+替换为[A-Z]{2}。