将 key=value 的字符串解析为 Map

2022-01-08 00:00:00 string xml regex hashmap java

我正在使用一个提供 XML 的 API，我需要从一个实际上是字符串的标签中获取地图.示例:

I'm using an API that gives me a XML and I need to get a map from one tag which is actually a string. Example:

拥有

Billable=7200,Overtime=false,TransportCosts=20$

我需要

["Billable"="7200","Overtime=false","TransportCosts"="20$"]

问题是字符串是完全动态的，所以，它可以像

The problem is that the string is totally dynamic, so, it can be like

Overtime=true,TransportCosts=one, two, three
Overtime=true,TransportCosts=1= 1,two, three,Billable=7200

所以我不能只用逗号分隔，然后用等号分隔.是否可以使用正则表达式将类似的字符串转换为地图?

So I can not just split by comma and then by equal sign. Is it possible to convert a string like those to a map using a regex?

到目前为止我的代码是:

My code so far is:

private Map<String, String> getAttributes(String attributes) {
    final Map<String, String> attr = new HashMap<>();
    if (attributes.contains(",")) {
        final String[] pairs = attributes.split(",");
        for (String s : pairs) {
            if (s.contains("=")) {
                final String pair = s;
                final String[] keyValue = pair.split("=");
                attr.put(keyValue[0], keyValue[1]);
            }
        }
        return attr;
    }
    return attr;
}

提前致谢

推荐答案

你可以用

(w+)=(.*?)(?=,w+=|$)

请参阅 正则表达式演示.

详情

  • (w+) - 第 1 组:一个或多个单词字符
  • = - 一个等号
  • (.*?) - 第 2 组:除换行符以外的任何零个或多个字符，尽可能少
  • (?=,w+=|$) - 需要 ,，然后是 1+ 个单词字符，然后是 =，或字符串的结尾紧挨当前位置的右侧.
  • (w+) - Group 1: one or more word chars
  • = - an equal sign
  • (.*?) - Group 2: any zero or more chars other than line break chars, as few as possible
  • (?=,w+=|$) - a positive lookahead that requires a ,, then 1+ word chars, and then =, or end of string immediately to the right of the current location.

Java 代码:

public static Map<String, String> getAttributes(String attributes) {
    Map<String, String> attr = new HashMap<>();
    Matcher m = Pattern.compile("(\w+)=(.*?)(?=,\w+=|$)").matcher(attributes);
    while (m.find()) {
        attr.put(m.group(1), m.group(2));
    }
    return attr;
}

Java 测试:

String s = "Overtime=true,TransportCosts=1= 1,two, three,Billable=7200";
Map<String,String> map = getAttributes(s);
for (Map.Entry entry : map.entrySet()) {
    System.out.println(entry.getKey() + "=" + entry.getValue());
}

结果:

Overtime=true
Billable=7200
TransportCosts=1= 1,two, three

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