将key = value的String解析为Map [英] Parse a String of key=value to a Map
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问题描述
我正在使用一个API,它为我提供了一个XML,我需要从一个标签获取一个实际上是字符串的地图。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:
Billable=7200,Overtime=false,TransportCosts=20$
我需要
["Billable"="7200","Overtime=false","TransportCosts"="20$"]
问题是字符串是完全动态的,所以,它可以像
The problem is that the string is totally dynamic, so, it can be like
Overtime=true,TransportCosts=one, two, three
Overtime=true,TransportCosts=1= 1,two, three,Billable=7200
所以我不能用逗号分割,然后用等号分开。
是否可以使用正则表达式将类似字符串转换为地图?
So I can not just split by comma and then by equal sign. Is it possible to convert a string like those to a map using a regex?
到目前为止我的代码是:
My code so far is:
private Map<String, String> getAttributes(String attributes) {
final Map<String, String> attr = new HashMap<>();
if (attributes.contains(",")) {
final String[] pairs = attributes.split(",");
for (String s : pairs) {
if (s.contains("=")) {
final String pair = s;
final String[] keyValue = pair.split("=");
attr.put(keyValue[0], keyValue[1]);
}
}
return attr;
}
return attr;
}
提前谢谢
推荐答案
您可以使用
(\w+)=(.*?)(?=,\w+=|$)
参见正则表达式演示。
详细信息
-
(\ w +)
- 第1组:一个或多个单词字符 -
=
- 等号 -
(。*?)
- 第2组:除了换行符之外的任何零个或多个字符,尽可能少 -
(?=,\ w + = | $)
- 一个积极的前瞻,需要一个,
,然后是1个字的字符,然后是=
,或紧接在当前位置右侧的字符串结尾。
(\w+)
- Group 1: one or more word chars=
- an equal sign(.*?)
- Group 2: any zero or more chars other than line break chars, as few as possible(?=,\w+=|$)
- 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;
}
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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