在以下情况下的正则表达式匹配问题

Question

I am developing an application. User will enter some of the setting value in the server. When I ask for the value to the server through the inbuilt API. I am getting values like as a whole string:
for example-

name={abc};display={xyz};addressname={123}

Here the properties are name, display and address and there respective values are abc, xyz and 123.
I used to split with ; as first delimeter and = as a second dleimeter.

String[] propertyValues=iPropertiesStrings.split(";");
        for(int i=0;i<propertyValues.length;i++)
        {
            if(isNullEmpty(propertyValues[i]))
                continue;

            String[] propertyValue=propertyValues[i].split("=");
            if(propertyValue.length!=2)
                mPropertyValues.put(propertyValue[0], "");
            else
                mPropertyValues.put(propertyValue[0], propertyValue[1]);
        }
    }

here mPropertyValues is hash map which is used for keeping property name and its value.

Problem is there can be string :

case 1:  name={abc};display={ xyz=deno; demo2=pol };addressname={123}
case 2:  name=;display={ xyz=deno; demo2=pol };addressname={123}

I want hashmap to be filled with :

case 1:

name ="abc" 
display = "xyz= demo; demo2 =pol"
addressname = "123"

for case 2:

name =""
display = "xyz= demo; demo2 =pol"
addressname = "123"

I am looking for a regular expression to split these strings;

Answer 1

以下正则表达式应符合您的条件，并使用命名的捕获组来获取您需要的三个值。

name=\{(?<name>[^}])\};display=\{(?<display>[^}]+)\};addressname=\{(?<address>[^}]\)}

Answer 2

Assuming your dataset can change, a better parser may be more dynamic, building a Map from whatever is found in that return type.

The regex for this is pretty simple, given the cases you list above (and no nesting of {} , as others have mentioned):

Matcher m = Pattern.compile("(\\w+)=(?:\\{(.*?)\\})?").matcher(source_string);
while (m.find()) {
    if (m.groupCount() > 1) {
        hashMap.put(m.group(1), m.group(2));
    }
}

There are, however, considerations to this:

1. If m.group(2) does not exist, "null" will be the value, (you can adjust that to be what you want with a tiny amount of logic).
2. This will account for varying data-sets - in case your data in the future changes.

What that regex does:

1. (\\\\w+) - This looks for one or more word characters in a row (A-z_) and puts them into a "capture group" ( group(1) )
2. = - The literal equals
3. (?:...)? - This makes the grouping not a capture group (will not be a .group(n) , and the trailing ? makes it an optional grouping.
4. \\\\{(.*?)\\\\} - This looks for anything between the literals { and } (note: if a stray } is in there, this will break). If this section exists, the contents between {} will be in the second "capture group" ( .group(2) ).

Answer 3

Assuming that there can't be nested {} this should do what you need

String data = "name=;display={ xyz=deno; demo2=pol };addressname={123}";

Pattern p = Pattern.compile("(?<name>\\w+)=(\\{(?<value>[^}]*)\\})?(;|$)");
Matcher m = p.matcher(data);

while (m.find()){
    System.out.println(m.group("name")+"->"+(m.group("value")==null?"":m.group("value").trim()));
}

Output:

name->
display->xyz=deno; demo2=pol
addressname->123

Explanation

(?<name>\\\\w+)=(\\\\{(?<value>[^}]*)\\\\})?(;|$) can be split into parts where

• (?<name>\\\\w+)= represents XXXX= and place XXXX in group named name (of property)
• (\\\\{(?<value>[^}]*)\\\\})? is optional {XXXX} part where X can't be } . Also it will place XXXX part in group named value .
• (;|$) represents ; OR end of data (represented by $ anchor) since formula is name=value; or in case of pair placed at the end of data name=value .