正则表达式以分割key = value

I have a string like this:

我有一个像这样的字符串:

KEY1=Value1, KE_Y2=[V@LUE2A, Value2B], Key3=, KEY4=V-AL.UE4, KEY5={Value5}

I need to split it to get a Map with key-value pairs. Values in [] should be passed as a single value (KE_Y2 is a key and [V@LUE2A, Value2B] is a value).

我需要将其拆分以获得具有键值对的Map。 []中的值应作为单个值传递(KE_Y2是键,[V @ LUE2A,Value2B]是一个值)。

What regular expression should I use to split it correctly?

我应该用什么正则表达式正确分割它?

4 个解决方案

#1

There's a magic regex for the first split:

第一次拆分有一个神奇的正则表达式:

String[] pairs = input.split(", *(?![^\\[\\]]*\\])");

Then split each of the key/values with simply "=":

然后用简单的“=”分割每个键/值:

for (String pair : pairs) {
    String[] parts = pair.split("=");
    String key = parts[0];
    String value = parts[1];
}

---

Putting it all together:

把它们放在一起:

Map<String, String> map = new HashMap<String, String>();
for (String pair : input.split(", *(?![^\\[\\]]*\\])")) {
    String[] parts = pair.split("=");
    map.put(parts[0], parts[1]);
}

Voila!

---

Explanation of magic regex:

The regex says "a comma followed by any number of spaces (so key names don't have leading blanks), but only if the next bracket encountered is not a close bracket"

正则表达式说“一个逗号后跟任意数量的空格(所以键名没有前导空格),但只有遇到下一个括号不是一个紧密的括号”

#2

How about this:

这个怎么样:

Map<String, String> map = new HashMap<String, String>();
Pattern regex = Pattern.compile(
    "(\\w+)        # Match an alphanumeric identifier, capture in group 1\n" +
    "=             # Match =                                             \n" +
    "(             # Match and capture in group 2:                       \n" +
    " (?:          # Either...                                           \n" +
    "  \\[         #  a [                                                \n" +
    "  [^\\[\\]]*  #  followed by any number of characters except [ or ] \n" +
    "  \\]         #  followed by a ]                                    \n" +
    " |            # or...                                               \n" +
    "  [^\\[\\],]* #  any number of characters except commas, [ or ]     \n" +
    " )            # End of alternation                                  \n" +
    ")             # End of capturing group", 
    Pattern.COMMENTS);
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
    map.put(regexMatcher.group(1), regexMatcher.group(2));
} 

#3

Start with @achintya-jha's answer. When you split a String, it will give you an array (or something that acts like it) so you can iterate throught the pair of key/value and then you do the second split which is supposed to give you another array of size 2; you then use the first element as the key and the second as the value.

从@ achintya-jha的回答开始。当你拆分一个String时,它会给你一个数组(或类似它的东西),所以你可以遍历这对键/值,然后你做第二次拆分,它应该给你另一个大小为2的数组;然后使用第一个元素作为键,第二个元素作为值。

EDIT:

I dind't found useful link for what I meant (see the comments on the question) in JAVA, (there is plenty of them for C/C++ though) so I wrote it:

我在JAVA中没有找到有用的链接(请参阅问题的评论)(虽然它有很多用于C / C ++)所以我写了它:

Map<String, String> map = new HashMap<String, String>();
String str = "KEY1=Value1, KE_Y2=[V@LUE2A, Value2B]], Key3=, KEY4=V-AL.UE4, KEY5={Value5}";     


final String openBrackets =  "({[<";
final String closeBrackets = ")}]>";

String buffer = "";
int state = 0;
int i = 0;      
Stack<Integer> stack = new Stack<Integer>(); //For the brackets

String key = "";


while(  i < str.length() ) {

    char c = str.charAt(i);


    //Skip any whitespace
    if( " \t\n\r".indexOf(c) > -1 ) {
        ++i;
        continue;
    }


    switch(state) {

    //Reading Key
    case 0:
        if( c != '=' ) {
            buffer += c;
        } else {
            //Go read a value.
            key = buffer;
            state = 1;
            buffer = "";
        }
        ++i;
        break;

    //Reading value
    case 1:

        //Opening bracket
        int pos = openBrackets.indexOf(c);
        if( pos != -1 ) {
            stack.push(pos);
            ++i;
            break;
        }

        //Closing bracket
        pos = closeBrackets.indexOf(c);
        if( pos != -1 ) {

            if( stack.size() == 0 ) {
                throw new RuntimeException("Syntax error: Unmatched closing bracket '" + c + "'" );
            }

            int pos2 = stack.pop();
            if( pos != pos2 ) {
                throw new RuntimeException("Syntax error: Unmatched closing bracket, expected a '"
                        + closeBrackets.charAt(pos2) + "' got '" + c );             
            }
            ++i;
            break;
        }

        //Handling separators 
        if( c == ',' ) {
            if( stack.size() == 0 ) {
                //Put the pair in the map.
                map.put(key, buffer);

                //Go read a new Key.
                state = 0;
                buffer = "";
                ++i;
                break;
            }                       
        }

        //else
            buffer += c;
            ++i;


        } //switch
} //while

#4

1. split the given string with String.split(",");

用String.split(“,”)拆分给定的字符串;

2. Now split each element of the array with String.split("=");

现在用String.split(“=”)分割数组的每个元素;