我有一个这样的字符串:
KEY1=Value1, KE_Y2=[V@LUE2A, Value2B], Key3=, KEY4=V-AL.UE4, KEY5={Value5}
我需要拆分它以获取带有键值对的 Map。中的值[]
应作为单个值传递(KE_Y2
是键和[V@LUE2A, Value2B]
值)。
我应该使用什么正则表达式来正确拆分它?
There's a magic regex for the first split:
String[] pairs = input.split(", *(?![^\\[\\]]*\\])");
Then split each of the key/values with simply "=":
for (String pair : pairs) {
String[] parts = pair.split("=");
String key = parts[0];
String value = parts[1];
}
Putting it all together:
Map<String, String> map = new HashMap<String, String>();
for (String pair : input.split(", *(?![^\\[\\]]*\\])")) {
String[] parts = pair.split("=");
map.put(parts[0], parts[1]);
}
Voila!
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"
How about this:
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));
}
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.
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:
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