字符串的正则表达式

Question

I need to use regular expression to get some values from the String. Thing is, that it is quite complicated for me.

For example i have a string like this:

oneWord [first, second, third]

My output should be:

first
second
third

So i need words which are between [ and ]. Plus there can be a different number of words between [].

Tried using some regex creator, but that wasn't very accurate:

String re1=".*?";   // Non-greedy match on filler
String re2="(?:[a-z][a-z]+)";   // Uninteresting: word
String re3=".*?";   // Non-greedy match on filler
String re4="((?:[a-z][a-z]+))"; // Word 1
String re5=".*?";   // Non-greedy match on filler
String re6="((?:[a-z][a-z]+))"; // Word 2
String re7=".*?";   // Non-greedy match on filler
String re8="((?:[a-z][a-z]+))"; // Word 3

Answer 1

I would do it like this, in just one line:

String[] words = str.replaceAll(".*\\[|\\].*", "").split(", ");

The first replaceAll() call strips off the leading and trailing wrapper, and the split() breaks up what's left into separate words.

Answer 2

You could try the below regex and get the words you want from group index 1.

(?:\[|(?<!^)\G),? *(\w+)(?=[^\[\]]*\])

DEMO

Java regex would be,

(?:\\[|(?<!^)\\G),? *(\\w+)(?=[^\\[\\]]*\\])

Example:

String s = "oneWord [first, second, third] foo bar [foobar]";
Pattern regex = Pattern.compile("(?:\\[|(?<!^)\\G),? *(\\w+)(?=[^\\[\\]]*\\])");
 Matcher matcher = regex.matcher(s);
 while(matcher.find()){
        System.out.println(matcher.group(1));
}

Output:

first
second
third
foobar

Answer 3

You should use this string.

String[] words = str.replaceAll(". \\[|\\]. ", "").split(", ");

Hope it helps.

Answer 4

You can do it easily with method split.

String string = [first, second, third];
String[] parts = string.split(",");
String part1 = parts[0]; // first
String part2 = parts[1]; // second
String part3 = parts[2]; // third

if it dont work for you, please tell me that I will debug your regular expression.