如何拆分字符串（基于各种分隔符）但不保留空格？

Question

I have Java strings which are boolean expressions with parentheses, & , | , and ! as operators, and I want to split them into tokens. For example:

((!A1)&(B2|C3)) should become "(","(",",","A1",")","&","(","B2","|","C3",")",")"

Following this answer I found that I can use Java's String.split() with a regex that includes lookahead and lookbehind clauses:

List<String> tokens = "((!A1)&(B2|C3))".split("((?<=[!&()|])|(?=[!&()|]))")

My only problem is that whitespace will be included in the list of tokens. For example if I were to write the expression as ( ( !A1 ) & ( B2 | C3 ) ) then my split() would produce at least four strings like " " and there'd be padding around my variables (eg " A1 " ).

How can I modify this split expression and regex to tokenize the string but not keep any of the witespace?

Answer 1

Instead of split you can use this this regex to match what you want:

[!&()]|[^!&()\h]+

RegEx Demo

RegEx Details:

• [!&()] : Match ! or & or ( or )
• | : OR
• [^!&()\h]+ : Match any characters that is NOT ! , & , ( , ) and a whitespace

Code:

final String regex = "[!&()]|[^!&()\\h]+";
final String string = "((!A1)&( B2 | C3 ))";

final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);

List<String> result = new ArrayList<>();
while (matcher.find()) {
    result.add(matcher.group(0));
}

System.out.println(result);