如何在数学运算符上将字符串拆分为定界符，但在引号内转义运算符（在Java中）？

Question

For example,

AM2 + 'G - D08 - 28 - 14 .xlsx]General Inputs' should be split as

AM2 and 'G - D08 - 28 - 14 .xlsx]General Inputs' .

Answer 1

For your type of given input example, I would probably match vs splitting.

String s  = "AM2 + 'G - D08 - 28 - 14 .xlsx]General Inputs'";
Pattern p = Pattern.compile("'[^']*'|[^ '+*/-]+");
Matcher m = p.matcher(s);
while (m.find()) {
  System.out.println(m.group());
}

Output

AM2
'G - D08 - 28 - 14 .xlsx]General Inputs'

Answer 2

I don't think you can do this with split --if you can, it would be very tricky and messy. split is good at looking for delimiters, but not so good when a pattern has to be applied to the stuff in between delimiters, which it would in this case.

Instead, I'd use a regex to look for the text that occurs between delimiters, and use the Matcher methods. The way I look at problems like this is to think of the non-operator text as a sequence of entities, where each entity is

• a quoted string;
• a single character that is not a quote, and is not an operator (or the start of an operator, if some operators are two or more characters).

If all your operators are one character, a regex that finds an "operand" might look like

('.*?'|[^'+\-*/])*

which says to look for any number of characters between quote marks, or for any single character that is not + , - , * , or / (note that the - has to be escaped inside the character class). The last * means to look for zero or more of this pattern.

To look for a case where an operator might be multiple characters, such as << or >> , you can use negative lookahead:

('.*?'|(?!\+|-|\*|/|<<|>>)[^'])*

which means to find either a quoted string, or a non-quote character at a point where we're not looking at + , - , * , / , << , or >> , and find this zero or more times.

The plan would be to use lookingAt() with a matcher to find the operand, then use lookingAt() to find the operator, and go back and forth. (Or if you don't need to keep the operators at all, use find() as in @hwnd's answer.)

I may have some details wrong, but this should give you an idea of the best approach.