CODEWITHSUNDEEP
javaregex
Franklin
Franklin

Reputation: 612

Regex matching unescaped commas in Java

Problem description

I am trying to split a into separate strings, with the split() method that the String class provides. The documentation tells me that it will split around matches of the argument, which is a regular expression. The delimiter that I use is a comma, but commas can also be escaped. Escaping character that I use is a forward slash / (just to make things easier by not using a backslash, because that requires additional escaping in string literals in both Java and the regular expressions).

For instance, the input might be this:

a,b/,b//,c///,//,d///,

And the output should be:

a
b,b/
c/,/
d/,

So, the string should be split at each comma, unless that comma is preceded by an odd number of slashes (1, 3, 5, 7, ..., ∞) because that would mean that the comma is escaped.

Possible solutions

My initial guess would be to split it like this:

String[] strings = longString.split("(?<![^/](//)*/),");

but that is not allowed because Java doesn't allow infinite look-behind groups. I could limit the recurrence to, say, 2000 by replacing the * with {0,2000}:

String[] strings = longString.split("(?<![^/](//){0,2000}/),");

but that still puts constraints on the input. So I decided to take the recurrence out of the look-behind group, and came up with this:

String[] strings = longString.split("(?<!/)(?:(//)*),");

However, its output is the following list of strings:

a
b,b (the final slash is lacking in the output)
c/, (the final slash is lacking in the output)
d/,

Why are those slashes omitted in the 2nd and 3rd string, and how can I solve it (in Java)?

Upvotes: 3

Views: 1562

Answers (4)

Jerry
Jerry

Reputation: 71538

If you don't mind another method with regex, I suggest using .matcher:

Pattern pattern = Pattern.compile("(?:[^,/]+|/.)+");
String test = "a,b/,b//,c///,//,d///,";
Matcher matcher = pattern.matcher(test);
while (matcher.find()) {
    System.out.println(matcher.group().replaceAll("/(.)", "$1"));
}

Output:

a
b,b/
c/,/
d/,

ideone demo

This method will match everything except the delimiting commas (kind of the reverse). The advantage is that it doesn't rely on lookarounds.

Upvotes: 1

Bohemian
Bohemian

Reputation: 424983

You can achieve the split using a positive look behind for an even number of slashes preceding the comma:

String[] strings = longString.split("(?<=[^/](//){0,999999999}),");

But to display the output you want, you need a further step of removing the remaining escapes:

String longString = "a,b/,b//,c///,//,d///,";
String[] strings = longString.split("(?<=[^/](//){0,999999999}),");
for (String s : strings)
    System.out.println(s.replaceAll("/(.)", "$1"));

Output:

a
b,b/
c/,/
d/,

Upvotes: 3

anubhava
anubhava

Reputation: 784998

You are pretty close. To overcome lookbehind error you can use this workaround:

String[] strings = longString.split("(?<![^/](//){0,99}/),")

Upvotes: 3

gexicide
gexicide

Reputation: 40058

I love regexes, but wouldn't it be easy to write the code manually here, i.e.

boolean escaped = false;
for(int i = 0, len = s.length() ; i < len ; i++){
    switch(s.charAt(i)){
    case "/": escaped = !escaped; break;            
    case ",": 
      if(!escaped){
         //found a segment, do something with it
      }
      //Fallthrough!
    default:
      escaped = false;
    }
}
// handle last segment

Upvotes: 0

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