Java: splitting a comma-separated string but ignoring commas in quotes

JavaRegexString

Java Problem Overview


I have a string vaguely like this:

foo,bar,c;qual="baz,blurb",d;junk="quux,syzygy"

that I want to split by commas -- but I need to ignore commas in quotes. How can I do this? Seems like a regexp approach fails; I suppose I can manually scan and enter a different mode when I see a quote, but it would be nice to use preexisting libraries. (edit: I guess I meant libraries that are already part of the JDK or already part of a commonly-used libraries like Apache Commons.)

the above string should split into:

foo
bar
c;qual="baz,blurb"
d;junk="quux,syzygy"

note: this is NOT a CSV file, it's a single string contained in a file with a larger overall structure

Java Solutions


Solution 1 - Java

Try:

public class Main { 
    public static void main(String[] args) {
        String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
        String[] tokens = line.split(",(?=(?:[^\"]*\"[^\"]*\")*[^\"]*$)", -1);
        for(String t : tokens) {
            System.out.println("> "+t);
        }
    }
}

Output:

> foo
> bar
> c;qual="baz,blurb"
> d;junk="quux,syzygy"

In other words: split on the comma only if that comma has zero, or an even number of quotes ahead of it.

Or, a bit friendlier for the eyes:

public class Main { 
    public static void main(String[] args) {
        String line = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
        
        String otherThanQuote = " [^\"] ";
        String quotedString = String.format(" \" %s* \" ", otherThanQuote);
        String regex = String.format("(?x) "+ // enable comments, ignore white spaces
                ",                         "+ // match a comma
                "(?=                       "+ // start positive look ahead
                "  (?:                     "+ //   start non-capturing group 1
                "    %s*                   "+ //     match 'otherThanQuote' zero or more times
                "    %s                    "+ //     match 'quotedString'
                "  )*                      "+ //   end group 1 and repeat it zero or more times
                "  %s*                     "+ //   match 'otherThanQuote'
                "  $                       "+ // match the end of the string
                ")                         ", // stop positive look ahead
                otherThanQuote, quotedString, otherThanQuote);

        String[] tokens = line.split(regex, -1);
        for(String t : tokens) {
            System.out.println("> "+t);
        }
    }
}

which produces the same as the first example.

EDIT

As mentioned by @MikeFHay in the comments:

> I prefer using Guava's Splitter, as it has saner defaults (see discussion above about empty matches being trimmed by String#split(), so I did: > > Splitter.on(Pattern.compile(",(?=(?:[^"]"[^"]")[^"]$)"))

Solution 2 - Java

While I do like regular expressions in general, for this kind of state-dependent tokenization I believe a simple parser (which in this case is much simpler than that word might make it sound) is probably a cleaner solution, in particular with regards to maintainability, e.g.:

String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
List<String> result = new ArrayList<String>();
int start = 0;
boolean inQuotes = false;
for (int current = 0; current < input.length(); current++) {
    if (input.charAt(current) == '\"') inQuotes = !inQuotes; // toggle state
    else if (input.charAt(current) == ',' && !inQuotes) {
        result.add(input.substring(start, current));
        start = current + 1;
    }
}
result.add(input.substring(start));

If you don't care about preserving the commas inside the quotes you could simplify this approach (no handling of start index, no last character special case) by replacing your commas in quotes by something else and then split at commas:

String input = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
StringBuilder builder = new StringBuilder(input);
boolean inQuotes = false;
for (int currentIndex = 0; currentIndex < builder.length(); currentIndex++) {
    char currentChar = builder.charAt(currentIndex);
    if (currentChar == '\"') inQuotes = !inQuotes; // toggle state
    if (currentChar == ',' && inQuotes) {
        builder.setCharAt(currentIndex, ';'); // or '♡', and replace later
    }
}
List<String> result = Arrays.asList(builder.toString().split(","));

Solution 3 - Java

Solution 4 - Java

I would not advise a regex answer from Bart, I find parsing solution better in this particular case (as Fabian proposed). I've tried regex solution and own parsing implementation I have found that:

  1. Parsing is much faster than splitting with regex with backreferences - ~20 times faster for short strings, ~40 times faster for long strings.
  2. Regex fails to find empty string after last comma. That was not in original question though, it was mine requirement.

My solution and test below.

String tested = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\",";
long start = System.nanoTime();
String[] tokens = tested.split(",(?=([^\"]*\"[^\"]*\")*[^\"]*$)");
long timeWithSplitting = System.nanoTime() - start;

start = System.nanoTime(); 
List<String> tokensList = new ArrayList<String>();
boolean inQuotes = false;
StringBuilder b = new StringBuilder();
for (char c : tested.toCharArray()) {
	switch (c) {
	case ',':
		if (inQuotes) {
			b.append(c);
		} else {
			tokensList.add(b.toString());
			b = new StringBuilder();
		}
		break;
	case '\"':
		inQuotes = !inQuotes;
	default:
		b.append(c);
	break;
	}
}
tokensList.add(b.toString());
long timeWithParsing = System.nanoTime() - start;
			
System.out.println(Arrays.toString(tokens));
System.out.println(tokensList.toString());
System.out.printf("Time with splitting:\t%10d\n",timeWithSplitting);
System.out.printf("Time with parsing:\t%10d\n",timeWithParsing);

Of course you are free to change switch to else-ifs in this snippet if you feel uncomfortable with its ugliness. Note then lack of break after switch with separator. StringBuilder was chosen instead to StringBuffer by design to increase speed, where thread safety is irrelevant.

Solution 5 - Java

You're in that annoying boundary area where regexps almost won't do (as has been pointed out by Bart, escaping the quotes would make life hard) , and yet a full-blown parser seems like overkill.

If you are likely to need greater complexity any time soon I would go looking for a parser library. For example this one

Solution 6 - Java

I was impatient and chose not to wait for answers... for reference it doesn't look that hard to do something like this (which works for my application, I don't need to worry about escaped quotes, as the stuff in quotes is limited to a few constrained forms):

final static private Pattern splitSearchPattern = Pattern.compile("[\",]"); 
private List<String> splitByCommasNotInQuotes(String s) {
	if (s == null)
		return Collections.emptyList();
	
	List<String> list = new ArrayList<String>();
	Matcher m = splitSearchPattern.matcher(s);
	int pos = 0;
	boolean quoteMode = false;
	while (m.find())
	{
		String sep = m.group();
		if ("\"".equals(sep))
		{
			quoteMode = !quoteMode;
		}
		else if (!quoteMode && ",".equals(sep))
		{
			int toPos = m.start(); 
			list.add(s.substring(pos, toPos));
			pos = m.end();
		}
	}
	if (pos < s.length())
		list.add(s.substring(pos));
	return list;
}

(exercise for the reader: extend to handling escaped quotes by looking for backslashes also.)

Solution 7 - Java

Try a lookaround like (?!\"),(?!\"). This should match , that are not surrounded by ".

Solution 8 - Java

The simplest approach is not to match delimiters, i.e. commas, with a complex additional logic to match what is actually intended (the data which might be quoted strings), just to exclude false delimiters, but rather match the intended data in the first place.

The pattern consists of two alternatives, a quoted string ("[^"]*" or ".*?") or everything up to the next comma ([^,]+). To support empty cells, we have to allow the unquoted item to be empty and to consume the next comma, if any, and use the \\G anchor:

Pattern p = Pattern.compile("\\G\"(.*?)\",?|([^,]*),?");

The pattern also contains two capturing groups to get either, the quoted string’s content or the plain content.

Then, with Java 9, we can get an array as

String[] a = p.matcher(input).results()
    .map(m -> m.group(m.start(1)<0? 2: 1))
    .toArray(String[]::new);

whereas older Java versions need a loop like

for(Matcher m = p.matcher(input); m.find(); ) {
    String token = m.group(m.start(1)<0? 2: 1);
    System.out.println("found: "+token);
}

Adding the items to a List or an array is left as an excise to the reader.

For Java 8, you can use the results() implementation of this answer, to do it like the Java 9 solution.

For mixed content with embedded strings, like in the question, you can simply use

Pattern p = Pattern.compile("\\G((\"(.*?)\"|[^,])*),?");

But then, the strings are kept in their quoted form.

Solution 9 - Java

A regular expression is not capable of handling escaped characters. For my application, I needed the ability to escape quotes and spaces (my separator is spaces, but the code is the same).

Here is my solution in Kotlin (the language from this particular application), based on the one from Fabian Steeg:

fun parseString(input: String): List<String> {
    val result = mutableListOf<String>()
    var inQuotes = false
    var inEscape = false
    val current = StringBuilder()
    for (i in input.indices) {
        // If this character is escaped, add it without looking
        if (inEscape) {
            inEscape = false
            current.append(input[i])
            continue
        }
        when (val c = input[i]) {
            '\\' -> inEscape = true // escape the next character, \ isn't added to result
            ',' -> if (inQuotes) {
                current.append(c)
            } else {
                result += current.toString()
                current.clear()
            }
            '"' -> inQuotes = !inQuotes
            else -> current.append(c)
        }
    }
    if (current.isNotEmpty()) {
        result += current.toString()
    }
    return result
}

I think this is not a place to use regular expressions. Contrary to other opinions, I don't think a parser is overkill. It's about 20 lines and fairly easy to test.

Solution 10 - Java

Rather than use lookahead and other crazy regex, just pull out the quotes first. That is, for every quote grouping, replace that grouping with __IDENTIFIER_1 or some other indicator, and map that grouping to a map of string,string.

After you split on comma, replace all mapped identifiers with the original string values.

Solution 11 - Java

what about a one-liner using String.split()?

String s = "foo,bar,c;qual=\"baz,blurb\",d;junk=\"quux,syzygy\"";
String[] split = s.split( "(?<!\".{0,255}[^\"]),|,(?![^\"].*\")" );

Solution 12 - Java

I would do something like this:

boolean foundQuote = false;

if(charAtIndex(currentStringIndex) == '"')
{
   foundQuote = true;
}

if(foundQuote == true)
{
   //do nothing
}

else 

{
  string[] split = currentString.split(',');  
}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJason SView Question on Stackoverflow
Solution 1 - JavaBart KiersView Answer on Stackoverflow
Solution 2 - JavaFabian SteegView Answer on Stackoverflow
Solution 3 - JavaJonathan FeinbergView Answer on Stackoverflow
Solution 4 - JavaMarcin KosinskiView Answer on Stackoverflow
Solution 5 - JavadjnaView Answer on Stackoverflow
Solution 6 - JavaJason SView Answer on Stackoverflow
Solution 7 - JavaMatthew SowdersView Answer on Stackoverflow
Solution 8 - JavaHolgerView Answer on Stackoverflow
Solution 9 - JavaSeanView Answer on Stackoverflow
Solution 10 - JavaStefan KendallView Answer on Stackoverflow
Solution 11 - JavaKaplanView Answer on Stackoverflow
Solution 12 - JavaWoot4MooView Answer on Stackoverflow