Develop Reference

Parsing a string that contains double quotes

i have a relatively simple java question. I have a string that looks like this:
"Anderson,T",CWS,SS
I need to parse it in a way that I have
Anderson,T
CWS
SS
all as separate strings.
Thanks!

Here's a solution that will capture quoted strings, remove spaces, and match empty items:
public static void main(String[] args) {
String quoted = "\"(.*?(?<!\\\\)(?:\\\\\\\\)*)\"";
Pattern regex = Pattern.compile(
"(?:^|(?<=,))\\s*(" + quoted + "|[^,]*?)\\s*(?:$|,)");
String line = "\"Anderson,T\",CWS,\"single quote\\\"\", SS ,,hello,,";
Matcher m = regex.matcher(line);
int count = 0;
while (m.find()) {
String s = m.group(2) == null ? m.group(1) : m.group(2);
System.out.println(s);
count++;
}
System.out.printf("(%d matches found)%n", count);
}
I split out the quoted part of the pattern to make it a bit easier to follow. Capturing group 1 is the quoted string, 2 is every other match.
To break down the overall pattern:
Look for start of line or previous comma (?:^|(?<=,)) (don't capture)
Ignore 0+ spaces \\s*
Look for quoted string or string without comma (" + quoted + "|[^,]*?)
(The non-comma match is non-greedy so it doesn't grab any following spaces)
Ignore 0+ spaces again \\s*
Look for end of line, or comma (?:$|,) (don't capture)
To break down the quote pattern:
Look for opening quote \"
Start group capture (
Get the minimum match of any character .*?
Match 0+ even number of backslashes (?<!\\\\)(?:\\\\\\\\)* (to avoid matching escaped quotes with or without preceding escaped backslashes)
Close capturing group )
Match closing quote \"

Assuming your string looks like this
String input = "\"Anderson,T\",CWS,SS";
You can use this solution found for a similar scenario.
String input = "\"Anderson,T\",CWS,SS";
List<String> result = new ArrayList<String>();
int start = 0; //start index. Used to determine where the word starts
boolean inQuotes = false;
for (int current = 0; current < input.length(); current++) { //iterate through characters
if (input.charAt(current) == '\"') //if found a quote
inQuotes = !inQuotes; // toggle state
if(current == (input.length() - 1))//if it is the last character
result.add(input.substring(start)); //add last word
else if (input.charAt(current) == ',' && !inQuotes) { //if found a comma not inside quotes
result.add(input.substring(start, current)); //add everything between the start index and the current character. (add a word)
start = current + 1; //update start index
}
}
System.out.println(result);
I have modified it a bit to improve readability. This code stores the strings you want in the list result.

Java regex: Replace all characters with `+` except instances of a given string

I have the following problem which states
Replace all characters in a string with + symbol except instances of the given string in the method
so for example if the string given was abc123efg and they want me to replace every character except every instance of 123 then it would become +++123+++.
I figured a regular expression is probably the best for this and I came up with this.
str.replaceAll("[^str]","+")
where str is a variable, but its not letting me use the method without putting it in quotations. If I just want to replace the variable string str how can I do that? I ran it with the string manually typed and it worked on the method, but can I just input a variable?
as of right now I believe its looking for the string "str" and not the variable string.
Here is the output its right for so many cases except for two :(
List of open test cases:
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
plusOut("abXYabcXYZ", "ab") → "ab++ab++++"
plusOut("abXYabcXYZ", "abc") → "++++abc+++"
plusOut("abXYabcXYZ", "XY") → "++XY+++XY+"
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
plusOut("--++ab", "++") → "++++++"
plusOut("aaxxxxbb", "xx") → "++xxxx++"
plusOut("123123", "3") → "++3++3"

Looks like this is the plusOut problem on CodingBat.
I had 3 solutions to this problem, and wrote a new streaming solution just for fun.
Solution 1: Loop and check
Create a StringBuilder out of the input string, and check for the word at every position. Replace the character if doesn't match, and skip the length of the word if found.
public String plusOut(String str, String word) {
StringBuilder out = new StringBuilder(str);
for (int i = 0; i < out.length(); ) {
if (!str.startsWith(word, i))
out.setCharAt(i++, '+');
else
i += word.length();
}
return out.toString();
}
This is probably the expected answer for a beginner programmer, though there is an assumption that the string doesn't contain any astral plane character, which would be represented by 2 char instead of 1.
Solution 2: Replace the word with a marker, replace the rest, then restore the word
public String plusOut(String str, String word) {
return str.replaceAll(java.util.regex.Pattern.quote(word), "#").replaceAll("[^#]", "+").replaceAll("#", word);
}
Not a proper solution since it assumes that a certain character or sequence of character doesn't appear in the string.
Note the use of Pattern.quote to prevent the word being interpreted as regex syntax by replaceAll method.
Solution 3: Regex with \G
public String plusOut(String str, String word) {
word = java.util.regex.Pattern.quote(word);
return str.replaceAll("\\G((?:" + word + ")*+).", "$1+");
}
Construct regex \G((?:word)*+)., which does more or less what solution 1 is doing:
\G makes sure the match starts from where the previous match leaves off
((?:word)*+) picks out 0 or more instance of word - if any, so that we can keep them in the replacement with $1. The key here is the possessive quantifier *+, which forces the regex to keep any instance of the word it finds. Otherwise, the regex will not work correctly when the word appear at the end of the string, as the regex backtracks to match .
. will not be part of any word, since the previous part already picks out all consecutive appearances of word and disallow backtrack. We will replace this with +
Solution 4: Streaming
public String plusOut(String str, String word) {
return String.join(word,
Arrays.stream(str.split(java.util.regex.Pattern.quote(word), -1))
.map((String s) -> s.replaceAll("(?s:.)", "+"))
.collect(Collectors.toList()));
}
The idea is to split the string by word, do the replacement on the rest, and join them back with word using String.join method.
Same as above, we need Pattern.quote to avoid split interpreting the word as regex. Since split by default removes empty string at the end of the array, we need to use -1 in the second parameter to make split leave those empty strings alone.
Then we create a stream out of the array and replace the rest as strings of +. In Java 11, we can use s -> String.repeat(s.length()) instead.
The rest is just converting the Stream to an Iterable (List in this case) and joining them for the result

This is a bit trickier than you might initially think because you don't just need to match characters, but the absence of specific phrase - a negated character set is not enough. If the string is 123, you would need:
(?<=^|123)(?!123).*?(?=123|$)
https://regex101.com/r/EZWMqM/1/
That is - lookbehind for the start of the string or "123", make sure the current position is not followed by 123, then lazy-repeat any character until lookahead matches "123" or the end of the string. This will match all characters which are not in a "123" substring. Then, you need to replace each character with a +, after which you can use appendReplacement and a StringBuffer to create the result string:
String inputPhrase = "123";
String inputStr = "abc123efg123123hij";
StringBuffer resultString = new StringBuffer();
Pattern regex = Pattern.compile("(?<=^|" + inputPhrase + ")(?!" + inputPhrase + ").*?(?=" + inputPhrase + "|$)");
Matcher m = regex.matcher(inputStr);
while (m.find()) {
String replacement = m.group(0).replaceAll(".", "+");
m.appendReplacement(resultString, replacement);
}
m.appendTail(resultString);
System.out.println(resultString.toString());
Output:
+++123+++123123+++
Note that if the inputPhrase can contain character with a special meaning in a regular expression, you'll have to escape them first before concatenating into the pattern.

You can do it in one line:
input = input.replaceAll("((?:" + str + ")+)?(?!" + str + ").((?:" + str + ")+)?", "$1+$2");
This optionally captures "123" either side of each character and puts them back (a blank if there's no "123"):

So instead of coming up with a regular expression that matches the absence of a string. We might as well just match the selected phrase and append + the number of skipped characters.
StringBuilder sb = new StringBuilder();
Matcher m = Pattern.compile(Pattern.quote(str)).matcher(input);
while (m.find()) {
for (int i = 0; i < m.start(); i++) sb.append('+');
sb.append(str);
}
int remaining = input.length() - sb.length();
for (int i = 0; i < remaining; i++) {
sb.append('+');
}

Absolutely just for the fun of it, a solution using CharBuffer (unexpectedly it took a lot more that I initially hoped for):
private static String plusOutCharBuffer(String input, String match) {
int size = match.length();
CharBuffer cb = CharBuffer.wrap(input.toCharArray());
CharBuffer word = CharBuffer.wrap(match);
int x = 0;
for (; cb.remaining() > 0;) {
if (!cb.subSequence(0, size < cb.remaining() ? size : cb.remaining()).equals(word)) {
cb.put(x, '+');
cb.clear().position(++x);
} else {
cb.clear().position(x = x + size);
}
}
return cb.clear().toString();
}

To make this work you need a beast of a pattern. Let's say you you are operating on the following test case as an example:
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
What you need to do is build a series of clauses in your pattern to match a single character at a time:
Any character that is NOT "X", "Y" or "Z" -- [^XYZ]
Any "X" not followed by "YZ" -- X(?!YZ)
Any "Y" not preceded by "X" -- (?<!X)Y
Any "Y" not followed by "Z" -- Y(?!Z)
Any "Z" not preceded by "XY" -- (?<!XY)Z
An example of this replacement can be found here: https://regex101.com/r/jK5wU3/4
Here is an example of how this might work (most certainly not optimized, but it works):
import java.util.regex.Pattern;
public class Test {
public static void plusOut(String text, String exclude) {
StringBuilder pattern = new StringBuilder("");
for (int i=0; i<exclude.length(); i++) {
Character target = exclude.charAt(i);
String prefix = (i > 0) ? exclude.substring(0, i) : "";
String postfix = (i < exclude.length() - 1) ? exclude.substring(i+1) : "";
// add the look-behind (?<!X)Y
if (!prefix.isEmpty()) {
pattern.append("(?<!").append(Pattern.quote(prefix)).append(")")
.append(Pattern.quote(target.toString())).append("|");
}
// add the look-ahead X(?!YZ)
if (!postfix.isEmpty()) {
pattern.append(Pattern.quote(target.toString()))
.append("(?!").append(Pattern.quote(postfix)).append(")|");
}
}
// add in the other character exclusion
pattern.append("[^" + Pattern.quote(exclude) + "]");
System.out.println(text.replaceAll(pattern.toString(), "+"));
}
public static void main(String [] args) {
plusOut("12xy34", "xy");
plusOut("12xy34", "1");
plusOut("12xy34xyabcxy", "xy");
plusOut("abXYabcXYZ", "ab");
plusOut("abXYabcXYZ", "abc");
plusOut("abXYabcXYZ", "XY");
plusOut("abXYxyzXYZ", "XYZ");
plusOut("--++ab", "++");
plusOut("aaxxxxbb", "xx");
plusOut("123123", "3");
}
}
UPDATE: Even this doesn't quite work because it can't deal with exclusions that are just repeated characters, like "xx". Regular expressions are most definitely not the right tool for this, but I thought it might be possible. After poking around, I'm not so sure a pattern even exists that might make this work.

The problem in your solution that you put a set of instance string str.replaceAll("[^str]","+") which it will exclude any character from the variable str and that will not solve your problem
EX: when you try str.replaceAll("[^XYZ]","+") it will exclude any combination of character X , character Y and character Z from your replacing method so you will get "++XY+++XYZ".
Actually you should exclude a sequence of characters instead in str.replaceAll.
You can do it by using capture group of characters like (XYZ) then use a negative lookahead to match a string which does not contain characters sequence : ^((?!XYZ).)*$
Check this solution for more info about this problem but you should know that it may be complicated to find regular expression to do that directly.
I have found two simple solutions for this problem :
Solution 1:
You can implement a method to replace all characters with '+' except the instance of given string:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
for(int i = 0; i < str.length(); i++){
// exclude any instance string of exWord from replacing process in str
if(str.substring(i, str.length()).indexOf(exWord) + i == i){
i = i + exWord.length()-1;
}
else{
str = str.substring(0,i) + "+" + str.substring(i+1);//replace each character with '+' symbol
}
}
Note : str.substring(i, str.length()).indexOf(exWord) + i this if statement will exclude any instance string of exWord from replacing process in str.
Output:
+++++++XYZ
Solution 2:
You can try this Approach using ReplaceAll method and it doesn't need any complex regular expression:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
str = str.replaceAll(exWord,"*"); // replace instance string with * symbol
str = str.replaceAll("[^*]","+"); // replace all characters with + symbol except *
str = str.replaceAll("\\*",exWord); // replace * symbol with instance string
Note : This solution will work only if your input string str doesn't contain any * symbol.
Also you should escape any character with a special meaning in a regular expression in phrase instance string exWord like : exWord = "++".

Optimizing several RegEx in Java Code

The below mentioned RegEx perform very poorly on a very large string or more than 2000 Lines. Basically the Java String is composed of PL/SQL script.
1- Replace each occurrence of delimiting character, for example ||, != or > sign with a space before and after the characters. This takes infinite time and never ends, so no time can be recorded.
// Delimiting characters for SQLPlus
private static final String[] delimiters = { "\\|\\|", "=>", ":=", "!=", "<>", "<", ">", "\\(", "\\)", "!", ",", "\\+", "-", "=", "\\*", "\\|" };
for (int i = 0; i < delimiters.length; i++) {
script = script.replaceAll(delimiters[i], " " + delimiters[i] + " ");
}
2- The following pattern looks for all occurances of forward slash / except the ones that are preceded by a *. That mean don't look for forward slash in a block comment syntax. This takes about 103 Seconds for a 2000 lines of String.
Pattern p = Pattern.compile("([^\\*])([\\/])([^\\*])");
Matcher m = p.matcher(script);
while (m.find()) {
script = script.replaceAll(m.group(2), " " + m.group(2) + " ");
}
3- Remove any white spaces from within date or date format
Pattern p = Pattern.compile("(?i)(\\w{1,2}) +/ +(\\w{1,2}) +/ +(\\w{2,4})");
// Create a matcher with an input string
Matcher m = p.matcher(script);
while (m.find()) {
part1 = script.substring(0, m.start());
part2 = script.substring(m.end());
script = part1 + m.group().replaceAll("[ \t]+", "") + part2;
m = p.matcher(script);
}
Is there any way to optimize all the three RegEx so that they take less time?
Thanks
Ali

I'll answer the first question.
You can combine all this into a single regex replace operation:
script = script.replaceAll("\\|\\||=>|[:!]=|<>|[<>()!,+=*|-]", " $0 ");
Explanation:
\|\| # Match ||
| # or
=> # =>
| # or
[:!]= # := or !=
| # or
<> # <>
| # or
[<>()!,+=*|-] # <, >, (, ), !, comma, +, =, *, | or -

Sure. Your second approach is "almost" good. The problem is that you do not use your pattern for replacement itself. When you are using str.replaceAll() you actually creating Pattern instance every time you are calling this method. Pattern.compile() is called for you and it takes 90% of time.
You should use Matcher.replaceAll() instead.
String script = "dfgafjd;fjfd;jfd;djf;jds\\fdfdf****\\/";
String result = script;
Pattern p = Pattern.compile("[\\*\\/\\\\]"); // write all characters you want to remove here.
Matcher m = p.matcher(script);
if (m.find()) {
result = m.replaceAll("");
}
System.out.println(result);

It isn't the regexes causing your performance problem, it's that fact that you're doing many passes over the text, and constantly creating new Pattern objects. And it's not just performance that suffers, as Tim pointed out; it's much too easy to mess up the results of prior passes when you do that.
In fact, I'm guessing that those extra spaces in the dates are just a side effect your other replacements. If so, here's a way you can do all the replacements in one pass, without adding unwanted characters:
static String doReplace(String input)
{
String regex =
"/\\*[^*]*(?:\\*(?!/)[^*]*)*\\*/|" // a comment
+ "\\b\\d{2}/\\d{2}/\\d{2,4}\\b|" // a date
+ "(/|\\|\\||=>|[:!]=|<>|[<>()!,+=*|-])"; // an operator
Matcher m = Pattern.compile(regex).matcher(input);
StringBuffer sb = new StringBuffer();
while (m.find())
{
// if we found an operator, replace it
if (m.start(1) != -1)
{
m.appendReplacement(sb, " $1 ");
}
}
m.appendTail(sb);
return sb.toString();
}
see the online demo
The trick is, if you don't call appendReplacement(), the match position is not updated, so it's as if the match didn't occur. Because I ignore them, the comments and dates get reinserted along with the rest of the unmatched text, and I don't have to worry about matching the slash characters inside them.
EDIT Make sure the "comment" part of the regex comes before the "operator" part. Otherwise, the leading / of every comment will be treated as an operator.

Splitting comma separated string, ignore commas in quotes, but allow strings with one double quotation

I have searched through several posts on stackoverflow on how to split a string on comma delimiter, but ignore splitting on comma in quotes (see: How do I split a string into an array by comma but ignore commas inside double quotes?) I am trying to achieve just similar results, but need to also allow for a string that contains one double quote.
IE. Need "test05, \"test, 05\", test\", test 05" to splits into
test05
"test, 05"
test"
test 05
I tried a similar method to one mentioned here:
Regex for splitting a string using space when not surrounded by single or double quotes
Using Matcher, instead of split(). however, that specific examples it splits on spaces, and not on commas. I've tried to adjust the pattern to account for commas, instead, but have not had any luck.
String str = "test05, \"test, 05\", test\", test 05";
str = str + " "; // add trailing space
int len = str.length();
Matcher m = Pattern.compile("((\"[^\"]+?\")|([^,]+?)),++").matcher(str);
for (int i = 0; i < len; i++)
{
m.region(i, len);
if (m.lookingAt())
{
String s = m.group(1);
if ((s.startsWith("\"") && s.endsWith("\"")))
{
s = s.substring(1, s.length() - 1);
}
System.out.println(i + ": \"" + s + "\"");
i += (m.group(0).length() - 1);
}
}

You have reached the point where regular expressions break down.
I would recommend that you write a simple splitter instead which handles your special cases as you wish. Test Driven Development is great for doing this.
It looks, however, like you are trying to parse CSV lines. Have you considered using a CSV-library for this?

I've had similar issues with this, and I've found no good .net solution so went DIY.
In my application I'm parsing a csv so my split credential is ",". this method I suppose only works for where you have a single char split argument.
So, I've written a function that ignores commas within double quotes. it does it by converting the input string into a character array and parsing char by char
public static string[] Splitter_IgnoreQuotes(string stringToSplit)
{
char[] CharsOfData = stringToSplit.ToCharArray();
//enter your expected array size here or alloc.
string[] dataArray = new string[37];
int arrayIndex = 0;
bool DoubleQuotesJustSeen = false;
foreach (char theChar in CharsOfData)
{
//did we just see double quotes, and no command? dont split then. you could make ',' a variable for your split parameters I'm working with a csv.
if ((theChar != ',' || DoubleQuotesJustSeen) && theChar != '"')
{
dataArray[arrayIndex] = dataArray[arrayIndex] + theChar;
}
else if (theChar == '"')
{
if (DoubleQuotesJustSeen)
{
DoubleQuotesJustSeen = false;
}
else
{
DoubleQuotesJustSeen = true;
}
}
else if (theChar == ',' && !DoubleQuotesJustSeen)
{
arrayIndex++;
}
}
return dataArray;
}
This function, to my application taste also ignores ("") in any input as these are unneeded and present in my input.

Unless you really need to DIY, you should consider the Apache Commons class org.apache.commons.csv.CSVParser
http://commons.apache.org/sandbox/csv/apidocs/org/apache/commons/csv/CSVParser.html

Split against this pattern:
(?<=\"?),(?!\")|(?<!\"),(?=\")
so it will be:
String[] splitArray = subjectString.split("(?<=\"?),(?!\")|(?<!\"),(?=\")");
UPD: according to recent changes in question logic, it's better not to use naked split, you should firstly separated text in comma from non-in-commas text, then make simple split(",") on the last one. Just use simple for loop and check how many quotes you've met, simultaneously saving characters you've read into a StringBuffer. At first you saving your characters into StringBuffer, until you met quotes, then you put your StringBuffer into array containing Strings that wasn't in quotes. Then you make new StringBuffer and saving next characters you read into it, after you've met second comma, you've stopping and putting your new StringBuffer into array containing strings that were in commas. Repeating until the end of the string. So you will have 2 arrays, one with Strings that were in commas, others with strings not in commas. Then you should split all elements of the second array.

Try this:
import java.util.regex.*;
public class Main {
public static void main(String[] args) throws Exception {
String text = "test05, \"test, 05\", test\", test 05";
Pattern p = Pattern.compile(
"(?x) # enable comments \n" +
"(\"[^\"]*\") # quoted data, and store in group #1 \n" +
"| # OR \n" +
"([^,]+) # one or more chars other than ',', and store it in #2 \n" +
"| # OR \n" +
"\\s*,\\s* # a ',' optionally surrounded by space-chars \n"
);
Matcher m = p.matcher(text);
while (m.find()) {
// get the match
String matched = m.group().trim();
// only print the match if it's group #1 or #2
if(m.group(1) != null || m.group(2) != null) {
System.out.println(matched);
}
}
}
}
For test05, "test, 05", test", test 05 it produces:
test05
"test, 05"
test"
test 05
and for test05, "test 05", test", test 05 it produces:
test05
"test 05"
test"
test 05

How to split a comma separated String while ignoring escaped commas?

I need to write a extended version of the StringUtils.commaDelimitedListToStringArray function which gets an additional parameter: the escape char.
so calling my:
commaDelimitedListToStringArray("test,test\\,test\\,test,test", "\\")
should return:
["test", "test,test,test", "test"]
My current attempt is to use String.split() to split the String using regular expressions:
String[] array = str.split("[^\\\\],");
But the returned array is:
["tes", "test\,test\,tes", "test"]
Any ideas?

The regular expression
[^\\],
means "match a character which is not a backslash followed by a comma" - this is why patterns such as t, are matching, because t is a character which is not a backslash.
I think you need to use some sort of negative lookbehind, to capture a , which is not preceded by a \ without capturing the preceding character, something like
(?<!\\),
(BTW, note that I have purposefully not doubly-escaped the backslashes to make this more readable)

Try:
String array[] = str.split("(?<!\\\\),");
Basically this is saying split on a comma, except where that comma is preceded by two backslashes. This is called a negative lookbehind zero-width assertion.

For future reference, here is the complete method i ended up with:
public static String[] commaDelimitedListToStringArray(String str, String escapeChar) {
// these characters need to be escaped in a regular expression
String regularExpressionSpecialChars = "/.*+?|()[]{}\\";
String escapedEscapeChar = escapeChar;
// if the escape char for our comma separated list needs to be escaped
// for the regular expression, escape it using the \ char
if(regularExpressionSpecialChars.indexOf(escapeChar) != -1)
escapedEscapeChar = "\\" + escapeChar;
// see http://stackoverflow.com/questions/820172/how-to-split-a-comma-separated-string-while-ignoring-escaped-commas
String[] temp = str.split("(?<!" + escapedEscapeChar + "),", -1);
// remove the escapeChar for the end result
String[] result = new String[temp.length];
for(int i=0; i<temp.length; i++) {
result[i] = temp[i].replaceAll(escapedEscapeChar + ",", ",");
}
return result;
}

As matt b said, [^\\], will interpret the character preceding the comma as a part of the delimiter.
"test\\\\\\,test\\\\,test\\,test,test"
-(split)->
["test\\\\\\,test\\\\,test\\,tes" , "test"]
As drvdijk said, (?<!\\), will misinterpret escaped backslashes.
"test\\\\\\,test\\\\,test\\,test,test"
-(split)->
["test\\\\\\,test\\\\,test\\,test" , "test"]
-(unescape commas)->
["test\\\\,test\\,test,test" , "test"]
I would expect being able to escape backslashes as well...
"test\\\\\\,test\\\\,test\\,test,test"
-(split)->
["test\\\\\\,test\\\\" , "test\\,test" , "test"]
-(unescape commas and backslashes)->
["test\\,test\\" , "test,test" , "test"]
drvdijk suggested (?<=(?<!\\\\)(\\\\\\\\){0,100}), which works well for lists with elements ending with up to 100 backslashes. This is far enough... but why a limit? Is there a more efficient way (isn't lookbehind greedy)? What about invalid strings?
I searched for a while for a generic solution, then I wrote the thing myself... The idea is to split following a pattern that matches the list elements (instead of matching the delimiter).
My answer does not take the escape character as a parameter.
public static List<String> commaDelimitedListStringToStringList(String list) {
// Check the validity of the list
// ex: "te\\st" is not valid, backslash should be escaped
if (!list.matches("^(([^\\\\,]|\\\\,|\\\\\\\\)*(,|$))+")) {
// Could also raise an exception
return null;
}
// Matcher for the list elements
Matcher matcher = Pattern
.compile("(?<=(^|,))([^\\\\,]|\\\\,|\\\\\\\\)*(?=(,|$))")
.matcher(list);
ArrayList<String> result = new ArrayList<String>();
while (matcher.find()) {
// Unescape the list element
result.add(matcher.group().replaceAll("\\\\([\\\\,])", "$1"));
}
return result;
}
Description for the pattern (unescaped):
(?<=(^|,)) forward is start of string or a ,
([^\\,]|\\,|\\\\)* the element composed of \,, \\ or characters wich are neither \ nor ,
(?=(,|$)) behind is end of string or a ,
The pattern may be simplified.
Even with the 3 parsings (matches + find + replaceAll), this method seems faster than the one suggested by drvdijk. It can still be optimized by writing a specific parser.
Also, what is the need of having an escape character if only one character is special, it could simply be doubled...
public static List<String> commaDelimitedListStringToStringList2(String list) {
if (!list.matches("^(([^,]|,,)*(,|$))+")) {
return null;
}
Matcher matcher = Pattern.compile("(?<=(^|,))([^,]|,,)*(?=(,|$))")
.matcher(list);
ArrayList<String> result = new ArrayList<String>();
while (matcher.find()) {
result.add(matcher.group().replaceAll(",,", ","));
}
return result;
}

split(/(?<!\\),/g) worked for me, but the accepted answer did not
> var x = "test,test\,test\,test,test"
undefined
> x.split(/(?<!\\),/g)
[ 'test', 'test\\,test\\,test', 'test' ]
> x.split("(?<!\\\\),")
[ 'test,test\\,test\\,test,test' ]

It's probably not "super fancy" solution, but possibly more time-efficient one. Escaping an escape character is also supported and it's working in browsers not supporting 'lookbehinds'.
function splitByDelimiterIfItIsNotEscaped (text, delimiter, escapeCharacter) {
const splittedText = []
let numberOfDelimitersBeforeOtherCharacter = 0
let nextSplittedTextPartIndex = 0
for (let characterIndex = 0, character = text[0]; characterIndex < text.length; characterIndex++, character = text[characterIndex]) {
if (character === escapeCharacter) {
numberOfDelimitersBeforeOtherCharacter++
} else if (character === delimiter && (!numberOfDelimitersBeforeOtherCharacter || !(numberOfDelimitersBeforeOtherCharacter % 2))) {
splittedText.push(text.substring(nextSplittedTextPartIndex, characterIndex))
nextSplittedTextPartIndex = characterIndex + 1
} else {
numberOfDelimitersBeforeOtherCharacter = 0
}
}
if (nextSplittedTextPartIndex <= text.length) {
splittedText.push(text.substring(nextSplittedTextPartIndex, text.length))
}
return splittedText
}
function onChange () {
console.log(splitByDelimiterIfItIsNotEscaped(inputBox.value, ',', '\\'))
}
addEventListener('change', onChange)
onChange()
After making a change unfocus the input box (use tab for example).
<input id="inputBox" value="test,test\,test\,test,test"/>