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Regular Expression in Java. Splitting a string using pattern and matcher

I am trying to get all the matching groups in my string.
My regular expression is "(?<!')/|/(?!')". I am trying to split the string using regular expression pattern and matcher. string needs to be split by using /, but '/'(surrounded by ') this needs to be skipped. for example "One/Two/Three'/'3/Four" needs to be split as ["One", "Two", "Three'/'3", "Four"] but not using .split method.
I am currently the below
// String to be scanned to find the pattern.
String line = "Test1/Test2/Tt";
String pattern = "(?<!')/|/(?!')";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.matches()) {
System.out.println("Found value: " + m.group(0) );
} else {
System.out.println("NO MATCH");
}
But it always saying "NO MATCH". where i am doing wrong? and how to fix that?
Thanks in advance

To get the matches without using split, you might use
[^'/]+(?:'/'[^'/]*)*
Explanation
[^'/]+ Match 1+ times any char except ' or /
(?: Non capture group
'/'[^'/]* Match '/' followed by optionally matching any char except ' or /
)* Close group and optionally repeat it
Regex demo | Java demo
String regex = "[^'/]+(?:'/'[^'/]*)*";
String string = "One/Two/Three'/'3/Four";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
Output
One
Two
Three'/'3
Four
Edit
If you do not want to split don't you might also use a pattern to not match / but only when surrounded by single quotes
[^/]+(?:(?<=')/(?=')[^/]*)*
Regex demo

Try this.
String line = "One/Two/Three'/'3/Four";
Pattern pattern = Pattern.compile("('/'|[^/])+");
Matcher m = pattern.matcher(line);
while (m.find())
System.out.println(m.group());
output:
One
Two
Three'/'3
Four

Here is simple pattern matching all desired /, so you can split by them:
(?<=[^'])\/(?=')|(?<=')\/(?=[^'])|(?<=[^'])\/(?=[^'])
The logic is as follows: we have 4 cases:
/ is sorrounded by ', i.e. `'/'
/ is preceeded by ', i.e. '/
/ is followed by ', i.e. /'
/ is sorrounded by characters other than '
You want only exclude 1. case. So we need to write regex for three cases, so I have written three similair regexes and used alternation.
Explanation of the first part (other two are analogical):
(?<=[^']) - positiva lookbehind, assert what preceeds is differnt frim ' (negated character class [^']
\/ - match / literally
(?=') - positiva lookahead, assert what follows is '\
Demo with some more edge cases

Try something like this:
String line = "One/Two/Three'/'3/Four";
String pattern = "([^/]+'/'\d)|[^/]+";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(line);
boolean found = false;
while(m.find()) {
System.out.println("Found value: " + m.group() );
found = true;
}
if(!found) {
System.out.println("NO MATCH");
}
Output:
Found value: One
Found value: Two
Found value: Three'/'3
Found value: Four

Using Pattern and Matcher to search for special characters (Example: $)

Apologies if this has already been answered.
I am using the following code to search for a substring:
String subject = "ABC"
String subString = "AB"
Pattern pattern = Pattern.compile(subString, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(subject);
while (matcher.find()){
//Matched
}
But when my subject string contains a $ in the beginning, it does not work since it is a special character.
String subject = "$ABC"
String subString = "$"
How does one handle that?

By escaping the special character in the subString. Like,
String subString = "\\$";
or telling the Pattern to match literals. Like,
Pattern pattern = Pattern.compile(subString, Pattern.LITERAL | Pattern.CASE_INSENSITIVE);

There are few meta characters in regex. And some of them which are supported by regex in java are
( ) [ ] { { \ ^ $ | ? * + . < > - = !
So $ is a indeed meta character here. The meta character conveys special meaning to the regex engine and hence can't be use literally. So in order to use them you have to combine them with escape character which is backslash \
So String subject = "\\$ABC"
String subString = "\\$"
would do. Java uses double backslash instead of single for escape character unlike the other regex engine.

Regex to find a word between $$ sign

I want regular expression to find a word between $$ sign only. It must start and end with $ sign. I have tried below expression
final String regex = "\\$\\w+\\$";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher("$abc$ cde$efg$hij pqr");
This should give me count as 1. But my regular expression also considering second occurrence of (cde$efg$hij) which it should not consider as it is not starting and ending with $$ sign.

You may use non-word boundaries:
final String regex = "\\B\\$\\w+\\$\\B";
The pattern will only match if the $abc$ is not preceded and followed with word chars. See the regex demo.
See Java demo:
String regex = "\\B\\$\\w+\\$\\B";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher("$abc$ cde$efg$hij pqr");
while (matcher.find()){
System.out.println(matcher.group(0));
} // => $abc$
Besides non-word boundaries, you may use whitespace boundaries if you only want to match in between whitespace chars or start/end of string:
String regex = "(?<!\\S)\\$\\w+\\$(?!\\S)";
Or, use unambiguous word boundaries (as I call them):
String regex = "(?<!\\w)\\$\\w+\\$(?!\\w)";
The (?<!\\w) negative lookbehind will fail the match if a word char is found immediately to the left of the current location, and the (?!\w) negative lookahead will fail the match if a word char is found immediately to the right of the current location.

The problem was extracting fields between dollar signs for me.
List<String> getFieldNames(#NotNull String str) {
final String regex = "\\$(\\w+)\\$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
List<String> fields = new ArrayList<>();
while (matcher.find()) {
fields.add(matcher.group(1));
}
return fields;
}
This will return list of words between dollar signs.

pattern matching to detect special characters in a word

I am trying to identify any special characters ('?', '.', ',') at the end of a string in java. Here is what I wrote:
public static void main(String[] args) {
Pattern pattern = Pattern.compile("{.,?}$");
Matcher matcher = pattern.matcher("Sure?");
System.out.println("Input String matches regex - "+matcher.matches());
}
This returns a false when it's expected to be true. Please suggest.

Use "sure?".matches(".*[.,?]").
String#matches(...) anto-anchors the regex with ^ and $, no need to add them manually.

This is your code:
Pattern pattern = Pattern.compile("{.,?}$");
Matcher matcher = pattern.matcher("Sure?");
System.out.println("Input String matches regex - "+matcher.matches());
You have 2 problems:
You're using { and } instead of character class [ and ]
You're using Matcher#matches() instead of Matcher#find. matches method matches the full input line while find performs a search anywhere in the string.
Change your code to:
Pattern pattern = Pattern.compile("[.,?]$");
Matcher matcher = pattern.matcher("Sure?");
System.out.println("Input String matches regex - " + matcher.find());

Try this
Pattern pattern = Pattern.compile(".*[.,?]");
...

RegEX: how to match string which is not surrounded

I have a String "REC/LESS FEES/CODE/AU013423".
What could be the regEx expression to match "REC" and "AU013423" (anything that is not surrounded by slashes /)
I am using /^>*/, which works and matches the string within slash's i.e. using this I am able to find "/LESS FEES/CODE/", but I want to negate this to find reverse i.e. REC and AU013423.
Need help on this. Thanks

If you know that you're only looking for alphanumeric data you can use the regex ([A-Z0-9]+)/.*/([A-Z0-9]+) If this matches you will have the two groups which contain the first & final text strings.
This code prints RECAU013423
final String s = "REC/LESS FEES/CODE/AU013423";
final Pattern regex = Pattern.compile("([A-Z0-9]+)/.*/([A-Z0-9]+)", Pattern.CASE_INSENSITIVE);
final Matcher matcher = regex.matcher(s);
if (matcher.matches()) {
System.out.println(matcher.group(1) + matcher.group(2));
}
You can tweak the regex groups as necessary to cover valid characters

Here's another option:
String s = "REC/LESS FEES/CODE/AU013423";
String[] results = s.split("/.*/");
System.out.println(Arrays.toString(results));
// [REC, AU013423]

^[^/]+|[^/]+$
matches anything that occurs before the first or after the last slash in the string (or the entire string if there is no slash present).
To iterate over all matches in a string in Java:
Pattern regex = Pattern.compile("^[^/]+|[^/]+$");
Matcher regexMatcher = regex.matcher(subjectString);
while (regexMatcher.find()) {
// matched text: regexMatcher.group()
// match start: regexMatcher.start()
// match end: regexMatcher.end()
}