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Regex for all Symbols except for |

Is there a way to use the patter \p{Punct} in a regex in Java without the Symbol |.
I tried \\p{Punct}&&[^|], but it didn't worked.

What you have is nearly correct. The correct syntax is:
[\\p{Punct}&&[^|]]

Did you see this question: Punctuation Regex in Java?
I think you can modify that answer slightly to use (?![|])\\p{Punct}

do you try with this class : [^\\P{Punct}|] (note the uppercase P)

Regex with -, ::, ( and )

I need to split the string
(age-is-25::OR::last_name-is-qa6)::AND::(age-is-20::OR::first_name-contains-test)
into
string[0] = (age-is-25::OR::last_name-is-qa6)
string[1] = AND
string[2] = (age-is-20::OR::first_name-contains-test)
I tried writing so many regex expressions, but nothing works as expected.
Using the following regex, Matcher.groupCount() which returns 2 but assigning results to an arraylist returns null as the elements.
Pattern pattern = Pattern.compile("(\\)::)?|(::\\()?");
I tried to split it using ):: or ::(.
I know the regex looks too stupid, but being a beginner this is the best I could write.

You can use positive lookahead and lookbehind to match the first and last parentheses.
String str = "(age-is-25::OR::last_name-is-qa6)::AND::(age-is-20::OR::first_name-contains-test)";
for (String s : str.split("(?<=\\))::|::(?=\\()"))
System.out.println(s);
Outputs:
(age-is-25::OR::last_name-is-qa6)
AND
(age-is-20::OR::first_name-contains-test)
Just a note however: It seems like you are parsing some kind of recursive language. Regular expressions are not good at doing this. If you are doing advanced parsing I would recommend you to look at other parsing methods.

To me it looks like a big part of your stress comes from the need for escaping special characters in your search term. I highly recommend to not do manual escaping of special characters, but instead to use Pattern.quote(...) for the escaping.

This should works
"(?<=\\))::|::(?=\\()"

This should work for you.
\)::|::\(

textString.split("\\)::|::\\(")
should work.

How do I write this regex in Java?

Basically, for this regex
{(\(\(("\w{1,}",{0,1}){2}\),\(("[^:=;#"\)\(\{\}\[\]]{1,}",{0,1}){2}"[LR]{1}"\)\),{0,1}){1,}}
Which I've tested on Regexpal for this input:
{(("st0","sy0"),("st1","sy3","L")),(("st0","sy0"),("st1","^","L"))}
I now need in Java. I can't seem to figure out how to convert it. Can somebody show me how to?

You need to escape the special chars - specifically the backslashes and the quote marks.

The regular expression could work on Java, the only thing that you have to do, is escape the backslash .

How do I write a regular expression to find the following pattern?

I am trying to write a regular expression to do a find and replace operation. Assume Java regex syntax. Below are examples of what I am trying to find:
12341+1
12241+1R1
100001+1R2
So, I am searching for a string beginning with one or more digits, followed by a "1+1" substring, followed by 0 or more characters. I have the following regex:
^(\d+)(1\\+1).*
This regex will successfully find the examples above, however, my goal is to replace the strings with everything before "1+1". So, 12341+1 would become 1234, and 12241+1R1 would become 1224. If I use the first grouped expression $1 to replace the pattern, I get the wrong result as follows:
12341+1 becomes 12341
12241+1R1 becomes 12241
100001+1R2 becomes 100001
Any ideas?

Your existing regex works fine, just that you are missing a \ before \d
String str = "100001+1R2";
str = str.replaceAll("^(\\d+)(1\\+1).*","$1");
Working link

IMHO, the regex is correct.
Perhaps you wrote it wrong in the code. If you want to code the regex ^(\d+)(1\+1).* in a string, you have to write something like String regex = "^(\\d+)(1\\+1).*".
Your output is the result of ^(\d+)(1+1).* replacement, as you miss some backslash in the string (e.g. "^(\\d+)(1\+1).*").

Your regex looks fine to me - I don't have access to java but in JavaScript the code..
"12341+1".replace(/(\d+)(1\+1)/g, "$1");
Returns 1234 as you'd expect. This works on a string with many 'codes' in too e.g.
"12341+1 54321+1".replace(/(\d+)(1\+1)/g, "$1");
gives 1234 5432.

Personally, I wouldn't use a Regex at all (it'd be like using a hammer on a thumbtack), I'd just create a substring from (Pseudocode)
stringName.substring(0, stringName.indexOf("1+1"))
But it looks like other posters have already mentioned the non-greedy operator.
In most Regex Syntaxes you can add a '?' after a '+' or '*' to indicate that you want it to match as little as possible before moving on in the pattern. (Thus: ^(\d+?)(1+1) matches any number of digits until it finds "1+1" and then, NOT INCLUDING the "1+1" it continues matching, whereas your original would see the 1 and match it as well).

Negating a set of words via java regex

I would like to negate a set of words using java regex.
Say, I want to negate cvs, svn, nvs, mvc. I wrote a regex which is ^[(svn|cvs|nvs|mvc)].
Some how that seems not to be working.

Try this:
^(?!.*(svn|cvs|nvs|mvc)).*$
this will match text if it doesn't contain one of svn, cvs, nvs or mvc.
This is a similar question: C# Regex to match a string that doesn't contain a certain string?

It's not that simple. If you want to negate a word you have to split it to letters and negate each letter.
so to negate
/svn/
you have to write
/[^s][^v][^n]/
So what you want to filter out will turn into really ugly regex and I think it's better idea to use this regex
/svn|cvs|nvs|mvc/
and when you test your string against it, just negate the result.
In JS this would look more less like that:
!/svn|cvs|nvs|mvc/.test("this is your test string");

Your regex is wrong. Between square brackets, you can put characters to require or to ignore. If you don't find ^(svn|cvs|nvs|mvc)$, you're fine.