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regex to filter out string

I'm filtering out string using below regex
^(?!.*(P1 | P2)).*groupName.*$
Here group name is specific string which I replace at run time. This regex is already running fine.
I've two input strings which needs to pass through from this regex. Can't change ^(?!.*(P1 | P2)) part of regex, so would like to change regex after this part only. Its a very generic regex which is being used at so many places, so I have only place to have changes is groupName part of regex. Is there any way where only 2 string could pass through this regex ?
1) ADMIN-P3-UI-READ-ONLY
2) ADMIN-P3-READ-ONLY
In regex groupName is a just a variable which will be replaced at run time with required string. In this case I want 2 string to be passed, so groupName part can be replaced with READ-ONLY but it will pass 1 string too.
Can anyone suggest on this how to make this work ?

You could use negative lookBehind:
(?<!UI-)READ-ONLY
so there must be no UI- before READ-ONLY

You can add another lookahead at the very start of your pattern to further restrict what it matches because your pattern is of the "match-everything-but" type.
So, it may look like
String extraCondition = "^(?!.*UI)";
String regex = "^(?!.*(P1|P2)).*READ-ONLY.*$";
String finalRegex = extraCondition + regex;
The pattern will look like
^(?!.*UI)^(?!.*(P1|P2)).*READ-ONLY.*$
matching
^(?!.*UI) - no UI after any zero or more chars other than line break chars as many as possible from the start of string
^(?!.*(P1|P2)) - no P1 nor P2 after any zero or more chars other than line break chars as many as possible from the start of string
.*READ-ONLY - any zero or more chars other than line break chars as many as possible and then READ-ONLY
.*$ - the rest of the string. Note you may safely remove $ here unless you want to make sure there are no extra lines in the input string.

Java String Regex replacement

Sample Input:
a:b
a.in:b
asds.sdsd:b
a:b___a.sds:bc___ab:bd
Sample Output:
a:replaced
a.in:replaced
asds.sdsd:replaced
a:replaced___a.sds:replaced___ab:replaced
String which comes after : should be replaced with custom function.
I have done the same without Regex. I feel it can be replaced with regex as we are trying to extract string out of specific pattern.
For first three cases, it's simple enough to extract String after :, but I couldn't find a way to deal with third case, unless I split the string ___ and apply the approach for first type of pattern and again concatenate them.

Just replace only the letters with exists next to : with the string replaced.
string.replaceAll("(?<=:)[A-Za-z]+", "replaced");
DEMO
or
If you also want to deal with digits, then add \d inside the char class.
string.replaceAll("(?<=:)[A-Za-z\\d]+", "replaced");

(:)[a-zA-Z]+
You can simply do this with string.replaceAll.Replace by $1replaced.See demo.
https://regex101.com/r/fX3oF6/18

Remove repeating set of characters in a string

I want to remove the sequesnce "-~-~-" if it repeats in a string, but only if they are together.
I have tried to create a regex based on the removing of multiple white spaces regex:
test.replaceAll("\\s+", " ");
Unfortunately I was unsuccessful. Can someone please help me write the correct regex? thanks.
Example:
string test = "hello-~-~--~-~--~-~-"
output:
hello-~-~-
Another example
string test = "-~-~--~-~--~-~-hello-~-~--~-~--~-~-"
output:
-~-~-hello-~-~-

The regex is:
test.replaceAll("(-~-~-){2,}", "-~-~-")
replaceAll replaces all occurrences matched by the regex (the first parameter) with the second parameter.
the () groups the expression -~-~- together, {2,} means two or more occurrences.
EDIT
Like #anubhava said, instead of using -~-~- for the replacement string, you could also use $1 which backreferences the first capturing group (i.e. the expression in the regex surrounded by ()).

test.replaceAll("(-~-~-)+", "-~-~-");

This is the regex you need:
(-~-~-){2}

Need some help getting some stuff off a string

I want to get some info out of my string but there's two possible "expressions" for the string. I want to get "a" & "b" out of the string. This is how they look:
Format one:
http://default.com/default/a/b
Format two:
http://default.com/#!default|1|a|b|1
How can I do this?

If the strings always looks like this, you could do the following:
Search for the #-char to decide, if you have type 1 or 2.
In case of type 1, split with delimiter '/' and always take the last and the one before. For type 2, also first split with '/' and then, split the last part again with delimiter '|' and take results[2] and results[3].

Use a regex to split the string.
Split on "default"
Regex Split

There are many ways you can do this - regular expressions is the most common.
In pseudo code:
if the string contains "/#!default" then:
Use the regular expression ^.*/([^/])/([^/])$
if the string contains "/default" then:
Use the regular expression ^.*|([^|])|([^|])|1$
Take the 1st and 2nd blocks from the matcher

Java Regex to remove start/end single quotes but leave inside quotes

I have data from a CSV file that is enclosed in single quotes, like:
'Company name'
'Price: $43.50'
'New York, New York'
I want to be able to replace the single quotes at the start/end of the value but leave quotes in the data, like:
'Joe's Diner' should become Joe's Diner
I can do
updateString = theString.replace("^'", "").replace("'$", "");
but I wanted to know if I could combine it to only do one replace.

You could use the or operator.
updateString = theString.replaceAll("(^')|('$)","");
See if that works for you :)

updateString = theString.replaceFirst("^'(.*)'$", "$1");
Note that the form you have no won't work because replace uses literal strings, not regexes.
This works by using a capturing group (.*), which is referred to with $1 in the replacement text. You could also do something like:
Pattern patt = Pattern.compile("^'(.*)'$"); // could be stored in a static final field.
Matcher matcher = patt.matcher(theString);
boolean matches = matcher.matches();
updateString = matcher.group(1);
Of course, if you're certain there's a single quote at the beginning and end, the simplest solution is:
updateString = theString.substring(1, theString.length() - 1);

You can use regex to remove double quotes around digits/numbers.
jsonString.replaceAll("\"(\\d+)\"","$1");
above will not work if negative numbers are present.
for negative numbers, the regex will be a little complex like below. But I haven't tried it.
"([0-9]+\.{0,1}[0-9]*)"