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Matching a whole word with leading or trailing special symbols like dollar in a string

I can replace dollar signs by using Matcher.quoteReplacement. I can replace words by adding boundary characters:
from = "\\b" + from + "\\b";
outString = line.replaceAll(from, to);
But I can't seem to combine them to replace words with dollar signs.
Here's an example. I am trying to replace "$temp4" (NOT $temp40) with "register1".
String line = "add, $temp4, $temp40, 42";
String to = "register1";
String from = "$temp4";
String outString;
from = Matcher.quoteReplacement(from);
from = "\\b" + from + "\\b"; //do whole word replacement
outString = line.replaceAll(from, to);
System.out.println(outString);
Outputs
"add, $temp4, $temp40, 42"
How do I get it to replace $temp4 and only $temp4?

Use unambiguous word boundaries, (?<!\w) and (?!\w), instead of \b that are context dependent:
from = "(?<!\\w)" + Pattern.quote(from) + "(?!\\w)";
See the regex demo.
The (?<!\w) is a negative lookbehind that fails the match if there is a non-word char immediately to the left of the current location and (?!\w) is a negative lookahead that fails the match if there is a non-word char immediately to the right of the current location. The Pattern.quote(from) is necessary to escape any special chars in the from variable.
See the Java demo:
String line = "add, $temp4, $temp40, 42";
String to = "register1";
String from = "$temp4";
String outString;
from = "(?<!\\w)" + Pattern.quote(from) + "(?!\\w)";
outString = line.replaceAll(from, to);
System.out.println(outString);
// => add, register1, $temp40, 42

Matcher.quoteReplacement() is for the replacement string (to), not the regex (from). To include a string literal in the regex, use Pattern.quote():
from = Pattern.quote(from);

$ has special meaning in regex (it means “end of input”). To remove any special meaning from characters in your target, wrap it in regex quote/unquote expressions \Q...\E. Also, because $ is not ”word” character, the word boundary won’t wiork, so use look arounds instead:
line = line.replaceAll("(?<!\\S)\\Q" + from + "\\E(?![^ ,])", to);

Normally, Pattern.quote is the way to go to escape characters that may be specially interpreted by the regex engine.
However, the regular expression is still incorrect, because there is no word boundary before the $ in line; space and $ are both non-word characters. You need to place the word boundary after the $ character. There is no need for Pattern.quote here, because you're escaping things yourself.
String from = "\\$\\btemp4\\b";
Or more simply, because you know there is a word boundary between $ and temp4 already:
String from = "\\$temp4\\b";
The from variable can be constructed from the expression to replace. If from has "$temp4", then you can escape the dollar sign and add a word boundary.
from = "\\" + from + "\\b";
Output:
add, register1, $temp40, 42

How to find and skip special characters at the start and end of the word

New to regex and using following code to find if a word contains special characters at the end/start.
String s = "K-factor:";
String regExp = "^[^<>{}\"/|;:.,~!?##$%^=&*\\]\\\\()\\[0-9_+]*$";
Matcher matcher = Pattern.compile(regExp).matcher(s);
while (matcher.find()) {
System.out.println("Start: "+ matcher.start());
System.out.println("End: "+ matcher.end());
System.out.println("Group: "+ matcher.group());
s = s.substring(0, matcher.start());
}
Would like to find if there's any special character(: in this sample code) at the start or end of the string. Trying to skip the character.
Neither compile time error nor output.

Note that your regex matches a whole string that does not contain the chars you defined in the character class. The string in question does not match that pattern since it contains :.
You might consider splitting the pattern into two parts to check for the unwanted chars at the start or end using an alternation group:
String regExp = "^[<>{}\"/|;:.,~!?##$%^=&*\\]\\\\()\\[0-9_+]|[<>{}\"/|;:.,~!?##$%^=&*\\]\\\\()\\[0-9_+]$";
Here, the pattern has a ^<special_char_class>|<special_char_class>$ structure, ^ anchors the match at start, $ anchors the match at the string end, and | is the alternation operator. Note I removed the ^ from the start of the character class to make them positive rather than negated, so that they could match those chars/ranges defined in the class.
Alternatively, since you seem to just match a string if it contains a non-letter at the start/end, you may use a
String regExp = "^\\P{L}|\\P{L}$";
that is Unicode letter aware or - ASCII only:
String regExp = "^\\P{Alpha}|\\P{Alpha}$";

Matching a word with pound (#) symbol in a regex

I have regexp for check if some text containing word (with ignoring boundary)
String regexp = ".*\\bSOME_WORD_HERE\\b.*";
but this regexp return false when "SOME_WORD" starts with # (hashtag).
Example, without #
String text = "some text and test word";
String matchingWord = "test";
boolean contains = text.matches(".*\\b" + matchingWord + "\\b.*");
// now contains == true;
But with hashtag `contains` was false. Example:
text = "some text and #test word";
matchingWord = "#test";
contains = text.matches(".*\\b" + matchingWord + "\\b.*");
//contains == fasle; but I expect true

The \b# pattern matches a # that is preceded with a word character: a letter, digit or underscore.
If you need to match # that is not preceded with a word char, use a negative lookbehind (?<!\w). Similarly, to make sure the trailing \b matches if a non-word char is there, use (?!\w) negative lookahead:
text.matches("(?s).*(?<!\\w)" + matchingWord + "(?!\\w).*");
Using Pattern.quote(matchingWord) is a good idea if your matchingWord can contain special regex metacharacters.
Alternatively, if you plan to match your search words in between whitespace or start/end of string, you can use (?<!\S) as the initial boundary and (?!\S) as the trailing one
text.matches("(?s).*(?<!\\S)" + matchingWord + "(?!\\S).*");
And one more thing: the .* in the .matches is not the best regex solution. A regex like "(?<!\\S)" + matchingWord + "(?!\\S)" with Matcher#find() will be processed in a much more optimized way, but you will need to initialize the Matcher object for that.

If you are looking for words with leading '#', just simple remove the leading '#' from the searchword and use following regex.
text.matches("#\\b" + matchingWordWithoutLeadingHash + "\\b");

Java Pattern / Matcher not finding word break

I am having trouble with Java Pattern and Matcher. I've included a very simplified example of what I'm trying to do.
I had expected the pattern ".\b" to find the last character of the first word (or "4" in the example), but as I step through the code, m.find() always returns false. What am I missing here?
Why does the following Java code always print out "Not Found"?
Pattern p = Pattern.compile(".\b");
Matcher m = p.matcher("102939384 is a word");
int ixEndWord = 0;
if (m.find()) {
ixEndWord = m.end();
System.out.println("Found: " + ixEndWord);
} else {
System.out.println("Not Found");
}

You need to escape special characters in the regex: ".\\b"
Basically, in a String the backslash has to be escaped. So "\\" becomes the character '\'.
So the String ".\\b" becomes the litteral String ".\b", which will be used by the Pattern.

To expand upton AntonH's comment, whenever you want the "\" character to appear in a regex expression, you have to escape it so that it first appears in the string you are passing in.
As is, ".\b" is the string of a dot . followed by the special backspace character represented by \b, compared to ".\\b", which is the regex .\b.

How to replace last letter to another letter in java using regular expression

i have seen to replace "," to "." by using ".$"|",$", but this logic is not working with alphabets.
i need to replace last letter of a word to another letter for all word in string containing EXAMPLE_TEST using java
this is my code
Pattern replace = Pattern.compile("n$");//here got the real problem
matcher2 = replace.matcher(EXAMPLE_TEST);
EXAMPLE_TEST=matcher2.replaceAll("k");
i also tried "//n$" ,"\n$" etc
Please help me to get the solution
input text=>njan ayman
output text=> njak aymak

Instead of the end of string $ anchor, use a word boundary \b
String s = "njan ayman";
s = s.replaceAll("n\\b", "k");
System.out.println(s); //=> "njak aymak"

You can use lookahead and group matching:
String EXAMPLE_TEST = "njan ayman";
s = EXAMPLE_TEST.replaceAll("(n)(?=\\s|$)", "k");
System.out.println("s = " + s); // prints: s = njak aymak
Explanation:
(n) - the matched word character
(?=\\s|$) - which is followed by a space or at the end of the line (lookahead)
The above is only an example! if you want to switch every comma with a period the middle line should be changed to:
s = s.replaceAll("(,)(?=\\s|$)", "\\.");

Here's how I would set it up:
(?=.\b)\w
Which in Java would need to be escaped as following:
(?=.\\b)\\w
It translates to something like "a character (\w) after (?=) any single character (.) at the end of a word (\b)".
String s = "njan ayman aowkdwo wdonwan. wadawd,.. wadwdawd;";
s = s.replaceAll("(?=.\\b)\\w", "");
System.out.println(s); //nja ayma aowkdw wdonwa. wadaw,.. wadwdaw;
This removes the last character of all words, but leaves following non-alphanumeric characters. You can specify only specific characters to remove/replace by changing the . to something else.
However, the other answers are perfectly good and might achieve exactly what you are looking for.

if (word.endsWith("char oldletter")) {
name = name.substring(0, name.length() - 1 "char newletter");
}