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Regex not matching against ampersand

I'm trying to match the following regex:
\b(?:mr|mrs|ms|miss|messrs|mmes|dr|prof|rev|sr|jr|&|and)\.?\b
In other words, a word boundary followed by any of the strings above (optionally followed by a period character) followed by a word boundary.
I'm trying to match this in Java, but the ampersand will not match. For example:
Pattern p = Pattern.compile(
"\\b(?:mr|mrs|ms|miss|messrs|mmes|dr|prof|rev|sr|jr|&|and)\\.?\\b",
Pattern.CASE_INSENSITIVE);
String result = p.matcher("mr one and mrs.two and three & four").replaceAll(" ");
System.out.println("["+result+"]");
The output of this is: [ one two three & four]
I've also tried this at regex101, and again the ampersand does not match: https://regex101.com/r/klkmwl/1
Escaping the ampersand does not make a difference, and I've tried using the hex escape sequence \x26 instead of ampersand (as suggested in this question). Why is this not matching?

Your regex will match an ampersand if it is located in between word chars, e.g. three&four, see this regex demo. This happens because \b before a non-word char requires a word char to appear immediately before it. Also, as there is a \b after an optional dot, both the dot and ampersand will only match if there is a word char immediately on the left.
You need to re-write the pattern so that the word boundaries are applied to the words rather than symbols:
Pattern p = Pattern.compile(
"(?:\\b(?:mr|mrs|ms|miss|messrs|mmes|dr|prof|rev|sr|jr|and)\\b|&)\\.?",
Pattern.CASE_INSENSITIVE);
See the regex demo online.

Problem is due to use of word boundaries. There are no word boundaries before or after a non-word character like &.
In place of word boundary you can use lookarounds:
(?<!\w)(?:[jsdm]r|mr?s|miss|messrs|mmes|prof|re|&|and)\.?(?!\w)
Updated RegEx Demo
(?<!\w): Make sure that previous character is not a word character
(?!\w): Make sure that next character is not a word character
Note some tweaks in your regex to make it shorter.

How to put [] in my regex [duplicate]

I have comma separated list of regular expressions:
.{8},[0-9],[^0-9A-Za-z ],[A-Z],[a-z]
I have done a split on the comma. Now I'm trying to match this regex against a generated password. The problem is that Pattern.compile does not like square brackets that is not escaped.
Can some please give me a simple function that takes a string like so: [0-9] and returns the escaped string \[0-9\].

For some reason, the above answer didn't work for me. For those like me who come after, here is what I found.
I was expecting a single backslash to escape the bracket, however, you must use two if you have the pattern stored in a string. The first backslash escapes the second one into the string, so that what regex sees is \]. Since regex just sees one backslash, it uses it to escape the square bracket.
\\]
In regex, that will match a single closing square bracket.
If you're trying to match a newline, for example though, you'd only use a single backslash. You're using the string escape pattern to insert a newline character into the string. Regex doesn't see \n - it sees the newline character, and matches that. You need two backslashes because it's not a string escape sequence, it's a regex escape sequence.

You can use Pattern.quote(String).
From the docs:
public static String quote​(String s)
Returns a literal pattern String for the specified String.
This method produces a String that can be used to create a Pattern that would match the string s as if it were a literal pattern.
Metacharacters or escape sequences in the input sequence will be given no special meaning.

You can use the \Q and \E special characters...anything between \Q and \E is automatically escaped.
\Q[0-9]\E

Pattern.compile() likes square brackets just fine. If you take the string
".{8},[0-9],[^0-9A-Za-z ],[A-Z],[a-z]"
and split it on commas, you end up with five perfectly valid regexes: the first one matches eight non-line-separator characters, the second matches an ASCII digit, and so on. Unless you really want to match strings like ".{8}" and "[0-9]", I don't see why you would need to escape anything.

java regex matching &[text(text - text text) !text]

I am currently working on creating a regex to split out all occurrences of Strings that match the following format: &[text(text - text text) !text]. Here text can be any char really. and the spacing is important. The text will be listed as shown.
I have tried the following regex but I cannot seem to get it to work:
&\\[([^\\]]*)\\]
Any help would be greatly appreciated.

You replace text with \w+ to capture 1 or more word characters.
Assuming everything else was a literal, the following regular expression should work:
&\[\w+\(\w+ - \w+ \w+\) !\w+\]
You could also use [a-zA-Z] in place of \w if you would like. It is sometimes easier to understand since it explicitly describes the characters to match, a-z and A-Z inclusive.
&\[[a-zA-Z]+\([a-zA-Z]+ - [a-zA-Z]+ [a-zA-Z]+\) ![a-zA-Z]+\]
And for one character only, remove the +
&\[\w\(\w - \w \w\) !\w\]
&\[[a-zA-Z]\([a-zA-Z] - [a-zA-Z] [a-zA-Z]\) ![a-zA-Z]\]
P.S - I cant remember if -, &, or ! are coutned as regex symbols and if they are you can make them literals by using \-, \&, or \!.
P.P.S - In java you have to escape \ so \w becomes \\w in a string.

If you want to extract text as groups to work with them after:
&\\[(\\w+)\\((\w+)\\s\\-\\s(\\w+)\\s(\\w+)\\)\\s!(\\w+)]
example

Regex Match word that include a Dot

I have a Question I have this Sentence for Example:
"HalloAnna daveca.nn dave anna ca. anna"
And I only wanna match the single Standing "ca." .
My RegEx is like that :
(?i)\b(ca\.)\b
But this doesn't work and I don't know why. Any ideas ?
//Update
I excecute it with:
testSource.replaceAll()
and with
pattern.matcher(testSource).replaceAll().
both doesn´t work.

You must escape the dot and assert a non-word following:
(?i)\bca\.(?=\W)
See live demo.

You should use it like this:
Pattern.compile("(?i)\\b(ca\\.)(?=\\W)").matcher(a).replaceAll("SOME TEXT");
Which if you omit the java escapes gives a regex: (?i)\b(ca\.)\W.
Every \ in normal regex has to be escaped in java - \\.
Also, before a word you have word boundary (\b), but it applies only to a part in String where you have a change from whitespace to a alphanumeric character or the other way around. But in your case you have a dot, which is not an alphanumeric character, so you can't use \b at the end. You can use \W which means that a non-word character is following the dot. But to use \W you need to ignore it in the capture group (so it won't be replaced) - (?=.
Another issue was that you used ., which matches any character, but you actually want to match the real dot, so to do that you have to escape it - \., which in java String becomes \\..

Regex for "* word"

Any Regex masters out there? I need a regular expression in Java that matches:
"RANDOMSTUFF SPECIFICWORD"
Including the quotation marks.
Thus I need
to match the first quote,
RANDOMSTUFF (any number of words with spaces between preceding SPECIFICWORD)
SPECIFICWORD (a specific word which I won't specify here.)
and the ending quote.
I don't want to match things such as:
RANDOMSTUFF SPECIFICWORD
"RANDOMSTUFF NOTTHESPECIFICWORD"
"RANDOMSTUFF SPECIFICWORD MORERANDOMSTUFF"

\".*\sSPECIFICWORD\"
If you don't want to allow quotes in between, use \"[^"]*\sSPECIFICWORD\"
. matches any character
* says 0 or more of the preceding character (in this case, 0 or more of any characters)
\s matches any whitespace character
SPECIFICWORD will be treated as a string literal, assuming there are no special characters (escape them if there are)
\" matches the quote
[^"] means any character except a quote (the ^ is what makes it 'except')
Also, this link could be useful. Regex's are powerful expressions and are applicable across virtually any language, so it would be a good thing to become comfortable with using them.
EDIT:
As several other posters have pointed out, adding ^ to the beginning and $ to the end will only match if the entire line matches.
^ matches the beginning of the line
$ matches the end of the line

^.*\s+SPECIFICWORD"$
'^' matches 'from the start of the line'
.* matches anything
\s+ matches 'any amount of whitespace, but at least some'
SPECIFICWORD" is a string literal
$ means 'this is the end of the line'
Note that ^ and $ are not always 'line'-based; most languages allow you to specify a 'multiline' mode that would cause them to match 'start of the string/end of the string' instead of one line at a time.

Will this string be matched as a line by line basis or will it be found within the text? If so, you can add anchors to ensure that it matches the string.
^(\".*\sSPECIFICWPRD\")$
Saying, at the start of the line, look for a double quote followed by zero or more random characters followed by a single whitespace, followed by the specific word, followed by a double quote at the end of the string.
Optionally, there are excellent tools for designing regex patterns and seeing what they match in real time.
Here are a couple of examples:
http://gskinner.com/RegExr/
http://regex101.com/r/zC3fM1

Try:
\"[\w\s]*SPECIFICWORD\"
Works like this:
\" matches opening quote
[\w\s]* matches zero or more of the characters from the following sets:
[a-zA-Z_0-9] (\w part)
[ \t\n\x0B\f\r] (\s part)
SPECIFICWORD matches the SPECIFICWORD
\" matches closing quote