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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.

Regular expression to match '\n' character

I am having a string "<?xml version=2.0><rss>Feed</rss>" I wrote a regex to match this string as
"<?xml.*<rss.*</rss>"
But if the input string contains \n like `"\nFeed" doesn't work for the above regex.
How to modify my regex to include \n character between strings.

The matching behavior of a dot can be controlled with a flag. It looks like in Java the default matching behavior for the dot is any character except the line terminators \r and \n.
I'm not a Java programmer, but usually using (?s) at beginning of a search string changes the matching behavior for a dot to any character including line terminators. So perhaps "(?s)<?xml.*<rss.*</rss>" works.
But better would be here to use "<?xml.*?<rss[\s\S]*?</rss>" as search string.
\s matches any whitespace character which includes line terminators and \S matches any non whitespace character. Both in square brackets results in matching any character.
For completness: [\w\W] matches also always any character.

You can combine it with (\\n)*. It is necessary to add an extra \ because it is a special character.
Another option is to execute replaceAll("\\n","") before executing the regex.

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

java regex pattern unclosed character class

I need some help. Im getting:
Caused by: java.util.regex.PatternSyntaxException: Unclosed character class near index 24
^[a-zA-Z└- 0-9£µ /.'-\]*$
^
at java.util.regex.Pattern.error(Pattern.java:1713)
at java.util.regex.Pattern.clazz(Pattern.java:2254)
at java.util.regex.Pattern.sequence(Pattern.java:1818)
at java.util.regex.Pattern.expr(Pattern.java:1752)
at java.util.regex.Pattern.compile(Pattern.java:1460)
at java.util.regex.Pattern.<init>(Pattern.java:1133)
at java.util.regex.Pattern.compile(Pattern.java:823)
Here is my code:
String testString = value.toString();
Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'-\\]*$");
Matcher m = pattern.matcher(testString);
I have to use the unicode value for some because I'm working with xhtml.
Any help would be great!

Assuming that you want to match \ and - and not ]:
Pattern pattern = Pattern.compile("^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'\\\\-]*$");
You need to double escape your backslashes, as \ is also an escape character in regex. Thus \\] escapes the backslash for java but not for regex. You need to add another java-escaped \ in order to regex-escape your second java-escaped \.
So \\\\ after java escaping becomes \\ which is then regex escaped to \.
Moving - to the end of the sequence means that it is used as a character, instead of a range operator as pointed out by Pshemo.

It is hard to say what are you trying to achieve, but I can see few strange things in your regex:
you have opened class of characters but never closed it. Instead you used \\] which makes ] normal character.
If you want to include ] in your characters class then you need additional ] at the end, like "^[a-zA-Z\300-\3770-9\u0153\346 \u002F.'-\\]]*$"
if you want to include \ in your characters class then you need to use \\\\ version, because you need to escape its special meaning two times, in regex engine, and in Javas String
you used - with ('-\\]) which in character class is used to specify range of characters like a-z or A-Z. To escape its special meaning you need to use \\-

Java regular expressions and dollar sign

I have Java string:
String b = "/feedback/com.school.edu.domain.feedback.Review$0/feedbackId");
I also have generated pattern against which I want to match this string:
String pattern = "/feedback/com.school.edu.domain.feedback.Review$0(.)*";
When I say b.matches(pattern) it returns false. Now I know dollar sign is part of Java RegEx, but I don't know how should my pattern look like. I am assuming that $ in pattern needs to be replaced by some escape characters, but don't know how many. This $ sign is important to me as it helps me distinguish elements in list (numbers after dollar), and I can't go without it.

Use
String escapedString = java.util.regex.Pattern.quote(myString)
to automatically escape all special regex characters in a given string.

You need to escape $ in the regex with a back-slash (\), but as a back-slash is an escape character in strings you need to escape the back-slash itself.
You will need to escape any special regex char the same way, for example with ".".
String pattern = "/feedback/com\\.navteq\\.lcms\\.common\\.domain\\.poi\\.feedback\\.Review\\$0(.)*";

In Java regex both . and $ are special. You need to escape it with 2 backslashes, i.e..
"/feedback/com\\.navtag\\.etc\\.Review\\$0(.*)"
(1 backslash is for the Java string, and 1 is for the regex engine.)

Escape the dollar with \
String pattern =
"/feedback/com.navteq.lcms.common.domain.poi.feedback.Review\\$0(.)*";
I advise you to escape . as well, . represent any character.
String pattern =
"/feedback/com\\.navteq\\.lcms\\.common\\.domain\\.poi\\.feedback\\.Review\\$0(.)*";

The ans by #Colin Hebert and edited by #theon is correct. The explanation is as follows. #azec-pdx
It is a regex as a string literal (within double quotes).
period (.) and dollar-sign ($) are special regex characters (metacharacters).
To make the regex engine interpret them as normal regex characters period(.) and dollar-sign ($), you need to prefix a single backslash to each. The single backslash ( itself a special regex character) quotes the character following it and thus escaping it.
Since the given regex is a string literal, another backslash is required to be prefixed to each to avoid confusion with the usual visible-ASCII escapes(character, string and Unicode escapes in string literals) and thus avoid compiler error.
Even if you use within a string literal any special regex construct that has been defined as an escape sequence, it needs to be prefixed with another backslash to avoid compiler error.For example, the special regex construct (an escape sequence) \b (word boundary) of regex would clash with \b(backspace) of the usual visible-ASCII escape(character escape). Thus another backslash is prefixed to avoid the clash and then \\b would be read by regex as word boundary.
To be always safe, all single backslash escapes (quotes) within string literals are prefixed with another backslash. For example, the string literal "\(hello\)" is illegal and leads to a compile-time error; in order to match the string (hello) the string literal "\\(hello\\)" must be used.
The last period (.)* is supposed to be interpreted as special regex character and thus it needs no quoting by a backslash, let alone prefixing a second one.