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How do i check if string contains char sequence and backslash "\"?

I'm trying to get true in the following test. I have a string with the backslash, that for some reason doesn't recognized.
String s = "Good news\\ everyone!";
Boolean test = s.matches("(.*)news\\.");
System.out.println(test);
I've tried a lot of variants, but only one (.*)news(.*) works. But that actually means any characters after news, i need only with \.
How can i do that?

Group the elements at the end:(.*)news\\(.*)

You can use this instead :
Boolean test = s.matches("(.*)news\\\\(.*)");

Try something like:
Boolean test = s.matches(".*news\\\\.*");
Here .* means any number of characters followed by news, followed by double back slashes (escaped in a string) and then any number of characters after that (can be zero as well).
With your regex what it means is:
.* Any number of characters
news\\ - matches by "news\" (see one slash)
. followed by one character.
which doesn't satisfies for String in your program "Good news\ everyone!"

You are testing for an escaped occurrence of a literal dot: ".".
Refactor your pattern as follows (inferring the last part as you need it for a full match):
String s = "Good news\\ everyone!";
System.out.println(s.matches("(.*)news\\\\.*"));
Output
true
Explanation
The back-slash is used to escape characters and the back-slash itself in Java Strings
In Java Pattern representations, you need to double-escape your back-slashes for representing a literal back-slash ("\\\\"), as double-back-slashes are already used to represent special constructs (e.g. \\p{Punct}), or escape them (e.g. the literal dot \\.).
String.matches will attempt to match the whole String against your pattern, so you need the terminal part of the pattern I've added

you can try this :
String s = "Good news\\ everyone!";
Boolean test = s.matches("(.*)news\\\\(.*)");
System.out.println(test);

How to replace all regex string in a String object?

For a string str_in = "instance (\\w+\\s+){0,8}deleted"; how can I extract instance and deleted by using the replaceAll function?
I tried str_in = str_in.replaceAll("(\\w+\\s+){0,8}", ""); but it didn't work.

If you, as your question states, really want to use replaceAll() instead of the (in my opinion more suitable) replace(), you can use the \Q and \E markers to match the string literally:
String str_in = "instance (\\w+\\s+){0,8}deleted";
System.out.println(str_in.replaceAll("\\Q(\\w+\\s+){0,8}\\E", ""));
prints
instance deleted

You will need to escape the single characters so that they lose their regex nature:
str_in.replaceAll("\\(\\\\w\\+\\\\s\\+\\)\\{0,8\\}", "")
Each escaping backslash needs to be escaped for itself because of the string literal.

Do you mean that (\\w+\\s+){0,8} is literally in the string, and you want to remove it? You will need to escape each \ again in your replaceAll, so that they are interpreted literally, not as part of a regex, and also the ( and {.

Use str_in.replace("(\\w+\\s+){0,8}", "");
replace()
Replaces each substring of this string that matches the literal target
sequence with the specified literal replacement sequence. The
replacement proceeds from the beginning of the string to the end, for
example, replacing "aa" with "b" in the string "aaa" will result in
"ba" rather than "ab"
replaceAll()
Replaces each substring of this string that matches the given regular expression with the given replacement

Java and string split

split this String using function split. Here is my code:
String data= "data^data";
String[] spli = data.split("^");
When I try to do that in spli contain only one string. It seems like java dont see "^" in splitting. Do anyone know how can I split this string by letter "^"?
EDIT
SOLVED :P

This is because String.split takes a regular expression, not a literal string. You have to escape the ^ as it has a different meaning in regex (anchor at the start of a string). So the split would actually be done before the first character, giving you the complete string back unaltered.
You escape a regular expression metacharacter with \, which has to be \\ in Java strings, so
data.split("\\^")
should work.

You need to escape it because it takes reg-ex
\\^

Special characters like ^ need to be escaped with \

This does not work because .split() expects its argument to be a regex. "^" has a special meaing in regex and so does not work as you expect. To get it to work, you need to escape it. Use \\^.

The reason is that split's parameter is a regular expression, so "^" means the beginning of a line. So you need to escape to ASCII-^: use the parameter "\\^".

How do I replace all "[", "]" and double quotes in Java

I'm am having difficulty using the replaceAll method to replace square brackets and double quotes. Any ideas?
Edit:
So far I've tried:
replace("\[", "some_thing") // returns illegal escape character
replace("[[", "some_thing") // returns Unclosed character class
replace("^[", "some_thing") // returns Unclosed character class

Don't use replaceAll, use replace. The former uses regular expressions, and [] are special characters within a regex.
String replaced = input.replace("]", ""); //etc
The double quote is special in Java so you need to escape it with a single backslash ("\"").
If you want to use a regex you need to escape those characters and put them in a character class. A character class is surrounded by [] and escaping a character is done by preceding it with a backslash \. However, because a backslash is also special in Java, it also needs to be escaped, and so to give the regex engine a backslash you have to use two backslashes (\\[).
In the end it should look like this (if you were to use regex):
String replaced = input.replaceAll("[\\[\\]\"]", "");

The replaceAll method is operating against Regular Expressions. You're probably just wanting to use the "replace" method, which despite its name, does replace all occurrences.
Looking at your edit, you probably want:
someString
.replace("[", "replacement")
.replace("]", "replacement")
.replace("\"", "replacement")
or, use an appropriate regular expression, the approach I'd actually recommend if you're willing to learn regular expressions (see Mark Peter's answer for a working example).

replaceAll() takes a regex so you have to escape special characters. If you don't want all the fancy regex, use replace().
String s = "[h\"i]";
System.out.println( s.replace("[","").replace("]","").replace("\"","") );
With double quotes, you have to escape them like so: "\""

In java:
String resultString = subjectString.replaceAll("[\\[\\]\"]", "");
this will replace []" with nothing.

Alternatively, if you wished to replace ", [ and ] with different characters (instead of replacing all with empty String) you could use the replaceEachRepeatedly() method in the StringUtils class from Commons Lang.
For example,
String input = "abc\"de[fg]hi\"";
String replaced = StringUtils.replaceEachRepeatedly(input,
new String[]{"[","]","\""},
new String[]{"<open_bracket>","<close_bracket>","<double_quote>"});
System.out.println(replaced);
Prints the following:
abc<double_quote>de<open_bracket>fg<close_bracket>hi<double_quote>

How to match \Q and \E in Java regex?

I want to match \Q and \E in a Java regex.
I am writing a program which will compute the length of the string, matching to the pattern (this program assumes that there is no any quantifier in regex except {some number}, that's why the length of the string is uniquely defined) and I want at first delete all expressions like \Qsome text\E.
But regex like this:
"\\Q\\Q\\E\\Q\\E\\E"
obviously doesn't work.

Use Pattern.quote(...):
String s = "\\Q\\Q\\E\\Q\\E\\E";
String escaped = Pattern.quote(s);

Just escape the backslashes. The sequence \\\\ matches a literal backslash, so to match a literal \Q:
"\\\\Q"
and to match a literal \E:
"\\\\E"
You can make it more readable for a maintainer by making it obvious that each sequence matches a single character using [...] as in:
"[\\\\][Q]"