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How to find and replace '[' in a String?

i am having a String "['MET', 'MISSED']". Here i want to replace "[ to [ and ]" to ]. I have used the escape sequences in my String like
wkJsonStr.replaceAll("\"\\[","[");
and
wkJsonStr.replaceAll("\\]\"","]");
but none of the above worked. In 'watch' i edited like
wkJsonStr.replaceAll("\"[","[");
and it worked. But in my Android Studio Editor this Expression is not allowed. I am getting "Unclosed character class".
I am expecting my String after replacing to be like ['MET', 'MISSED']. I want to remove the first and last quotation alone and i would like to achieve it by replaceAll method.

Remember that string are immutable in java...
just calling the replace method will take no effect, you need to assign the return value, otherwise will get lost.
Example:
public static void main(String[] args) {
String wkJsonStr = "\"['MET', 'MISSED']\"";
System.out.println(wkJsonStr);
wkJsonStr = wkJsonStr.replaceAll("\"\\[", "[").replaceAll("\\]\"", "]");
System.out.println(wkJsonStr);
}
this will print.
"['MET', 'MISSED']"
['MET', 'MISSED']

You can use replace instead of replaceAll, but if you really need to use replaceAll then you can do:
wkJsonStr.replaceAll("\"\\Q[\\E","[").replaceAll("\\Q]\\E\"","]")
Here,
\" literally denotes "
The part between \\Q and \\E is literally treated., i.e,
\\Q[\\E denotes [
\\Q]\\E denotes ]

I agree with #Wiktor Stribiżew's comment. And you should declare your String like this:
String str = "\"['MET', 'MISSED']\""; // like this your editor will not give an error
str = str.replace("\"[","[");
str = str.replace("]\"","]");
textview.setText(str);

I finally realized that the isssue is not with the Regex character. Its the Variable which i am using. when i am replacing one by one it is not forwarding the result to the next step. So i have assigned the value to a variable and now its working.
String firstResult = wkJsonStr.replaceAll("\"\\[","[");
and
String result = firstResult.replaceAll("\\]\"","]");
is working for me. Or This step will do a trick.
String result= wkJsonStr.replaceAll("\"\\[","[").replaceAll("\\]\"","]");

How can i put a string in one line

i'm doing some conversion, from Hex to Ascii, when i convert the string, i got the following example:
F23C040100C1
100D200000000000
0000
I know that the string is coming like this, because of the base 16, but i want too put it in just one line, like this:
F23C040100C1100D2000000000000000
How can i do that?
I have tried:
mensagem.replaceAll("\r\n", " ");

There are multiple problems you could be running into, so I'll cover all them in this answer.
First, any methods on String which appear to modify it actually return a new instance of String. That means if you do this:
String something = "Hello";
something.replaceAll("l", "");
System.out.println(something); //"Hello"
You'll want to do
something = something.replaceAll("l", "");
Or in your case
mensagem = mensagem.replaceAll("\r\n", " ");
Secondly, there might not be any \r in the newline, but there is a \n, or vice versa. Because of that, you want to say that
if \r exists, remove it. if \n exists, also remove it
You can do so like this:
mensagem = mensagem.replaceAll("\r*\n*", " ");
The * operator in a regular expression says to match zero or more of the preceding symbol.

mensagem.replaceAll("\r\n|\r|\n", "");

Java String.replace/replaceAll not working

So, I'm trying to parse a String input in Java that contains (opening) square brackets. I have str.replace("\\[", ""), but this does absolutely nothing. I've tried replaceAll also, with more than one different regex, but the output is always unchanged. Part of me wonders if this is possibly caused by the fact that all my back-slash characters appear as yen symbols (ever since I added Japanese to my languages), but it's been that way for over a year and hasn't caused me any issues like this before.
Any idea what I might be doing wrong here?

Strings are immutable in Java. Make sure you re-assign the return value to the same String variable:
str = str.replaceAll("\\[", "");
For the normal replace method, you don't need to escape the bracket:
str = str.replace("[", "");

public String replaceAll(String regex, String replacement)
As shown in the code above, replaceAll method expects first argument as regular expression and hence you need to escape characters like "(", ")" etc (with "\") if these exists in your replacement text which is to be replaced out of the string. For example :
String oldString = "This is (stringTobeReplaced) with brackets.";
String newString = oldString.replaceAll("\\(stringTobeReplaced\\)", "");
System.out.println(newString); // will output "This is with brackets."
Another way of doing this is to use Pattern.quote("str") :
String newString = oldString.replaceAll(Pattern.quote("(stringTobeReplaced)"), "");
This will consider the string as literal to be replaced.

As always, the problem is not that "xxx doesn't work", it is that you don't know how to use it.
First things first:
a String is immutable; if you read the javadoc of .replace() and .replaceAll(), you will see that both specify that a new String instance is returned;
replace() accepts a string literal as its first argument, not a regex literal.
Which means that you probably meant to do:
str = str.replace("[", "");
If you only ever do:
str.replace("[", "");
then the new instance will be created but you ignore it...
In addition, and this is a common trap with String (the other being that .matches() is misnamed), in spite of their respective names, .replace() does replace all occurrences of its first argument with its second argument; the only difference is that .replaceAll() accepts a regex as a first argument, and a "regex aware" expression as its second argument; for more details, see the javadoc of Matcher's .replaceAll().

For it to work it has to be inside a method.
for example:
public class AnyClass {
String str = "gtrg4\r\n" + "grtgy\r\n" + "grtht\r\n" + "htrjt\r\n" + "jtyjr\r\n" + "kytht";
public String getStringModified() {
str.replaceAll("\r\n", "");
return str;
}
}

having trouble with arrays and maybe split

String realstring = "&&&.&&&&";
Double value = 555.55555;
String[] arraystring = realstring.split(".");
String stringvalue = String.valueof(value);
String [] valuearrayed = stringvalue.split(".");
System.out.println(arraystring[0]);
Sorry if it looks bad. Rewrote on my phone. I keep getting ArrayIndexOutOfBoundsException: 0 at the System.out.println. I have looked and can't figure it out. Thanks for the help.

split() takes a regexp as argument, not a literal string. You have to escape the dot:
string.split("\\.");
or
string.split(Pattern.quote("."));
Or you could also simply use indexOf('.') and substring() to get the two parts of your string.
And if the goal is to get the integer part of a double, you could also simply use
long truncated = (long) doubleValue;

split uses regex as parameter and in regex . means "any character except line separators", so you could expect that "a.bc".split(".") would create array of empty strings like ["","","","",""]. Only reason it is not happening is because (from split javadoc)
This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.
so because all strings are empty you get empty array (and that is because you see ArrayIndexOutOfBoundsException).
To turn off removal mechanism you would have to use split(regex, limit) version with negative limit.
To split on . literal you need to escape it with \. (which in Java needs to be written as "\\." because \ is also Strings metacharacter) or [.] or other regex mechanism.

Dot (.) is a special character so you need to escape it.
String realstring = "&&&.&&&&";
String[] partsOfString = realstring.split("\\.");
String part1 = partsOfString[0];
String part2 = partsOfString[1];
System.out.println(part1);
this will print expected result of
&&&
Its also handy to test if given string contains this character. You can do this by doing :
if (string.contains(".")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain .");
}

Regular expression for string with apostrophes

I'm trying to build regex which will filter form string all non-alphabetical characters, and if any string contains single quotes then I want to keep it as an exception to the rule.
So for example when I enter
car's34
as a result I want to get
car's
when I enter
*&* Lisa's car 0)*
I want to get
Lisa's
at the moment I use this:
string.replaceAll("[^A-Za-z]", "")
however, it gives me only alphabets, and removed the desired single quotas.

This will also remove apostrophes that are not "part if words":
string = string.replaceAll("[^A-Za-z' ]+|(?<=^|\\W)'|'(?=\\W|$)", "")
.replaceAll(" +", " ").trim();
This first simply adds an apostrophe to the list of chars you want to keep, but uses look arounds to find apostrophes not within words, so
I'm a ' 123 & 'test'
would become
I'm a test
Note how the solitary apostrophe was removed, as well as the apostrophes wrapping test, but I'm was preserved.
The subsequent replaceAll() is to replace multiple spaces with a single space, which will result if there's a solitary apostrophe in the input. A further call to trim() was added in case it occurs at the end of the input.
Here's a test:
String string = "I'm a ' 123 & 'test'";
string = string.replaceAll("[^A-Za-z' ]+|(?<=^|\\W)'|'(?=\\W|$)", "").replaceAll(" +", " ").trim();
System.out.println(string);
Output:
I'm a test

Isn't this working ?
[^A-Za-z']

The obvious solution would be:
string.replaceAll("[^A-Za-z']", "")
I suspect you want something more.

You can try the regular expression:
[^\p{L}' ]
\p{L} denote the category of Unicode letters.
In ahother hand, you need to use a constant of Pattern for avoid recompiled the expression every time, something like that:
private static final Pattern REGEX_PATTERN =
Pattern.compile("[^\\p{L}' ]");
public static void main(String[] args) {
String input = "*&* Lisa's car 0)*";
System.out.println(
REGEX_PATTERN.matcher(input).replaceAll("")
); // prints " Lisa's car "
}

#Bohemian has a good idea but word boundaries are called for instead of lookaround:
string.replaceAll("([^A-Za-z']|\B'|'\B)+", " ");