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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;
}
}

Regex do not change anything

Please have a look at the below code
public class PuctuationRemover {
public PuctuationRemover()
{
String str = ":The red; third.fox is hungry!!! but, is he angry? doesn't! (yeah!). Call 911! system. can't access it! what the , hell . is this. people of my country, really? 123465 can^be,found.... OK . you got it? ";
String str2 = str.replaceAll("[^a-zA-Z'\\s]+", str);
System.out.println(str2);
}
public static void main(String[]args)
{
new PuctuationRemover();
}
}
The expected output is
The red thirdfox is hungry but is he angry doesn't yeah Call system can't access it what the hell is this people of my country really canbefound OK you got it
The output I get is
:The red; third.fox is hungry!!! but, is he angry? doesn't! (yeah!). Call 911! system. can't access it! what the , hell . is this. people of my country, ..............
The original working regex is here.
What has gone wrong here?

If you need to remove the punctuation, supply an empty string as the second argument instead of the original string itself. The second argument to replaceAll is not the original string, but what to replace the match with. Change
String str2 = str.replaceAll("[^a-zA-Z'\\s]+", str);
with
String str2 = str.replaceAll("[^a-zA-Z'\\s]+", "");

Use String str2 = str.replaceAll("[^a-zA-Z'\\s]+", "");
You are doing String str2 = str.replaceAll("[^a-zA-Z'\\s]+", str); .You are replacing the whole original String.
Please see this javadoc for more on String replaceAll.
Here in public String replaceAll(String regex, String replacement) second argument, i.e replacement is the string to be substituted for each match.
Note: Its better to catch PatternSyntaxException .

You're replacing your old string with your old string, which means you keep the same string even after replacing it. What you need to do is replace all of the characters specified in the regex with a no-character "".
EDIT: Darn, you people are fast typers! =P

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 .");
}

How to insert backslash into my string in java?

I have string, and I want to replace one of its character with backslash \
I tried the following, but no luck.
engData.replace("'t", "\\'t")
and
engData = engData.replace("'t", String.copyValueOf(new char[]{'\\', 't'}));
INPUT : "can't"
EXPECTED OUTPUT : "can\'t"
Any idea how to do this?

Try this..
String s = "can't";
s = s.replaceAll("'","\\\\'");
System.out.println(s);
out put :
can\'t
This will replace every ' occurences with \' in your string.

Try like this
engData.replace("'", "\\\'");
INPUT : can't
EXPECTED OUTPUT : can\'t

String is immutable in Java. You need to assign back the modified string to itself.
engData = engData.replace("'t", "\\'t"); // assign the modified string back.

This is possible with regex:
engData = engData.replaceAll("('t)","\\\\$1");
The ( and ) specify a group. The 't will match any string containing 't. Finally, the second part replaced such a string with a backslash character: \\\\ (four because this), and the first group: $1. Thus you are replacing any substring 't with \'t
The same thing is possible without regex, what you tried (see this for output):
engData = engData.replace("'t","\\'t"); //note the assignment; Strings are immutable
See String.replace(CharSequence, CharSequence)

For String instances you can use, str.replaceAll() will return a new String with the changes requested:
String str = "./";
String s_modified = s.replaceAll("\\./", "");

The following works for me:
class Foobar {
public static void main(String[] args) {
System.err.println("asd\\'t".replaceAll("\\'t", "\\\'t"));
}
}

String manipulation in java using replaceAll()

I have string of the following form:
भन्‍‌ने [-0.4531954191090929, 0.7931147934270654, -0.3875088408737827, -0.09427394940704822, 0.10065554475134718, -0.22044284832864797, 0.3532556916833505, -1.8256229909222224, 0.8036832111904731, 0.3395868096795993]
Whereever [ or ] or , char are present , I just want to remove them and i want each of the word and float separated by a space. It is follows:
भन्‍‌ने -0.4531954191090929 0.7931147934270654 -0.3875088408737827 -0.09427394940704822 0.10065554475134718 -0.22044284832864797 0.3532556916833505 -1.8256229909222224 0.8036832111904731 0.3395868096795993
I am representing each of these string as line. i did following:
line.replaceAll("([|]|,)$", " ");
But it didn't work for me. There was nothing change in the input line. Any help is really appreciated.

Strings are immutable. Try
line = line.replaceAll("([|]|,)$", " ");
Or to be a bit more verbose, but avoiding regular expressions:
char subst = ' ';
line = line.replace('[', subst).replace(']', subst).replace(',', subst);

In Java, strings are immutable, meaning that the contents of a string never change. So, calling
line.replaceAll("([|]|,)$", " ");
won't change the contents of line, but will return a new string. You need to assign the result of the method call to a variable. For instance, if you don't care about the original line, you can write
line = line.replaceAll("([|]|,)$", " ");
to get the effect you originally expected.

[ and ] are special characters in a regular expression. replaceAll is expected a regular expression as its first input, so you have to escape them.
String result = line.replaceAll("[\\[\\],]", " ");
Cannot tell what you were trying to do with your original regex, why you had the $ there etc, would need to understand what you were expecting the things you put there to do.

Try
line = "asdf [foo, bar, baz]".replaceAll("(\\[|\\]|,)", "");
The regex syntax uses [] to define groups like [a-z] so you have to mask them.