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Escaping * character in java

I am trying to execute the following operation on a String.
if (combatLog.contains("//*name//*")) {
combatLog.replaceAll("//*name//*",glad.target.name);
}
The slashes are my attempt to escape the *, as it doesn't work without them. I have also tried one slash, and slashes on contains or replaceAll individually. Thanks

replaceAll() (counter-intuitively) takes a regex, not a string.
To escape a character for a regex, you need a double-backslash (doubled to escape the backslash from the string literal).
However, you don't want a regex. You should simply call replace() instead, which won't need any escaping.

You're using forward slashes. The backslash is the escape character. Furthermore, unless the string is being used for regex or something similar, you need not escape the *, or the / if thats what you're trying to escape.
If combatLog is a String, its contains method checks for a sequence of characters only. If you're looking for *name* in the string, you only need call combatLog.contains("*name*").

You are using forward slashes use the backslash: \ to escape characters
[edit]
also as slaks said you need to use replace() which accepts a string as input rather than a regex.

Don't forget about immutability of strings, and reassign the newly created string. Also, if your if block doesn't contain any more code, you don't need the if check at all.
You have 3 options:
if (combatLog.contains("*name*")) { // don't escape in contains()
combatLog = combatLog.replaceAll("\\*name\\*", replacement);// correct escape
}
// another regex based solution
if (combatLog.contains("*name*")) {
combatLog = combatLog.replaceAll("[*]name[*]", replacement);// character class
}
or without a regex
if (combatLog.contains("*name*")) {
combatLog = combatLog.replace("*name*", replacement);// literal string
}

String.replaceAll(...) of Java not working for \\ and \

I want to convert the directory path from:
C:\Users\Host\Desktop\picture.jpg
to
C:\\Users\\Host\\Desktop\\picture.jpg
I am using replaceAll() function and other replace functions but they do not work.
How can I do this?
I have printed the statement , it gives me the one which i wanted ie
C:\Users\Host\Desktop\picture.jpg
but now when i pass this variable to open the file, i get this exception why?
java.io.FileNotFoundException: C:\Users\Host\Desktop\picture.jpg

EDIT: Changed from replaceAll to replace - you don't need a regex here, so don't use one. (It was a really poor design decision on the part of the Java API team, IMO.)
My guess (as you haven't provided enough information) is that you're doing something like:
text.replace("\\", "\\\\");
Strings are immutable in Java, so you need to use the return value, e.g.
String newText = oldText.replace("\\", "\\\\");
If that doesn't answer your question, please provide more information.
(I'd also suggest that usually you shouldn't be doing this yourself anyway - if this is to include the information in something like a JSON response, I'd expect the wider library to perform escaping for you.)
Note that the doubling is required as \ is an escape character for Java string (and character) literals. Note that as replace doesn't treat the inputs as regular expression patterns, there's no need to perform further doubling, unlike replaceAll.
EDIT: You're now getting a FileNotFoundException because there isn't a filename with double backslashes in - what made you think there was? If you want it as a valid filename, why are you doubling the backslashes?

You have to use :
String t2 = t1.replaceAll("\\\\", "\\\\\\\\");
or (without pattern) :
String t2 = t1.replace("\\", "\\\\");
Each "\" has to be preceeded by an other "\". But it's also true for the preceeding "\" so you have to write four backslashes each time you want one in regex.

In strings \ is bydefault used as escape character therefore in order to select "\" in a string you have to use "\" and for "\" (i.e blackslack two times) use "\\". This will solve your problem and thos will also apply to other symbols also like "

Two explanations:
1. Replace double backslashes to one (not what you asked)
You have to escape the backslash by backslashes. Like this:
String newPath = oldPath.replaceAll("\\\\\\\\", "\\");
The first parameter needs to be escaped twice. Once for the Java Compiler and once because you use regular expressions. So you want to replace two backslashes by one. So, since we have to escape a backslash add one backslash. Now you have \\. This will be compiled to \. BUT!! you have to escape the backslash once again because the first parameter of the replaceAll method uses regular expressions. So to escape it, add a backslash, but that backslash needs to be escaped, so we get \\\\. These for backslashes represents one backslash in the regex. But you want to replace the double backslash to one. So use 8 backslashes.
The second parameter of the replaceAll method isn't using regular expressions, but it has to be escaped as well. So, you need to escape it once for the Java Compiler and once for the replace method: \\\\. This is compiled to two backslashes, which are being interpreted as 1 backslash in the replaceAll method.
2. Replace single backslash to a pair of backslashes (what you asked)
String newPath = oldPath.replaceAll("\\\\", "\\\\\\\\");
Same logic as above.
3. Use replace() instead of replaceAll().
String newPath = oldPath.replace("\\", "\\\\");
The difference is that the replace() method doesn't use regular expressions, so you don't have to escape every backslash twice for the first parameter.
Hopefully, I explained well...
-- Edit: Fixed error, as pointed out by xehpuk --

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>

Split string with dot as delimiter

I am wondering if I am going about splitting a string on a . the right way? My code is:
String[] fn = filename.split(".");
return fn[0];
I only need the first part of the string, that's why I return the first item. I ask because I noticed in the API that . means any character, so now I'm stuck.

split() accepts a regular expression, so you need to escape . to not consider it as a regex meta character. Here's an example :
String[] fn = filename.split("\\.");
return fn[0];

I see only solutions here but no full explanation of the problem so I decided to post this answer
Problem
You need to know few things about text.split(delim). split method:
accepts as argument regular expression (regex) which describes delimiter on which we want to split,
if delim exists at end of text like in a,b,c,, (where delimiter is ,) split at first will create array like ["a" "b" "c" "" ""] but since in most cases we don't really need these trailing empty strings it also removes them automatically for us. So it creates another array without these trailing empty strings and returns it.
You also need to know that dot . is special character in regex. It represents any character (except line separators but this can be changed with Pattern.DOTALL flag).
So for string like "abc" if we split on "." split method will
create array like ["" "" "" ""],
but since this array contains only empty strings and they all are trailing they will be removed (like shown in previous second point)
which means we will get as result empty array [] (with no elements, not even empty string), so we can't use fn[0] because there is no index 0.
Solution
To solve this problem you simply need to create regex which will represents dot. To do so we need to escape that .. There are few ways to do it, but simplest is probably by using \ (which in String needs to be written as "\\" because \ is also special there and requires another \ to be escaped).
So solution to your problem may look like
String[] fn = filename.split("\\.");
Bonus
You can also use other ways to escape that dot like
using character class split("[.]")
wrapping it in quote split("\\Q.\\E")
using proper Pattern instance with Pattern.LITERAL flag
or simply use split(Pattern.quote(".")) and let regex do escaping for you.

Split uses regular expressions, where '.' is a special character meaning anything. You need to escape it if you actually want it to match the '.' character:
String[] fn = filename.split("\\.");
(one '\' to escape the '.' in the regular expression, and the other to escape the first one in the Java string)
Also I wouldn't suggest returning fn[0] since if you have a file named something.blabla.txt, which is a valid name you won't be returning the actual file name. Instead I think it's better if you use:
int idx = filename.lastIndexOf('.');
return filename.subString(0, idx);

the String#split(String) method uses regular expressions.
In regular expressions, the "." character means "any character".
You can avoid this behavior by either escaping the "."
filename.split("\\.");
or telling the split method to split at at a character class:
filename.split("[.]");
Character classes are collections of characters. You could write
filename.split("[-.;ld7]");
and filename would be split at every "-", ".", ";", "l", "d" or "7". Inside character classes, the "." is not a special character ("metacharacter").

As DOT( . ) is considered as a special character and split method of String expects a regular expression you need to do like this -
String[] fn = filename.split("\\.");
return fn[0];
In java the special characters need to be escaped with a "\" but since "\" is also a special character in Java, you need to escape it again with another "\" !

String str="1.2.3";
String[] cats = str.split(Pattern.quote("."));

Wouldn't it be more efficient to use
filename.substring(0, filename.indexOf("."))
if you only want what's up to the first dot?

Usually its NOT a good idea to unmask it by hand. There is a method in the Pattern class for this task:
java.util.regex
static String quote(String s)

The split must be taking regex as a an argument... Simply change "." to "\\."

The solution that worked for me is the following
String[] fn = filename.split("[.]");

Note: Further care should be taken with this snippet, even after the dot is escaped!
If filename is just the string ".", then fn will still end up to be of 0 length and fn[0] will still throw an exception!
This is, because if the pattern matches at least once, then split will discard all trailing empty strings (thus also the one before the dot!) from the array, leaving an empty array to be returned.

Using ApacheCommons it's simplest:
File file = ...
FilenameUtils.getBaseName(file.getName());
Note, it also extracts a filename from full path.

split takes a regex as argument. So you should pass "\." instead of "." because "." is a metacharacter in regex.