Develop Reference

How to check and replace a sequence of characters in a String?

Here what the program is expectiong as the output:
if originalString = "CATCATICATAMCATCATGREATCATCAT";
Output should be "I AM GREAT".
The code must find the sequence of characters (CAT in this case), and remove them. Plus, the resulting String must have spaces in between words.
String origString = remixString.replace("CAT", "");
I figured out I have to use String.replace, But what could be the logic for finding out if its not cat and producing the resulting string with spaces in between the words.

First off, you probably want to use the replaceAll method instead, to make sure you replace all occurrences of "CAT" within the String. Then, you want to introduce spaces, so instead of an empty String, replace "CAT" with " " (space).
As pointed out by the comment below, there might be multiple spaces between words - so we use a regular expression to replace multiple instances of "CAT" with a single space. The '+' symbol means "one or more",.
Finally, trim the String to get rid of leading and trailing white space.
remixString.replaceAll("(CAT)+", " ").trim()

You can use replaceAll which accepts a regular expression:
String remixString = "CATCATICATAMCATCATGREATCATCAT";
String origString = remixString.replaceAll("(CAT)+", " ").trim();
Note: the naming of replace and replaceAll is very confusing. They both replace all instances of the matching string; the difference is that replace takes a literal text as an argument, while replaceAll takes a regular expression.

Maybe this will help
String result = remixString.replaceAll("(CAT){1,}", " ");

how to convert one line containing several sentences into lines according to dot(.) [duplicate]

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.

how to ignore newlines for split function

I am splitting the string using ^ char. The String which I am reading, is coming from some external source. This string contains some \n characters.
The string may look like:
Hi hello^There\nhow are\nyou doing^9987678867abc^popup
when I am splitting like below, why the array length is coming as 2 instead of 4:
String[] st = msg[0].split("^");
st.length //giving "2" instead of "4"
It look like, split is ignoring after \n.
How can I fix it without replacing \n to some other character.

the string parameter for split is interpreted as regular expression.
So you have to escape the char and use:
st.split("\\^")
see this answer for more details

Escape the ^ character. Use msg[0].split("\\^") instead.
String.split considers its argument as regular expression. And as ^ has a special meaning when it comes to regular expressions, you need to escape it to use its literal representation.

If you want to split by ^ only, then
String[] st = msg[0].split("\\^");
If I read your question correctly, you want to split by ^ and \n characters, so this would suffice.
String[] st = msg[0].split("[\\^\\\\n]");
This considers that \n literally exists as 2 characters in a string.

"^" it's know as regular expression by the JDK.
To avoid this confusion you need to modify the code as below
old code = msg[0].split("^")
new code = msg[0].split("\\^")

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 "\\^".

String replace a Backslash

How can I do a string replace of a back slash.
Input Source String:
sSource = "http://www.example.com\/value";
In the above String I want to replace "\/" with a "/";
Expected ouput after replace:
sSource = "http://www.example.com/value";
I get the Source String from a third party, therefore I have control over the format of the String.
This is what I have tried
Trial 1:
sSource.replaceAll("\\", "/");
Exception
Unexpected internal error near index 1
\
Trial 2:
sSource.replaceAll("\\/", "/");
No Exception, but does not do the required replace. Does not do anything.
Trial 3:
sVideoURL.replace("\\", "/");
No Exception, but does not do the required replace. Does not do anything.

sSource = sSource.replace("\\/", "/");
String is immutable - each method you invoke on it does not change its state. It returns a new instance holding the new state instead. So you have to assign the new value to a variable (it can be the same variable)
replaceAll(..) uses regex. You don't need that.

Try replaceAll("\\\\", "") or replaceAll("\\\\/", "/").
The problem here is that a backslash is (1) an escape chararacter in Java string literals, and (2) an escape character in regular expressions – each of this uses need doubling the character, in effect needing 4 \ in row.
Of course, as Bozho said, you need to do something with the result (assign it to some variable) and not throw it away. And in this case the non-regex variant is better.

Try
sSource = sSource.replaceAll("\\\\", "");
Edit : Ok even in stackoverflow there is backslash escape... You need to have four backslashes in your replaceAll first String argument...
The reason of this is because backslash is considered as an escape character for special characters (like \n for instance).
Moreover replaceAll first arg is a regular expression that also use backslash as escape sequence.
So for the regular expression you need to pass 2 backslash. To pass those two backslashes by a java String to the replaceAll, you also need to escape both backslashes.
That drives you to have four backslashes for your expression! That's the beauty of regex in java ;)

s.replaceAll ("\\\\", "");
You need to mask a backslash in your source, and for regex, you need to mask it again, so for every backslash you need two, which ends in 4.
But
s = "http://www.example.com\\/value";
needs two backslashes in source as well.

This will replace backslashes with forward slashes in the string:
source = source.replace('\\','/');

you have to do
sSource.replaceAll("\\\\/", "/");
because the backshlash should be escaped twice one for string in source one in regular expression

To Replace backslash at particular location:
if ((stringValue.contains("\\"))&&(stringValue.indexOf("\\", location-1)==(location-1))) {
stringValue=stringValue.substring(0,location-1);
}

sSource = StringUtils.replace(sSource, "\\/", "/")