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how to split a string which contains of ( \n : , .)

so how can split this combination in android ?
Thanks in advance :)`
i'm trying like thatString lines[] = String.split("\\r?\\n", -1);
but how can split all data in one time

You can use Pattern for regex split
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of
Birth], joindate[Date of Joining]";
Pattern pattern = Pattern.compile("\\[.+\\]+?,?\\s*" );
String[] split = pattern.split(fields);
References: How to split this string using Java Regular Expressions

Regex split not working

I want split my string using regex.
String Str = " Dřevo5068Hlína5064Železo5064Obilí4895";
String reg = "(\\D+)(\\d+)(\\D+)(\\d+)(\\D+)(\\d+)(\\D+)(\\d+)";
if (Str.matches(reg)) {
String[] l = Str.split(reg);
System.out.println(Arrays.toString(l));
}
But, output is []. Where is problem?
Edit: I want split to:
Dřevo
5068
Hlína
5064
Železo
5064
Obilí
4895
Then I want get numbers from this String.

if your engine permits look-around, split using this pattern
(?<=\D)(?=\d)|(?<=\d)(?=\D)
Demo

Parsing String and Int in java

Im using java and I have a String that I would like to parse which contains the following
Logging v0.12.4
and would like to split it into a String containing
Logging
and an integer array containing
0.12.4
where
array[i][0] = 0
and
array[i][1] = 12
and so on. I have been stuck on this for a while now.

split your string on space to get Logging and v0.12.4
remove (substring) v from v0.12.4
split 0.12.4 on dot (use split("\\.") since dot is special in regex)
you can also parse each "0" "12" "4" to integer (Integer.parseInt can be helpful).

You can use a regex or just normal String splitting
String myString = "Logging v0.12.4";
String[] parts = myString.split(" v");
// now parts[0] will be "Logging" and
// parts[1] will be "0.12.4";
Then do the same for the version part:
String[] versionParts = parts[1].split("\\.");
// versionParts[0] will be "0"
// versionParts[1] will be "12"
// versionParts[2] will be "4"
You can "convert" these to integers by using Integer.parseInt(...)

Here ya go buddy, because I'm feeling generous today:
String string = "Logging v0.12.4";
Matcher matcher = Pattern.compile("(.+?)\\s+v(.*)").matcher(string);
if (matcher.matches()) {
String name = matcher.group(1);
int[] versions = Arrays.stream(matcher.group(2).split("\\.")).mapToInt(Integer::parseInt).toArray();
}

How to find the delimiter encountered in a string in JAVA

I have written simple program in Java which does manipulation of a given string.
The input string has some delimiters which are non-alphabets. I have used String Tokenizer to read and manipulate the individual words in a string.
Now I need to reconstruct this manipulated string with the same set of delimiters. Appreciate if any one can suggest me how to identify the delimiter.
In other words, this is what input is:
Text1 Delimiter1 Text2 Delimiter2 Text3 Delimiter3 Text4 Delimiter4
This is what my code does:
NewText1 NewText2 NewText3 NewText4
I made use of string tokenizer to identify the next token in this manner:
StringTokenizer st = new StringTokenizer(str, ", 0123456789(*&^%$##!-_)");
But now I would like to identify the delimiter that was encountered so that I can build my new string.
This is what I actually want:
NewText1 Delimiter1 NewText2 Delimiter2 NewText3 Delimiter3 NewText4 Delmiter4

You can proceed according to this:
String dels = "-, 0123456789(*&^%$##!_)";
String specs = "[" + dels + "]+";
String letts = "[^" + dels + "]+";
String text = "one, two - three! four";
String[] words = text.split( specs );
String[] delim = text.split( letts );
Note that in dels the hyphen must be up front. If you ever add [ or ] or ^ more care must be taken - check the javadoc in java.util.regex.Pattern.
There is no particular problem with composing the original string.
The disadvantage with StringTokenizer with a third argument is that it returns each delimiter as a separate token of length 1.

Splitting String using split method

I want split a string like this:
C:\Program\files\images\flower.jpg
but, using the following code:
String[] tokens = s.split("\\");
String image= tokens[4];
I obtain this error:
11-07 12:47:35.960: E/AndroidRuntime(6921): java.util.regex.PatternSyntaxException: Syntax error U_REGEX_BAD_ESCAPE_SEQUENCE near index 1:

try
String s="C:\\Program\\files\\images\\flower.jpg"
String[] tokens = s.split("\\\\");
In java(regex world) \ is a meta character. you should append with an extra \ or enclose it with \Q\E if you want to treat a meta character as a normal character.
below are some of the metacharacters
<([{\^-=$!|]})?*+.>
to treat any of the above listed characters as normal characters you either have to escape them with '\' or enclose them around \Q\E
like:
\\\\ or \\Q\\\\E

You need to split with \\\\, because the original string should have \\. Try it yourself with the following test case:
#Test
public void split(){
String s = "C:\\Program\\files\\images\\flower.jpg";
String[] tokens = s.split("\\\\");
String image= tokens[4];
assertEquals("flower.jpg",image);
}

There is 2 levels of interpreting the string, first the language parser makes it "\", and that's what the regex engine sees and it's invalid because it's an escape sequence without the character to escape.
So you need to use s.split("\\\\"), so that the regex engine sees \\, which in turn means a literal \.
If you are defining that string in a string literal, you must escape the backslashes there as well:
String s = "C:\\Program\\files\\images\\flower.jpg";

String[] tokens=s.split("\\\\");

Try this:
String s = "C:/Program/files/images/flower.jpg";
String[] tokens = s.split("/");
enter code hereString image= tokens[4];

Your original input text should be
C:\\Program\\files\\images\\flower.jpg
instead of
C:\Program\files\images\flower.jpg

This works,
public static void main(String[] args) {
String str = "C:\\Program\\files\\images\\flower.jpg";
str = str.replace("\\".toCharArray()[0], "/".toCharArray()[0]);
System.out.println(str);
String[] tokens = str.split("/");
System.out.println(tokens[4]);
}