Develop Reference

Separate string by whitespace, but keep newlines in split array

I'm trying to split a string in Java, but keep the newline characters as elements in the array.
For example, with input: "Hello \n\n\nworld!"
I want the output to be: ["Hello", "\n", "\n", "\n", "world", "!"]
The regex I have in place right now is this:
String[] parsed = input.split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
This gets me the punctuation separation I want, but its output looks like this:["Hello", "\n\n\nworld", "!"]
Is there a way to unclump the newlines in Java?

You could first replace all \n with \n (newline and a space) and then do a simple split on the space character.
String input = "Hello \n\n\nworld!";
String replacement = input.replace("\n", "\n ");
String[] result = replacement.split(" ");
input: "Hello \n\n\nworld!"
replacement: "Hello \n \n \n world!"
result: ["Hello", "\n", "\n", "\n", "world!"]
Note: my example does not handle the final exclamation mark - but it seems you already know how to handle that.

The trick is to add whitespace after each "\n" and then apply your regex.
String line = "Hello \n\n\nworld!";
line = line.replaceAll("\n", "\n "); // here we replace all "\n" to "\n "
String[] items = line.split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
or shorter version:
String line = "Hello \n\n\nworld!";
String[] items = line.replaceAll("\n", "\n ").split(" +|(?=\\p{Punct})|(?<=\\p{Punct})");
So, in this context the result is: ["Hello", "\n", "\n", "\n", "world", "!"]

Using the find method makes things easier:
String str = "Hello \n\n\nworld!";
List<String> myList = new ArrayList<String>();
Pattern pat = Pattern.compile("\\w+|\\H");
Matcher m = pat.matcher(str);
while (m.find()) {
myList.add(m.group(0));
}
If you use Java 7, change \\H to [\\S\\n].
Note that using this approach, you obtain a pattern easier to write and to edit since you don't need to use lookarounds.

How to ignore white spaces in Java [duplicate]

This question already has answers here:
Remove all occurrences of char from string
(13 answers)
Closed 5 years ago.
I was wondering how to ignore spaces in Java. This program allows you to enter your first, middle and surname which then outputs your initials. I'm now trying to make it ignore any white spaces. Thanks in advance!
String fullName;
char firstName;
char secondName;
char surname;
int space1;
int space2;
System.out.println("Please enter your first name, your second name and your surname: ");
fullName = kybd.nextLine();
firstName = fullName.charAt(0);
space1 = fullName.indexOf(" ");
secondName = fullName.charAt(space1 + 1);
space2 = fullName.lastIndexOf(" ");
surname = fullName.charAt(space2 + 1);
System.out.println("Initials: " + firstName + secondName + surname);

Explanation
You can implicitly ignore them by just removing them from your input text.
Therefore replace all occurrences with "" (empty text):
fullName = fullName.replaceAll(" ", "");
After that call fullName won't contain a whitespace anymore.
However you'll then get a problem with your logic as you split on whitespaces.
Solution
An alternative could be to first trim the text (removing leading and trailing whitespaces). Then do your split and after that you can remove all other whitespaces:
fullName = kybd.nextLine();
// Remove leading and trailing whitespaces
fullName = fullName.trim();
// Bounds
firstSpace = fullName.indexOf(" ");
lastSpace = fullName.lastIndexOf(" ");
// Extract names
String fullFirstName = fullName.substring(0, firstSpace);
String fullSecondName = fullName.substring(firstSpace + 1, lastSpace);
String fullSurname = fullName.substring(lastSpace + 1);
// Trim everything
fullFirstName = fullFirstName.trim(); // Not needed
fullSecondName = fullSecondName.trim();
fullSurname = fullSurname.trim();
// Get initials
firstName = fullFirstName.charAt(0);
secondName = fullSecondName.charAt(0);
surname = fullSurname.charAt(0);
Example
Let's take a look at an example input (_ stands for whitespace):
__John___Richard_Doe_____
We will first trim fullName and thus get:
John___Richard_Doe
Now we identify the first and the last whitespace and split on them:
First name: John
Second name: ___Richard
Surname: _Doe
Last we also trim everything and get:
First name: John
Second name: Richard
Surname: Doe
With charAt(0) we access the initials:
First name: J
Second name: R
Surname: D
More dynamic
Another more dynamic approach would be to merge all successive whitespaces into a single whitespace. Therefore you would need to traverse the text from left to right and start recording once you see a whitespace, end recording if visiting a non-whitespace character, then replace that section by a single whitespace.
Our example then is:
_John_Richard_Doe_
And after an additional trim you can use your regular approach again:
John_Richard_Doe
Or you can use split(" ") and then reject every empty String:
Iterator<String> elements = Pattern.compile(" ").splitAsStream(fullName)
.filter(e -> !e.isEmpty()) // Reject empty elements
.collect(Collectors.toList()) // Collect to list
.iterator() // Iterator
firstName = elements.next().charAt(0);
secondName = elements.next().charAt(0);
surname = elements.next().charAt(0);
Using the example again the Stream first consists of
"", "", "John", "", "", "Richard", "Doe", "", "", "", "", ""
after the filtering it's
"John", "Richard", "Doe"
Minus Sign
As you said you also want
Richard Jack Smith-Adams
output RJS-A, you can simply split on - after splitting on the whitespace.
Pattern spacePatt = Pattern.compile(" ");
Pattern minusPatt = Pattern.compile("-");
String result = spacePatt.splitAsStream(fullName) // Split on " "
.filter(e -> !e.isEmpty()) // Reject empty elements
.map(minusPatt::splitAsStream) // Split on "-"
.map(stream ->
stream.map(e -> e.substring(0, 1))) // Get initials
.map(stream ->
stream.collect(Collectors.joining("-"))) // Add "-"
.collect(Collectors.joining("")); // Concatenate
Which outputs RJS-A.
This approach is a bit more complicated as we need to maintain the information of the sub-streams, we can't just flatMap everything together, otherwise we wouldn't know where to add the - again. So in the middle part we are indeed operating on Stream<Stream<String>> objects.

I think what you're after here is the split method in String
Which you could use like this:
String fullName = "John Alexander Macdonald";
String[] split = fullName.split(" "); // ["John", "Alexander", "Macdonald"]
The other thing you might want is the trim method which removes spaces from the front and the back of a string.
String withSpaces = " a b c ";
String trimmed = withSpace.trim(); // "a b c"

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.

Java Regex ReplaceAll with grouping

I want to surround all tokens in a text with tags in the following manner:
Input: " abc fg asd "
Output:" <token>abc</token> <token>fg</token> <token>asd</token> "
This is the code I tried so far:
String regex = "(\\s)([a-zA-Z]+)(\\s)";
String text = " abc fg asd ";
text = text.replaceAll(regex, "$1<token>$2</token>$3");
System.out.println(text);
Output:" <token>abc</token> fg <token>asd</token> "
Note: for simplicity we can assume that the input starts and ends with whitespaces

Use lookaround:
String regex = "(?<=\\s)([a-zA-Z]+)(?=\\s)";
...
text = text.replaceAll(regex, "<token>$1</token>");

If your tokens are only defined with a character class you don't need to describe what characters are around. So this should suffice since the regex engine walks from left to right and since the quantifier is greedy:
String regex = "[a-zA-Z]+";
text = text.replaceAll(regex, "<token>$0</token>");

// meaning not a space, 1+ times
String result = input.replaceAll("([^\\s]+)", "<token>$1</token>");
this matches everything that isn't a space. Prolly the best fit for what you need. Also it's greedy meaning it will never leave out a character that it shouldn't ( it will never find the string "as" in the string "asd" when there is another character with which it matches)

How to remove spaces in between the String

I have below String
string = "Book Your Domain And Get\n \n\n \n \n \n Online Today."
string = str.replace("\\s","").trim();
which returning
str = "Book Your Domain And Get Online Today."
But what is want is
str = "Book Your Domain And Get Online Today."
I have tried Many Regular Expression and also googled but got no luck. and did't find related question, Please Help, Many Thanks in Advance

Use \\s+ instead of \\s as there are two or more consecutive whitespaces in your input.
string = str.replaceAll("\\s+"," ")

You can use replaceAll which takes a regex as parameter. And it seems like you want to replace multiple spaces with a single space. You can do it like this:
string = str.replaceAll("\\s{2,}"," ");
It will replace 2 or more consecutive whitespaces with a single whitespace.

First get rid of multiple spaces:
String after = before.trim().replaceAll(" +", " ");

If you want to just remove the white space between 2 words or characters and not at the end of string
then here is the
regex that i have used,
String s = " N OR 15 2 ";
Pattern pattern = Pattern.compile("[a-zA-Z0-9]\\s+[a-zA-Z0-9]", Pattern.CASE_INSENSITIVE);
Matcher m = pattern.matcher(s);
while(m.find()){
String replacestr = "";
int i = m.start();
while(i<m.end()){
replacestr = replacestr + s.charAt(i);
i++;
}
m = pattern.matcher(s);
}
System.out.println(s);
it will only remove the space between characters or words not spaces at the ends
and the output is
NOR152

Eg. to remove space between words in a string:
String example = "Interactive Resource";
System.out.println("Without space string: "+ example.replaceAll("\\s",""));
Output:
Without space string: InteractiveResource

If you want to print a String without space, just add the argument sep='' to the print function, since this argument's default value is " ".

//user this for removing all the whitespaces from a given string for example a =" 1 2 3 4"
//output: 1234
a.replaceAll("\\s", "")

String s2=" 1 2 3 4 5 ";
String after=s2.replace(" ", "");
this work for me

String string_a = "AAAA BBB";
String actualTooltip_3 = string_a.replaceAll("\\s{2,}"," ");
System.out.println(String actualTooltip_3);
OUTPUT will be:AAA BBB