What is the best way to extract the first word from a string in Java?

Trying to write a short method so that I can parse a string and extract the first word. I have been looking for the best way to do this.

I assume I would use st开发者_开发知识库r.split(","), however I would like to grab just the first first word from a string, and save that in one variable, and and put the rest of the tokens in another variable.

Is there a concise way of doing this?

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The second parameter of the split method is optional, and if specified will split the target string only N times.

For example:

String mystring = "the quick brown fox";
String arr[] = mystring.split(" ", 2);

String firstWord = arr[0];   //the
String theRest = arr[1];     //quick brown fox

Alternatively you could use the substring method of String.

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You should be doing this

String input = "hello world, this is a line of text";

int i = input.indexOf(' ');
String word = input.substring(0, i);
String rest = input.substring(i);

The above is the fastest way of doing this task.

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To simplify the above:

text.substring(0, text.indexOf(' ')); 

Here is a ready function:

private String getFirstWord(String text) {
  int index = text.indexOf(' ');
  if (index > -1) { // Check if there is more than one word.
    return text.substring(0, index).trim(); // Extract first word.
  } else {
    return text; // Text is the first word itself.
  }
}

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The simple one I used to do is

str.contains(" ") ? str.split(" ")[0] : str

Where str is your string or text bla bla :). So, if

1. str is having empty value it returns as it is.
2. str is having one word, it returns as it is.
3. str is multiple words, it extract the first word and return.

Hope this is helpful.

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import org.apache.commons.lang3.StringUtils;

...
StringUtils.substringBefore("Grigory Kislin", " ")

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You can use String.split with a limit of 2.

    String s = "Hello World, I'm the rest.";
    String[] result = s.split(" ", 2);
    String first = result[0];
    String rest = result[1];
    System.out.println("First: " + first);
    System.out.println("Rest: " + rest);

    // prints =>
    // First: Hello
    // Rest: World, I'm the rest.

• API docs for: split

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for those who are searching for kotlin

var delimiter = " "  
var mFullname = "Mahendra Rajdhami"  
var greetingName = mFullname.substringBefore(delimiter)

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like this:

final String str = "This is a long sentence";
final String[] arr = str.split(" ", 2);
System.out.println(Arrays.toString(arr));

arr[0] is the first word, arr[1] is the rest

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You could use a Scanner

http://download.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html

The scanner can also use delimiters other than whitespace. This example reads several items in from a string:

     String input = "1 fish 2 fish red fish blue fish";
     Scanner s = new Scanner(input).useDelimiter("\\s*fish\\s*");
     System.out.println(s.nextInt());
     System.out.println(s.nextInt());
     System.out.println(s.next());
     System.out.println(s.next());
     s.close(); 

prints the following output:

     1
     2
     red
     blue

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None of these answers appears to define what the OP might mean by a "word". As others have already said, a "word boundary" may be a comma, and certainly can't be counted on to be a space, or even "white space" (i.e. also tabs, newlines, etc.)

At the simplest, I'd say the word has to consist of any Unicode letters, and any digits. Even this may not be right: a String may not qualify as a word if it contains numbers, or starts with a number. Furthermore, what about hyphens, or apostrophes, of which there are presumably several variants in the whole of Unicode? All sorts of discussions of this kind and many others will apply not just to English but to all other languages, including non-human language, scientific notation, etc. It's a big topic.

But a start might be this (NB written in Groovy):

String givenString = "one two9 thr0ee four"
// String givenString = "oňňÜÐæne;:tŵo9===tĥr0eè? four!"
// String givenString = "mouse"
// String givenString = "&&^^^%"

String[] substrings = givenString.split( '[^\\p{L}^\\d]+' )

println "substrings |$substrings|"

println "first word |${substrings[0]}|"

This works OK for the first, second and third givenStrings. For "&&^^^%" it says that the first "word" is a zero-length string, and the second is "^^^". Actually a leading zero-length token is String.split's way of saying "your given String starts not with a token but a delimiter".

NB in regex \p{L} means "any Unicode letter". The parameter of String.split is of course what defines the "delimiter pattern"... i.e. a clump of characters which separates tokens.

NB2 Performance issues are irrelevant for a discussion like this, and almost certainly for all contexts.

NB3 My first port of call was Apache Commons' StringUtils package. They are likely to have the most effective and best engineered solutions for this sort of thing. But nothing jumped out... https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html ... although something of use may be lurking there.

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You could also use http://download.oracle.com/javase/6/docs/api/java/util/StringTokenizer.html

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I know this question has been answered already, but I have another solution (For those still searching for answers) which can fit on one line: It uses the split functionality but only gives you the 1st entity.

String test = "123_456";
String value = test.split("_")[0];
System.out.println(value);

The output will show:

123

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The easiest way I found is this:

void main() 
  String input = "hello world, this is a line of text";
  
  print(input.split(" ").first);
}

Output: hello

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Assuming Delimiter is a blank space here:

Before Java 8:

 private String getFirstWord(String sentence){
     String delimiter = " "; //Blank space is delimiter here
    String[] words = sentence.split(delimiter);
    return words[0];
    }

After Java 8:

private String getFirstWord(String sentence){
    String delimiter = " "; //Blank space is delimiter here
    String firstWord = Arrays.stream(sentence.split(delimiter))
           .findFirst()
           .orElse("No word found");
}

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String anotherPalindrome = "Niagara. O roar again!"; 
String roar = anotherPalindrome.substring(11, 15); 

You can also do like these