Java convert string to string array by spaces

Question

I would like to convert string to string array.. I have done something like this:

String name = "name";
String[] letters = name.split("(?<=.)");
System.out.println("letters = " + Arrays.toString(letters));

But now I want something like this:

String name = "John Mark Lester Andrew Todd";

To print array like this:

[John, Mark, Lest, Andrew, Todd]

So I want to look for spaces in my string and then put them into a string array. Do you have ideas to do this? Help is much appreciated. Thanks.

this print exactly what do you want System.out.println(Arrays.toString(name.split(" "))); — Baby
– Baby, Commented Jan 9, 2014 at 5:54

Rahul · Accepted Answer · 2014-01-09 05:53:19Z

8

You can simply split() your string on whitespace.

String[] words = name.split(" ");

answered Jan 9, 2014 at 5:53

Rahul

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Comments

Ruchira Gayan Ranaweera · Accepted Answer · 2014-01-09 05:53:37Z

2

You can use split() and create an array.

String name = "John Mark Lester Andrew Todd";
String[] names=name.split(" ");

answered Jan 9, 2014 at 5:53

Ruchira Gayan Ranaweera

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Comments

Charles Stevens · Accepted Answer · 2014-01-09 05:59:45Z

2

What ever criteria or letter on the basis of which a string is to be split,that need to be provided under

  string.split(" ");

so,u can try using these updated code:::

 String name = "John Mark Lester Andrew Todd";
 String[] letters = name.split(" ");
 System.out.println("letters = " + Arrays.toString(letters));

and the ouput will be::

  letters = [John, Mark, Lester, Andrew, Todd]

And for better understanding for splitting the strings in java,you can go to this link. http://www.tutorialspoint.com/java/java_string_split.htm

edited Jan 9, 2014 at 5:59

answered Jan 9, 2014 at 5:54

Charles Stevens

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2 Comments

TheLostMind Over a year ago

I think its better to use "\\s+" instead of " ".

TheLostMind Over a year ago

We've all been there bro.. :)

user2672373 · Accepted Answer · 2014-01-09 05:56:46Z

1

Perhaps you can use StringTokenizer

answered Jan 9, 2014 at 5:56

user2672373

2 Comments

Rahul Over a year ago

You forgot to mention this about StringTokenizer - StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead.

user2672373 Over a year ago

My bad. The split method is even mentioned in the docs. Anyway just pointed out that this is also possible. It IS deprecated though.

Sujith PS · Accepted Answer · 2014-01-09 06:02:25Z

1

You can use :

String values=Arrays.toString(name.split(" "));

or

String values[] = name.split("\\s")

answered Jan 9, 2014 at 6:02

Sujith PS

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Comments

Radiodef · Accepted Answer · 2014-01-09 06:58:37Z

1

The most flexible is to split on all whitespace:

String names = "John  \t\t  Mark  Lester \n  Andrew Todd \r";

System.out.println(Arrays.toString(names.split("\\s+")));

Outputs:

[John, Mark, Lester, Andrew, Todd]

\\s+ is regex for "any length of any whitespace characters". Similarly if you only care about spaces:

String names = "John  Mark  Lester   Andrew Todd  ";

System.out.println(Arrays.toString(names.split(" +")));

(Outputs same as above.)

Of course both of those will split single spaces too.

edited Jan 9, 2014 at 6:58

answered Jan 9, 2014 at 5:59

Radiodef

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8 Comments

Baby Over a year ago

you changed the actual value of names, and OP want the output in format [John, Mark, Lest, Andrew, Todd]. but I agree the use of "\\s+" instead of " "

Radiodef Over a year ago

@RafaEl All I did was mangle it to show the regex handles any whitespace. That would be a stupid reason to downvote. The regex splits the OP's single space input too.

Baby Over a year ago

then you might want to explain in the answer, why you add \t\t, \n, \r there.

Radiodef Over a year ago

@RafaEl Because they are whitespace characters. \t is a tab, \n is a new line and \r is a carriage return.

Radiodef Over a year ago

@RafaEl I think it's pretty clear what the regex does. I think somebody just had a vacuous reason for voting. Thanks for trying to help.

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