Behaviour of String.split in java 1.6?

问题:

My code is:

String s = "1;;;; 23;;";
System.out.println(s.split(";").length);

and gives as output 5. The source code of split is:

public String[] split(String regex) {
        return split(regex, 0);
    }

and the documentation says:

This method works as if by invoking the two-argument split(java.lang.String,int) method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.

The string "boo:and:foo", for example, yields the following results with these expressions:

Regex Result
: { "boo", "and", "foo" }
o { "b", "", ":and:f" }

If I print the strings I have:

1



 23

Shouldn't I get from this 1;;;; 23;; something like {"1", "", "", "", " 23", ""} ?

回答1:

No, five is correct, as your quoted docs state:

Trailing empty strings are therefore not included in the resulting array.

Which is why the empty strings at the end of the array are omitted. If you want the empty strings, do as Evgeniy Dorofeev's answer says and specify a limit of -1.

回答2:

Since limit = 0 trailing empty strings are not included. Try

System.out.println(s.split(";", -1).length);

and you will get 7

回答3:

It will split the string when ever ';' present and put into array.