Tokenizer java (string)

Question

For example, Ive got Some stringa.b.c.d`, I have separated it by tokens. but now I want to do a specific thing with a, then something else with b, and something else with c. How can I do that?

String value="a.b.c.d";
StringTokenizer tokenize = new StringTokenizer(value, ".");
while(tokenize.hasMoreTokens()){
    String separated1= tokenize.?????;  
    String separated2= tokenize.?????;                            
    String Val1 = someMethod1(separated1); 
    String Val2 = someMethod2(separated2); 
}
//tokenize next line is not solution

Why not use String.split()? StringTokenizer is a legacy class. — user3437460, Feb 02 '17 at 11:38
it did not solve my problem, I can only choose how many of them I want to "print out ", but I still cant choose which one. — mica1234, Feb 02 '17 at 11:55
There is no law that says you have to use a loop. You can just use several `if` statements one after the other to check if there is still a token and get the next token and do something with it. — RealSkeptic, Feb 02 '17 at 11:56

score 3 · Answer 1 · answered Feb 02 '17 at 12:10

Just to spell out what @RealSkeptic already said, avoid the loop:

    String value = "a.b.c.d";
    String[] separated = value.split("\\.");
    if (separated.length >= 2) {
        String val1 = someMethod1(separated[0]); 
        String val2 = someMethod2(separated[1]); 
    } else {
        // other processing here
    }

You could do something similar with a StringTokenizer if you liked, only you would need at least two if statements to check whether there were enough tokens. But as @user3437460 said, StringTokenizer is now legacy; String.split() is recommended in new code.

Will · Answer 2 · 2017-02-02T11:51:00.150

1

Using String.split you can split about the . into an array.

How about something like this which will print each token:

        String value = "a.b.c.d";
        String[] tokens = value.split("\\.");
        for (String token : tokens) {
            System.out.println(token);
        }

The . needs to be escaped as the split function takes a regexp and . is a wildcard

edited Feb 02 '17 at 11:51

answered Feb 02 '17 at 11:45

Will

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score 0 · Answer 3 · answered Feb 02 '17 at 11:39

0

Why don't use split ? Easier to use and makes more sense here

answered Feb 02 '17 at 11:39

Sylvain Attoumani

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Tokenizer java (string)

3 Answers3