Why am I storing null values in an array using regex in java?

Question

My code

    Scanner reader = new Scanner(System.in);
    String input = reader.nextLine();

    Pattern regex = Pattern.compile("^(<[a-z][0-9]>)|(<[\\/][a-z][0-9]>)");
    Matcher m = regex.matcher(input);

    while (m.find()) {
        listOfTags[i] = m.group();
        i++;
    }
    checkOpeningTag = listOfTags[0];
    checkClosingTag = listOfTags[1];
    for(int k = 0;k<listOfTags.length;k++){
        System.out.println("["+k+"] = "+listOfTags[k]);
    }

When I try to input <h1></h1> the output is

[0] = <h1>
[1] = </h1>

But when I input <h1><h1> the output is

[0] = <h1>
[1] = null

Why is it storing null value?

@rollback but how is it only one match? You mean if I have 10 h1 tags, it will only count as one? How can I count the rest of the same match? Is it a thing? — Yupyep, Oct 13 '17 at 07:43
I suspect you want to match opening/closing tag anywhere in the string. Remove `^`. Just use `"?[a-z][0-9]>"`. — Wiktor Stribiżew, Oct 13 '17 at 07:44
@Yupyep also look at my answer, its bad practice to not check for nulls and length — FrankK, Oct 13 '17 at 07:49
A possible duplicate of [Carets in Regular Expressions](https://stackoverflow.com/questions/16944357/carets-in-regular-expressions) — Wiktor Stribiżew, Oct 13 '17 at 07:52

FrankK · Answer 1 · 2017-10-13T07:47:02.173

0

to make sure you aren't exceeding the length of your listOfTags you might want to do this:

if(listOfTags.length >= 2){
    checkOpeningTag = listOfTags[0];
    checkClosingTag = listOfTags[1];
}

to prevent storing nulls in your list:

int i = 0;
while(m.find()){
    if(m.group() != null){
        listOfTags[i] = m.group();
        i++;
    }
}

edited Oct 13 '17 at 07:47

answered Oct 13 '17 at 07:41

FrankK

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score 0 · Accepted Answer · answered Oct 13 '17 at 07:56

0

The ^ caret symbol in the current regex matches the start of string. So, the only opening tag that will get matched will be at the beginning of the string.

Remove the caret:

(<[a-z][0-9]>)|(</[a-z][0-9]>)

Or even merge the alternatives into if you do not need to differentiate between opening and closing tags:

</?[a-z][0-9]>

answered Oct 13 '17 at 07:56

Wiktor Stribiżew

607,720
39
448
563

Thanks for posting this answer, Im new to using regex – Yupyep Oct 13 '17 at 08:25

Why am I storing null values in an array using regex in java?

2 Answers2