CODEWITHSUNDEEP
regex
user3210665
user3210665

Reputation: 3

Skipping first character regex

I'm currently using the following code to remove certain characters from a string in an array

myArray[x] = myArray[x].replaceAll("[aeiou]","");

which works fine, but I need to ignore the first character of the string so for example if an array element was Alan it would be stripped to Aln.

I'm not sure if using a replaceAll is the best way about doing it but the only other way I can think of is removing the first character, applying the above regex to the string, appending the character back on and then inserting back into the array, which seems a long winded way of doing it.

Upvotes: 0

Views: 758

Answers (2)

Miraage
Miraage

Reputation: 3464

What about something like ..

myArray[x] = myArray[x].replaceAll("(^.)[aeiou]", "\\1");

// upd

Negative lookbehind is your solution, like Boris answered.

Upvotes: 1

Boris the Spider
Boris the Spider

Reputation: 61128

You can use a negative lookbehind to assert that the pattern is not preceded by the line start marker (^):

public static void main(String[] args) throws Exception {
    final String[] input = {"abe", "bae"};
    for(final String s: input) {
        System.out.println(s.replaceAll("(?<!^)[aeiou]", ""));
    }
}

Output:

ab
b

Upvotes: 1

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