CODEWITHSUNDEEP
javaregex
user2917629
user2917629

Reputation: 942

java regex optional and pipe not working

Ahoy everyone, I have the following regex pattern

String pattern1 = "([0-9+\\s|\\s])integer";

that I'm using on a string below. It suppose to select "integer" and a number followed by empty space before it, or just empty space if no number before it. I think I must be close, but I already spent all morning on it. Any help is much appreciated.
"BRTR val !http://www.w3.org/2001/XMLSchema# 78928458695 integer integer 33 integer"

The trouble is that it could be an emptySpace+integer or number+integer, in no particular order. The closest I came to is

"\\d+\\s*integer(.*)"

but that doesn't make it a pattern - it only works if there is a number before the "integer" and just gets the rest of the string, but I'd like to have it as a pattern.

Upvotes: 0

Views: 337

Answers (2)

Bubletan
Bubletan

Reputation: 3863

Move ] after the 9:

"([0-9]+\\s|\\s)integer"

This matches with "78928458695 integer", " integer" and "33 integer", assuming that I got what you wanted.

An example:

String s = "BRTR val !http://www.w3.org/2001/XMLSchema# 78928458695 integer integer 33 integer";
Pattern pattern = Pattern.compile("([0-9]+\\s|\\s)integer");
Matcher matcher = pattern.matcher(s);
List<String> list = new ArrayList<>();
while (matcher.find()) {
    list.add('"' + matcher.group() + '"');
}
System.out.println(list); // ["78928458695 integer", " integer", "33 integer"]

Upvotes: 2

IMP1
IMP1

Reputation: 280

Given that either way you are checking for a space before the integer, you could remove that from the parentheses.

And now you're just looking for an optional integer, followed by a space, followed by 'integer', so you're final regex can be this:

"[0-9]*\\sinteger"

This matches 0 or more digits followed by a space, followed by the word integer.

Upvotes: 0

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