CODEWITHSUNDEEP
javaregex
Yeynno
Yeynno

Reputation: 331

Regular expression, omit few words

How can I write a regular expression to match sth like that :

he is capable of

he is not capable of

etc

general pattern "he is" + up to few words + "of"

I know how to solve it without regular expression, but maybe there is easier way

Upvotes: 0

Views: 72

Answers (3)

Tamas Rev
Tamas Rev

Reputation: 7166

I'd go with this:

\bhe is\b.*\bof\b

I'm using \b a lot to make sure I'm matching words. E.g. this won't match She is capable of, neither he isa wizard capable of

This is a little more complicated:

\bhe is\b( +\w+ *)*\bof\b

Here we have the ( +\w+ *)* in the middle. This makes sure that it matches words after each other.

You can play with the demo here.

Upvotes: 0

Tim Pietzcker
Tim Pietzcker

Reputation: 336128

A trivial solution would be to use

\bhe is(?: \w+){1,3} of\b

which allows between one and three "words" between he is and of.

\w+ means "a sequence of letters/digits/underscores", so it doesn't exactly match a word, but you can substitute your own word-matching regex if that one is too unspecific.

The \b word boundary anchors are used to only match he and of and not the or often.

Upvotes: 2

kimy82
kimy82

Reputation: 4475

If you really want to check if there is the capable or the not capatable.

"he is\\s+(not\\s+)?(capable\\s+)?of"

Upvotes: 1

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