90

Let's say I have

1 ABC Street
1 A ABC Street

With \d, it matches 1 (what I expect), with \d \w, it matches 1 A (expected). When I combine the patterns together \d|\d \w, it matches only the first one but ignores the second one.

My question is how to use "or" condition correctly in this particular case?

PS: The condition is wrapping the number only when there is no single letter after that, otherwise wrap the number and the single letter.

Example: 1 ABC Street match number 1 only, but when 1 A ABC Street wrap the 1 A

Matthias Braun
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Hoan Dang
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3 Answers3

93

Try

\d \w |\d

or add a positive lookahead if you don't want to include the trailing space in the match

\d \w(?= )|\d

When you have two alternatives where one is an extension of the other, put the longer one first, otherwise it will have no opportunity to be matched.

MikeM
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55

A classic "or" would be |. For example, ab|de would match either side of the expression.

However, for something like your case you might want to use the ? quantifier, which will match the previous expression exactly 0 or 1 times (1 times preferred; i.e. it's a "greedy" match). Another (probably more relyable) alternative would be using a custom character group:

\d+\s+[A-Z\s]+\s+[A-Z][A-Za-z]+

This pattern will match:

  • \d+: One or more numbers.
  • \s+: One or more whitespaces.
  • [A-Z\s]+: One or more uppercase characters or space characters
  • \s+: One or more whitespaces.
  • [A-Z][A-Za-z\s]+: An uppercase character followed by at least one more character (uppercase or lowercase) or whitespaces.

If you'd like a more static check, e.g. indeed only match ABC and A ABC, then you can combine a (non-matching) group and define the alternatives inside (to limit the scope):

\d (?:ABC|A ABC) Street

Or another alternative using a quantifier:

\d (?:A )?ABC Street
Mario
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    This seems a bit silly `\s+[A-Z\s]+\s+` - why do you need to match white space in three separate places? Surely either `[A-Z\s]+` or `\s+[A-Z]+\s+` make more sense... – Boris the Spider Apr 13 '13 at 09:30
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    Hm, yeah, thinking about it, it's a bit overkill after I removed the "first or last have to be letter" requirements. :) Editing... or not. It is still useful, because that way I force leading/trailing whitespaces, while allowing the brackets to match multiple words. – Mario Apr 13 '13 at 12:13
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    Surely that's a case for lookaround? – Boris the Spider Apr 13 '13 at 12:42
20

I think what you need might be simply:

\d( \w)?

Note that your regex would have worked too if it was written as \d \w|\d instead of \d|\d \w.

This is because in your case, once the regex matches the first option, \d, it ceases to search for a new match, so to speak.

Roney Michael
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