REGEX to print every alternate word in a sentence

Suppose i have a string:

$s= "The quick brown fox jumps over the lazy dog"

I want to use php re开发者_开发百科gex functions to retrieve every alternate word from the sentence. Like for the above sentence, the output should be:

The brown jumps the dog

Can anyone help me with the REGEX for this?

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You can replace every two words by the first one. So, replace

(\w+) \w+

by $1:

preg_replace('/(\w+) \w+/', "$1", 'The quick brown fox jumps over the lazy dog')

Quick test:

php -r 'echo preg_replace("/(\w+) \w+/", "$1", "The quick brown fox jumps over the lazy dog");'
The brown jumps the dog

If you want the second, fourth, etc. words retained, then you can adapt the regex to

\w+ (\w+)

which will put the second word in a capture group. However. This will retain the very last word, even when the number of words is odd:

php -r "echo preg_replace('/\\w+ (\\w+)/', '\\1', 'The quick brown fox jumps over the lazy dog'),\"\\n\";"
quick fox over lazy dog

See that stray »dog« at the end? To solve that you need to remove the very last word if there isn't one following it:

\w+(?: (\w+))?

Demo:

php -r "echo preg_replace('/\\w+(?: (\\w+))?/', '\\1', 'The quick brown fox jumps over the lazy dog'),\"\\n\";"
quick fox over lazy

The (?:...) part is a so-called non-capturing group. It will group parts of the regex without capturing its contents for backreferences. This is here mainly so you can still replace by $1 and not $2.

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Or, if you want to capture the matches in an array:

preg_match_all('/(?|(\w+) \w+|(\w+)$)/', $s, $matches);
var_dump($matches[1]);

And to grab the opposite positions:

preg_match_all('/\w+ (\w+)/', $s, $matches);
var_dump($matches[1]);