Choosing just the alphanumeric words with regex

I'm trying to find the regular expression to find just the alphanumeric words from a string i.e the words that are a combination of alphabets or numbers. If a word is pure numb开发者_运维知识库ers or pure characters I need to discard it.

Try this regular expression:

\b([a-z]+[0-9]+[a-z0-9]*|[0-9]+[a-z]+[a-z0-9]*)\b

Or more compact:

\b([a-z]+[0-9]+|[0-9]+[a-z]+)[a-z0-9]*\b

This matches all words (note the word boundaries \b) that either start with one or more letters followed by one or more digits or vice versa that may be followed by one or more letters or digits. So the condition of at least one letter and at least one digit is always fulfilled.

With lookaheads:

'/\b(?![0-9]+\b)(?![a-z]+\b)[0-9a-z]+\b/i'

A quick test that also shows example usage:

$str = 'foo bar F0O 8ar';
$arr = array();
preg_match_all('/\b(?![0-9]+\b)(?![a-z]+\b)[0-9a-z]+\b/i', $str, $arr);
print_r($arr);

Output:

F0O
8ar

This will return all individual alphanumeric words, which you can loop through. I don't think regex can do the whole job by itself.

\b[a-z0-9]+\b

Make sure you mark that as case-insensitive.

\b(?:[a-z]+[0-9]+|[0-9]+[a-z]+)[[:alnum:]]*\b

'\b([a-zA-Z]+[0-9]+ | [0-9]+[a-zA-Z]+ | [a-zA-Z]+[0-9]+[a-zA-Z]*)\b'