PHP strpos匹配多个干草堆中的所有针

Question

I want to check if ALL words from $words exist in one or more $sentences, word order is not important.

Words will only contain [a-z0-9].

Sentences will only contain [a-z0-9-].

My code so far, it almost work as expected:

$words = array("3d", "4");
$sentences = array("x-3d-abstract--part--282345", "3d-speed--boat-430419", "beautiful-flower-462451", "3d-d--384967");

foreach ($words as $word) {
    $sentences_found = array_values(array_filter($sentences, function($find_words) use ($word) {return strpos($find_words, $word);}));
}
print_r($sentences_found);

If you run this code here http://3v4l.org/tD5t5 , you'll get 4 results, but in reality it should be 3 results

Array
(
    [0] => x-3d-abstract--part--282345
    [1] => 3d-speed--boat-430419
    [2] => beautiful-flower-462451   // this one is wrong, no "3d" in here, only "4"
    [3] => 3d-d--384967
)

How can I do this?

Also is there any better way to do this than strpos?

Regex?

Regex is maybe slow for this job because sometimes there will be 1000's of $sentences (don't ask why).

Answer 1

You could use an intersection of found sentences per word:

$found = array();

foreach ($words as $word) {
    $found[$word] = array_filter($sentences, function($sentence) use ($word) {
        return strpos($sentence, $word) !== false;
    });
}

print_r(call_user_func_array('array_intersect', $found));

Or, approach from $sentences:

$found = array_filter($sentences, function($sentence) use ($words) {
    foreach ($words as $word) {
        if (strpos($sentence, $word) === false) {
            return false;
        }
    }
    // all words found in sentence 
    return true;
});

print_r($found);

One important thing to mention is that your search criteria was wrong; instead of strpos($sentence, $word) you should explicitly compare against false, otherwise you will miss a match at the start of a sentence.