如何在php中使用in_array并将数组作为needle，但在至少有一个值匹配时返回true

Here is my in_array code

$array = array('a', 'b', 'c');

if(in_array(array('p', 'c'), $array)){
    echo "found";
}else{
    echo "not found";
}

it returns not found, but actually I want it to return found, because there is one value match c .

Your idea can be realized by using array_intersect and count functions.
If there's at least one matched item between two arrays - count will return the number of matched items (1 or more):

$needle = array('p', 'c');
$haystack = array('a', 'b', 'c');

echo (count(array_intersect($needle, $haystack))) ? "found" : "not found";
// will output: "found"

http://php.net/manual/en/function.array-intersect.php

use array_interset():-

$search = array('p', 'c');
$array = array('a', 'b', 'c');

$result = !empty(array_intersect($search , $array ));

var_dump($result); // print result

//OR
if(count($result) >=1){echo 'found';}else{'not found';}

Output:-https://eval.in/599429

Reference:-

http://php.net/manual/en/function.array-intersect.php

Another approach by creating a user function

function found_in_array($needle, $haystack) {
    foreach ($needle as $array) {
        if(in_array($array, $haystack)){
            return "found";
        }
    }

    return "not found";
}

$haystack = array('a', 'b', 'c');
$needle = array('p', 'c');

echo found_in_array($needle, $haystack);