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

问题描述：

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 .

这是我的in_array code>代码 p>

  $ array = array（'a'，'b'，'c'）; 
 
if（in_array（array（'p'，'c'），$ array））{
 echo“found”;  
}其他{
 echo“not found”; 
} 
  code>  pre> 
 
 它返回未找到，但实际上我希望它返回 found ，因为有一个值匹配 c  code>。   p> 
  div>

答

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

答

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

答

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);

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

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