How can the issue of "Warning: in_array() expects parameter 2 to be array" be resolved in PHP?

The issue of "Warning: in_array() expects parameter 2 to be array" can be resolved by ensuring that the second parameter passed to the in_array() function is indeed an array. This warning occurs when a non-array variable is passed as the second parameter, causing the function to fail. To fix this, make sure to pass an array as the second parameter. 

// Example code to resolve the issue
$my_array = array(1, 2, 3, 4, 5);
$search_value = 3;

if(is_array($my_array) && in_array($search_value, $my_array)) {
    echo "Value found in the array!";
} else {
    echo "Value not found in the array!";
}

Keywords

in_array parameter array PHP warning

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