php - 如何捕获 Uncaught TypeError fatal error ？

Question

我有一个方法接受 string 类型的参数，但它们也可以是 null。我怎样才能捕捉到这个 fatal error ？ fatal error :未捕获类型错误:传递给 Employee::insertEmployee() 的参数 5 必须是字符串类型，给定为 null， 在以下代码中:

public function insertEmployee(string $first_name, string $middle_name, string $last_name, string $department, string $position, string $passport)
{
    if($first_name == null || $middle_name == null || $last_name == null || $department == null || $position == null || $passport == null) {
        throw new InvalidArgumentException("Error: Please, check your input syntax.");
    }
    $stmt = $this->db->prepare("INSERT INTO employees (first_name, middle_name, last_name, department, position, passport_id) VALUES (?, ?, ?, ?, ?, ?)");
    $stmt->execute([$first_name, $middle_name, $last_name, $department, $position, $passport]);
    echo "New employee ".$first_name.' '.$middle_name.' '.$last_name.' saved.';
}


try {
    $app->insertEmployee($first_name, $middle_name, $last_name, $department, $position, $passport_id);
} catch (Exception $ex) {
    echo $ex->getMessage();
}

Answer 1

TypeError延伸Error它实现了 Throwable .这不是 Exception .因此，您需要捕获 TypeError 或 Error:

try {
    $app->insertEmployee($first_name, $middle_name, $last_name, $department, $position, $passport_id);
} catch (TypeError $ex) {
    echo $ex->getMessage();
}