Question

我在for-each循环中遇到此错误：

“类型不匹配：无法从元素类型对象转换为员工”

private static HashMap<Integer,Employee> employeeDatabase = new HashMap<Integer,Employee>();

public HashMap getEmployeeDatabase() {
    return employeeDatabase;
}

for(Employee e: c.getEmployeeDatabase().values()) 
    {
        e.print();
    }

如果有关系-'Employee'包含int（id），String（name），double（salary）。填充HashMap时，将复制int（id）用作我的HashMap的Integer键。

编辑：问题在于getEmployeeDatabase访问器返回原始类型，谢谢那些回答的人。

对于那些想知道'c'变量的人：

Company c = new Company();

那是它的来历。 Company类的默认构造函数使用私有方法从Scanner填充employeeDatabase。

Answer 1

您的方法返回类型为raw type。不要使用原始类型。

private static HashMap<Integer,Employee> employeeDatabase = new HashMap<Integer,Employee>();

public HashMap getEmployeeDatabase() {
    return employeeDatabase;
}

应该像

private static Map<Integer, Employee> employeeDatabase = new HashMap<>();

public Map<Integer, Employee> getEmployeeDatabase() {
    return employeeDatabase;
}

Answer 2

当您不指定泛型时，假定为对象，然后

// Utility function to get the shipping method Id from order object
function wc_get_shipping_method_id( $order ){
    foreach ( $order->get_shipping_methods() as $shipping_method ) {
        return $shipping_method->get_method_id();
    }
}

// Add coditionally a "reply to" based on shipping methods IDs for specific email notifications
add_filter( 'woocommerce_email_headers', 'add_headers_replay_to_conditionally', 10, 3 );
function add_headers_replay_to_conditionally( $headers, $email_id, $order ) {
    // Avoiding errors
    if ( ! is_a( $order, 'WC_Order' ) || ! isset( $email_id ) )
        return $headers;

    // The defined emails notifications to customer
    $allowed_email_ids = array('customer_on_hold_order', 'customer_processing_order', 'customer_completed_order');

    // Only for specific email notifications to the customer
    if( in_array( $email_id, $allowed_email_ids ) ) {
        // Local Pickup Plus shipping method
        if( wc_get_shipping_method_id( $order ) === 'local_pickup_plus' ){
            $headers .= "Reply-to: reply1@webshop.com". "\r\n"; // Email adress 1
        } 
        // Other shipping methods
        else {
            $headers .= "Reply-to: reply2@webshop.com". "\r\n"; // Email adress 2
        }
    }

    return $headers;
}

实际上是

public HashMap getEmployeeDatabase()

所以

public HashMap<Object, Object> getEmployeeDatabase()

不正确，因为values（）将返回

for(Employee e : c.getEmployeeDatabase().values()) {
    e.print();
}

使用每个循环的类型转换错误

2 个答案: