Question

我有一个简单的脚本，可以在正常情况下正常运行。我正在尝试使其抛出异常，以进行测试。这是一个PHP PDO插入。如果没有插入新行，我希望它引发异常。

这是我的代码：

$query = "INSERT INTO reservations (requestor, location)
    VALUES (:requestor, :location)";
$statement = $pdo->prepare($query);

try {
    $statement->execute([
            'requestor' => $requestor_id,
            'location' => $request_location
    ]);
    echo "Success.";
}
catch(Exception $e) {
    echo "Sorry, but your request was not completed properly.<br>";
    echo "Message: " . $e->getMessage();
}

这很好用。但是，如果我通过修改表列名或绑定参数来使代码不起作用，它仍然会给出成功消息。我没有按预期插入新行。见下文：

$query = "INSERT INTO reservations (requestorzzz, locationzzz)
    VALUES (:requestor, :location)";
$statement = $pdo->prepare($query);

try {
    $statement->execute([
            'requestorzzz' => $requestor_id,
            'locationzzz' => $request_location
    ]);
    echo "Thank you! Your request was received.";
}
catch(Exception $e) {
    echo "Sorry, but your request was not completed properly.<br>";
    echo "Message: " . $e->getMessage();
}

我确实收到php错误消息，但未引发异常。（自然）不会创建新的列，但是我需要它去捕获以通知用户他们的请求没有发生。我做错了什么吗？

Answer 1

2分。

首先由@Funk Forty Niner指出。默认情况下，PDO发送警告，也不例外。您必须设置错误模式：

public class Orders { 
    public static void main(String[] s) {
        Orders jp   = new Orders();

        jp.method1();           
        List<OrderItems> lstOrdItms =  jp.getListOrderItems();
        jp.processOrderItems( lstOrdItms );
    }

    public void method1(List<OrderItem> lstOrderItem) {
        .........
    }

    public List<OrderItem> getListOrderItems() {
        ............
    }

    public void processOrderItems() {
        ............
    }
 }

 public class OrdersTest { 
    @Mocks
    Orders orders;

    .....
    @Before
    public void setUp() {
    ........
    }

    @Test
    public void testMain() throws SQLException {
         // Not sure how to test here
         // This will actually execute the main method instead of mock.
         orders.main(new String[]{});              
   }
}

第二：在您的execute方法中为数组使用“命名参数”而不是“列名”

$pdo->setAttribute(\PDO::ATTR_ERRMODE, \PDO::ERRMODE_EXCEPTION);

PDO插入应该引发异常，但不会

1 个答案: