如何模拟 amqp.Dial 等库中的函数

Question

我正在开发一个小型 AMQP 消费者，我想测试我的消费者代码，但我很难模拟amqp.Dial. 我添加了一些接口，以便我可以模拟Connection并Channel添加一个属性，以便我可以控制拨号功能：

//consumer.go
type AmqpChannel interface {
    ExchangeDeclare(name, kind string, durable, autoDelete, internal, noWait bool, args amqp.Table) error
    QueueDeclare(name string, durable, autoDelete, exclusive, noWait bool, args amqp.Table) (amqp.Queue, error)
    QueueBind(name, key, exchange string, noWait bool, args amqp.Table) error
    Consume(queue, consumer string, autoAck, exclusive, noLocal, noWait bool, args amqp.Table) (<-chan amqp.Delivery, error)
    Publish(exchange, key string, mandatory, immediate bool, msg amqp.Publishing) error
}

type AmqpConnection interface {
    Channel() (AmqpChannel, error)
    Close() error
}

type AmqpDial func(url string) (AmqpConnection, error)

type MyConsumer struct {
    HostDsn    string
    channel    AmqpChannel
    queue      amqp.Queue
    connection AmqpConnection
    DialFunc   AmqpDial
}

func (c *MyConsumer) Connect() error {
    var err error
    c.connection, err = c.DialFunc(c.HostDsn)
...

这似乎接近我想要实现的目标，我可以像这样指定我的测试：

func TestConsumer(t *testing.T) {
    mockCtrl := gomock.NewController(t)
    defer mockCtrl.Finish()

    var myConsumer = consumer.MyConsumer{
        HostDsn: "test",
        DialFunc: func(url string) (consumer.AmqpConnection, error) {
            return mocks.NewMockAmqpConnection(mockCtrl), nil
        },
    }
    _ = myConsumer.Connect()
}

amqp.Dial但我无法在主例程中将原始数据作为 dial-func 传递：

myConsumer := consumer.MyConsumer{
        HostDsn: fmt.Sprintf(
            "amqp://%s:%s@rabbitmq:5672/?heartbeat=5s",
            os.Getenv("RABBITMQ_USER"),
            url.QueryEscape(os.Getenv("RABBITMQ_PASSWORD")),
        ),
        DialFunc: amqp.Dial,
    }

给出

./main.go:28:9: cannot use amqp.Dial (type func(string) (*amqp.Connection, error)) as type consumer.AmqpDial in field value

我希望/认为，随着接口amqp.Connection的实现AmqpConnection，这会起作用：/模拟方法的正确方法是什么amqp.Dial？

PS：我知道https://github.com/NeowayLabs/wabbit但我更愿意在较低级别上解决这个问题:)

更新：@mkopriva 建议使用另一种包装方法，所以我尝试了：

./main.go:28:9: cannot use amqp.Dial (type func(string) (*amqp.Connection, error)) as type consumer.AmqpDial in field value

但这会导致：

cannot use connection (type *amqp.Connection) as type consumer.AmqpConnection in return argument:
    *amqp.Connection does not implement consumer.AmqpConnection (wrong type for Channel method)
        have Channel() (*amqp.Channel, error)
        want Channel() (consumer.AmqpChannel, error)
make: *** [Makefile:4: build-consumer] Error 2

Answer 1

给定以下类型：

type AmqpChannel interface {
    ExchangeDeclare(name, kind string, durable, autoDelete, internal, noWait bool, args amqp.Table) error
    QueueDeclare(name string, durable, autoDelete, exclusive, noWait bool, args amqp.Table) (amqp.Queue, error)
    QueueBind(name, key, exchange string, noWait bool, args amqp.Table) error
    Consume(queue, consumer string, autoAck, exclusive, noLocal, noWait bool, args amqp.Table) (<-chan amqp.Delivery, error)
    Publish(exchange, key string, mandatory, immediate bool, msg amqp.Publishing) error
}

type AmqpConnection interface {
    Channel() (AmqpChannel, error)
    Close() error
}

type AmqpDial func(url string) (AmqpConnection, error)

您可以创建委托给实际代码的简单包装器：

func AmqpDialWrapper(url string) (AmqpConnection, error) {
    conn, err := amqp.Dial(url)
    if err != nil {
        return nil, err
    }
    return AmqpConnectionWrapper{conn}, nil
}

type AmqpConnectionWrapper struct {
    conn *amqp.Connection
}

// If *amqp.Channel does not satisfy the consumer.AmqpChannel interface
// then you'll need another wrapper, a AmqpChannelWrapper, that implements
// the consumer.AmqpChannel interface and delegates to *amqp.Channel.
func (w AmqpConnectionWrapper) Channel() (AmqpChannel, error) {
    return w.conn.Channel()
}

func (w AmqpConnectionWrapper) Close() error {
    return w.conn.Close()
}