如何创建Single.just(Void) [英] How to create Single.just(Void)

问题描述

我正在为我的应用程序编写一些单元测试用例.我想模拟 MongoClient update 方法，但是更新返回 Single< Void> .

I am writing some unit test cases for my application. I want to mock MongoClient update method, but the update returns Single<Void>.

when(mongoClient.rxUpdate(anyString(), any(JsonObject.class), any(JsonObject.class)))
.thenReturn(Single.just(Void))

现在 Single.just(Void)不起作用，正确的方法是什么?

Now Single.just(Void) doesn't work, what is the correct way of doing it?

-更新-

因此，我正在为 updateUserProfile 方法编写单元测试，并且为此我模拟了 service .但是 service.updateAccount 方法返回是我无法模拟的.

So I am writing unit test for updateUserProfile method and for that I have mocked service. But the service.updateAccount method return is something I am not able to mock.

//Controller class
public void updateUserProfile(RoutingContext routingContext){
        // some code
    service.updateAccount(query, update)
            .subscribe(r -> routingContext.response().end());
}

//Service Class
public Single<Void> updateAccount(JsonObject query, JsonObject update){
    return mongoClient.rxUpdate("accounts", query, update);
}

由于 mongoClient.rxUpdate 的返回类型为 Single< Void> ，因此我无法模拟该部分.

Because the return type of mongoClient.rxUpdate is Single<Void>, I am not able to mock that part.

目前，我已经确定的解决方法是:

For now the workaround which I have figured out is:

public Single<Boolean> updateAccount(JsonObject query, JsonObject update){
    return mongoClient.rxUpdate("accounts", query, update).map(_void -> true);
}

但这只是一个很简单的方法，我想知道如何才能准确地创建 Single< Void>

But this is just a hacky way of doing it, I want to know how can I exactly create Single<Void>

推荐答案

有一种返回 Single< Void> 的方法可能会引起一些担忧，因为一些用户已经在评论中表达了对此的看法.

Having a method returning Single<Void> may raise some concerns, as some users have already expressed their view on this in the comments.

但是，如果您坚持这样做，而您确实需要对其进行模拟(无论出于何种原因)，肯定有多种方法可以创建 Single< Void> 实例，例如，您可以使用Single类的create方法:

But if you are stuck with this and you really need to mock it (for whatever reason), there are definitely ways to create a Single<Void> instance, for example you could use the create method of the Single class:

Single<Void> singleVoid = Single.create(singleSubscriber -> {});

when(test.updateAccount(any(JsonObject.class), any(JsonObject.class))).thenReturn(singleVoid);

Single<Void> result = test.updateAccount(null, null);

result.subscribe(
        aVoid -> System.out.println("incoming!") // This won't be executed.
);

请注意:由于无法在没有反射的情况下实例化Void，因此您将无法实际发射Single项.

Please note: you won't be able to actually emmit a Single item, since Void can't be instantiated without reflection.

在某些情况下最终可以使用的技巧是省略泛型类型参数，而改为发出Object，但这很容易导致ClassCastException.我不建议使用此:

A trick that could eventually work in some cases is to ommit the generic type argument and emmit an Object instead, but this could lead easily to a ClassCastException. I would not recommend to use this:

Single singleObject = Single.just(new Object());

when(test.updateAccount(any(JsonObject.class), any(JsonObject.class))).thenReturn(singleObject);

Single<Void> result = test.updateAccount(null, null);

// This is going to throw an exception:
// "java.base/java.lang.Object cannot be cast to java.base/java.lang.Void"
result.subscribe(
        aVoid -> System.out.println("incoming:" + aVoid)
);

当然，您也可以使用反射(如Minami Namikaze所建议的那样):

And of course you could use reflection as well (as already suggested by Minato Namikaze):

Constructor<Void> constructor = Void.class.getDeclaredConstructor(new Class[0]);
constructor.setAccessible(true);
Void instance = constructor.newInstance();

Single<Void> singleVoidMock = Single.just(instance);

when(test.updateAccount(any(JsonObject.class), any(JsonObject.class))).thenReturn(singleVoidMock);

Single<Void> result = test.updateAccount(null, null);

result.subscribe(
        aVoid -> System.out.println("incoming:" + aVoid) // Prints: "incoming:java.lang.Void@4fb3ee4e"
);

这篇关于如何创建Single.just(Void)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！