Nsubstitute的延迟回报

发布于 2025-02-09 12:41:13 字数 930 浏览 1 评论 0原文

我有一个接口idiscosclient，用于测试/演示目的时，我希望在.getSingle＆lt; t＆gt;（）方法的随机延迟在1到5秒之间。这主要是这样，我可以看到我所有的各种负载旋转器组件以及什么都没有起作用。

因此，我认为我能够做这样的事情：

Fixture fixture = new();
fixture.Customize(new DiscosModelFixtureCustomizationNoLinks());

builder.Services.AddTransient(_ =>
                              {
                                  IDiscosClient client = Substitute.For<IDiscosClient>();
                                  DiscosObject  obj    = fixture.Create<DiscosObject>();
                                  client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(Task.Delay(Random.Shared.Next(1000,5000)).ContinueWith(_ => obj));
                                  return client;
                              });

但是，虽然当我第一次调用该方法时似乎有延迟每次我称其为idiscosclient实例。

有足够简单的方法来实现这一目标吗？

原文

I have an interface IDiscosClient, for testing/demo purposes while I'm developing the app, I want a mock to return a new model when the .GetSingle<T>() method is called with a random delay of between 1 and 5 seconds. This is mostly so I can see that all of my various loading spinner components and whatnot work.

So, I thought I'd be able to do something like this:

Fixture fixture = new();
fixture.Customize(new DiscosModelFixtureCustomizationNoLinks());

builder.Services.AddTransient(_ =>
                              {
                                  IDiscosClient client = Substitute.For<IDiscosClient>();
                                  DiscosObject  obj    = fixture.Create<DiscosObject>();
                                  client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(Task.Delay(Random.Shared.Next(1000,5000)).ContinueWith(_ => obj));
                                  return client;
                              });

However, while there seems to be a delay when I first call the method, once this has resolved, it just seems to return the completed task with the same model in it every time I call it for that IDiscosClient instance.

Is there a simple enough way to accomplish this?

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贪恋 2025-02-16 12:41:13

因此，问题在于，上面的代码仅创建一个新的任务第一次返回了每个后续时间（已经完成的）。

为了解决此问题，我们可以将上述代码更改为：

client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(_ => Task.Delay(Random.Shared.Next(1000,5000)).ContinueWith(_ => obj));

或者，出于遗产，我们可以将其提取到方法中并制作整个代码块：

builder.Services.AddTransient(_ =>
                              {
                                  IDiscosClient client = Substitute.For<IDiscosClient>();
                                  client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(GetDiscosObject);
                                  return client;
                              });

async Task<DiscosObject> GetDiscosObject(CallInfo _)
{
    await Task.Delay(Random.Shared.Next(1000, 5000));
    return fixture.Create<DiscosObject>();
}

So the issue is that the code above only creates a fresh Task the first time and then returns the same one (which has already completed) each subsequent time.

To fix this, we can either change the code above to:

client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(_ => Task.Delay(Random.Shared.Next(1000,5000)).ContinueWith(_ => obj));

Or, for legibilities sake, we can extract it into a method and make the whole code block:

builder.Services.AddTransient(_ =>
                              {
                                  IDiscosClient client = Substitute.For<IDiscosClient>();
                                  client.GetSingle<DiscosObject>(Arg.Any<string>()).Returns(GetDiscosObject);
                                  return client;
                              });

async Task<DiscosObject> GetDiscosObject(CallInfo _)
{
    await Task.Delay(Random.Shared.Next(1000, 5000));
    return fixture.Create<DiscosObject>();
}

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