如何在 Sinon 中存根一系列方法? [英] How do I stub a chain of methods in Sinon?

问题描述

我知道如何使用存根替换一个函数.

I know how to use stub to replace one function.

sandbox.stub(Cars, "findOne",
            () => {return car1 });

但是现在我的函数中有一行我想测试我需要存根看起来像这样

But now I have a line in my function I want to test that I need to stub that looks like this

Cars.find().fetch()

所以这里有一个函数链，我不确定我需要做什么.如何存根查找"以返回可用于存根获取"的内容?

So there is a chain of function here and I'm unsure what I need to do. How do I stub "find" to return something that I can use to stub "fetch"?

推荐答案

试试这个:

sandbox.stub(Cars, "find", () => {
    return {
        fetch: sinon.stub().returns(anything);
    };
});

这篇关于如何在 Sinon 中存根一系列方法?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！