如何使用sinon监视内部存根返回(How to spy inside stub returns with sinon)

如何使用sinon监视内部存根返回(How to spy inside stub returns with sinon)

嗨，我想窥探下面的函数，因为函数返回一个promise，所以我需要返回promise，让其他代码继续运行。

但它从未通过测试，我做错了吗？ 我怎么能正确地做到这一点？ 先谢谢你。

抱歉，我可能没有清楚地描述问题， another_fn将调用object.method.origin_fn ，并且orgin_fn返回一个promise; 我想窥探origin_fn被调用。 我怎样才能做到这一点？

let spy_fn = sinon.spy(); sinon.stub(object, 'method').returns({ origin_fn(args) { spy_fn(args); return Promise.resolve(); } }); another_fn('test_args'); spy_fn.should.have.been.calledWith('test_args');

Hi I want to spy the function as below, and because the function returns a promise, so I need to return the promise let other code continue running.

But it never passes the test, am I doing it wrong? how can I do this correctly? Thank you in advance.

Sorry I might not describe the problem clearly, another_fn will invoke object.method.origin_fn, and the orgin_fn returns a promise; I want to spy the origin_fn is get called. How can I do this?

let spy_fn = sinon.spy(); sinon.stub(object, 'method').returns({ origin_fn(args) { spy_fn(args); return Promise.resolve(); } }); another_fn('test_args'); spy_fn.should.have.been.calledWith('test_args');

最满意答案

如果您只想确保使用正确的参数调用origin_fn则可以使用以下内容：

let spy = sinon.stub(object, 'method').returns({ origin_fn(arg) { arg.should.equal('test_args'); } }) let fn = function() { another_fn('test_args'); }; fn.should.not.throw();

If you only want to make sure that origin_fn is called with the correct argument, you can use something like this:

let spy = sinon.stub(object, 'method').returns({ origin_fn(arg) { arg.should.equal('test_args'); } }) let fn = function() { another_fn('test_args'); }; fn.should.not.throw();