[英]Javascript testing - function called with specfic argument
I am trying to write a unit test for a function but cannot figure out how to check if it makes a call to a nested function with a specific argument.我正在尝试为 function 编写单元测试,但无法弄清楚如何检查它是否使用特定参数调用嵌套的 function。 I am assuming I will need to use sinon alongside chai and mocha for this, but I could really use some help.
我假设我需要将 sinon 与 chai 和 mocha 一起使用,但我真的可以使用一些帮助。
The function I would like to test looks like:我想测试的 function 看起来像:
function myFunc(next, value) {
if (value === 1) {
const err = new Error('This sets an error');
next(err);
} else {
next();
}
}
I would like to test if next is called with or without the err variable.我想测试是否在有或没有 err 变量的情况下调用 next。 From what I read so far I should use a spy for this (I think) but how would I use that spy?
从我目前阅读的内容来看,我应该为此使用间谍(我认为),但我将如何使用该间谍? Looking at this example from the Sinon docs it is unclear to me where PubSub comes from:
从 Sinon 文档中查看这个示例,我不清楚 PubSub 来自哪里:
"test should call subscribers with message as first argument" : function () {
var message = "an example message";
var spy = sinon.spy();
PubSub.subscribe(message, spy);
PubSub.publishSync(message, "some payload");
sinon.assert.calledOnce(spy);
sinon.assert.calledWith(spy, message);
}
Source: https://sinonjs.org/releases/latest/assertions/来源: https://sinonjs.org/releases/latest/assertions/
If you have a function like this如果您有这样的 function
function myFunc(next, value) {
if (value === 1) {
const err = new Error('This sets an error');
next(err);
} else {
next();
}
}
The test could look like this测试可能看起来像这样
it ('should call the callback with an Error argument', function (done) {
const callback = (err) => {
if (err && err instanceof Error && err.message === 'This sets an error'){
// test passed, called with an Error arg
done();
} else {
// force fail the test, the `err` is not what we expect it to be
done(new Error('Assertion failed'));
}
}
// with second arg equal to `1`, it should call `callback` with an Error
myFunc(callback, 1);
});
so you don't necessarily need sinon
for that所以你不一定需要
sinon
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.