测试 returns next() / next("route") 的快速中间件

Question

我创建了这个中间件，当路由的唯一参数由 queryItems 定义时，它将 return next()。

我最近发现 node-mocks-http 伪造表达 req 和 res 对象。但是它不会伪造 next。我想知道应该怎么做？下面是我打开 next 回调并在其中定义我的 expect 语句的示例。

middleware.hasOnlyQuery = function(queryItems){
  return function(req, res, next){
    if(typeof queryItems == "string") queryItems = [queryItems]
    if(_.hasOnly(req.query, queryItems)) return next()
    return next("route")
  }
}

这是测试。

it("should only have shop query", function(done){
  var req = httpMocks.createRequest({
      method: 'GET',
      query: {
        foo: "bar"
      }
  });
  var res = httpMocks.createResponse()
  var fn = middleware.hasOnlyQuery(["foo"])(req, res, function(err){
    expect(err).to.equal()
    return done()
  })
})

it("should not only have shop query", function(done){
  var req = httpMocks.createRequest({
      method: 'GET',
      query: {
        foo: "bar",
        bar: "foo"
      }
  });
  var res = httpMocks.createResponse()
  var fn = middleware.hasOnlyQuery(["foo"])(req, res, function(err){
    expect(err).to.equal("route")
    return done()
  })
})

这是执行此操作的正确方法吗？有什么方法可以使它更简单/更容易，也许可以将其转换为承诺，以便我可以使用 chai-as-promised？

Note: _.hasOnly is a custom underscore mixin.

Answer 1

使这更顺畅的一种方法是创建下一个可以有的三种响应。

function nextNormal(done){
  return function(err){
    expect(err).to.equal(undefined)
    return done()
  }
}

function nextRoute(done){
  return function(err){
    expect(err).to.equal("route")
    return done()
  }
}

function nextError(done){
  return function(err){
    expect(err).to.be.an('object')
    return done()
  }
}

然后一行就可以用了

middleware.hasOnlyQuery(["foo", "bar"])(req, res, nextNormal(done))

Answer 2

将中间件转换为与 express-promise-router 一起使用的承诺。

middleware.hasOnlyQuery = function(queryItems){
  return function(req, res){
    if(typeof queryItems == "string") queryItems = [queryItems]
    if(_.hasOnly(req.query, queryItems)) return Promise.resolve('next')
    return Promise.resolve('route')
  }
}

新测试使用chai-as-promised and bluebird：

it("should fire correct next callback", function(done){
  Promise.all([
    middleware.hasOnlyQuery(["foo"])(reqFoo, res).should.eventually.equal("next"),
    middleware.hasOnlyQuery(["foo", "bar"])(reqFoo, res).should.eventually.equal("route"),
    middleware.hasOnlyQuery(["foo", "alpha"])(reqFoo, res).should.eventually.equal("route"),
    middleware.hasOnlyQuery(["foo"])(reqFooBar, res).should.eventually.equal("route"),
    middleware.hasOnlyQuery(["foo", "bar"])(reqFooBar, res).should.eventually.equal("next"),
    middleware.hasOnlyQuery(["foo", "alpha"])(reqFooBar, res).should.eventually.equal("route")
  ]).then(function(){
    done()
  })
})

Answer 3

我喜欢 Sinon 这样的测试：

// first test
...
var res = httpMocks.createResponse()
var spy = sinon.spy();
middleware.hasOnlyQuery(["foo"])(req, res, spy);
expect(spy.calledWithExactly()).to.be.true;

// second test
...
expect(spy.calledWithExactly('route')).to.be.true;

测试 returns next() / next("route") 的快速中间件

Testing express middleware that returns next() / next("route")

javascript

unit-testing

express

mocha.js

middleware