测试返回 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，它将返回next()。

我最近发现了node-mock -http，它假装表达req和res对象。然而，它不假next。我想知道应该怎么做?下面我有一个例子，我打开next回调并在其中定义我的expect语句。

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()
  })
})

这样做合适吗?有没有办法让这个更简单/更容易，也许把它转换成一个承诺，所以我可以使用chai-as-promise ?

注意:_。hasOnly是一个自定义下划线mixin。

Answer 1

我喜欢Sinon做这样的测试:

// 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;

Answer 2

让这更流畅的一种方法是创建' next '可以具有的三种响应。

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()
  }
}

那么你可以在一行中使用

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

Answer 3

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')
  }
}

使用chai-as-promised和bluebird的新测试:

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()
  })
})