我有一个数据服务具有以下功能一个stub与sinon的承诺如何?
function getInsureds(searchCriteria) {
var deferred = $q.defer();
insuredsSearch.get(searchCriteria,
function (insureds) {
deferred.resolve(insureds);
},
function (response) {
deferred.reject(response);
});
return deferred.promise;
}
我想测试以下功能:
function search()
{
dataService
.getInsureds(vm.searchCriteria)
.then(function (response) {
vm.searchCompleted = true;
if (response.insureds.length > 100) {
vm.searchResults = response.insureds.slice(0, 99);
} else {
vm.searchResults = response.insureds;
}
});
}
我将如何存根的承诺,这样,当我把getInsureds它将解决的承诺和回报我马上得到结果。我开始像这样(茉莉花测试),但我被卡住了,因为我不知道如何解决承诺并传递所需的论据。
it("search returns over 100 results searchResults should contain only 100 records ", function() {
var results103 = new Array();
for (var i = 0; i < 103; i++) {
results103.push(i);
}
var fakeSearchForm = { $valid: true };
var isSearchValidStub = sinon.stub(sut, "isSearchCriteriaValid").returns(true);
var deferred = $q.defer();
var promise = deferred.promise;
var dsStub = sinon.stub(inSearchDataSvc, "getInsureds").returns(promise);
var resolveStub = sinon.stub(deferred, "resolve");
//how do i call resolve and pass in results103
sut.performSearch(fakeSearchForm);
sinon.assert.calledOnce(isSearchValidStub);
sinon.assert.calledOnce(dsStub);
sinon.assert.called(resolveStub);
expect(sut.searchResults.length).toBe(100);
});
存根myClass.myFunction不同意我们现在使用的承诺。 [Promises/A + say](https://promisesaplus.com/)“只有当执行上下文堆栈只包含平台代码时,才能调用onFulfilled或onRejected”。 – Kos
实际上,这是我做的:var resolveStub = sinon.stub(deferred,“resolve”)。returns({then:function(){}});' –