如何使用Jasmine模拟链式承诺？

Question

I am writing a unit test to a method that contains this piece of code:

Name.get($scope.nameId).then(function(name){
    Return name;
}).then(doSomething);

The function doSomething(name) looks like this.

function doSomething(name){
    addNametoList(name);
}

I don't need to test this part of the code. Since I can't just ignore it in my test (or can I?), I need to mock it. I mocked the first promise

 spyOn(mockName, 'get').and.returnValues($q.resolve({"Mike"})); 

and thought that it would propagate through the second then(doSomething) but name is undefined in function addNametoList .

I think I have to also mock doSomething but I have no idea how to chain them together.

Answer 1

I don't need to test this part of the code. Since I can't just ignore it in my test (or can I?), I need to mock it. I mocked the first promise

IMO, there is no need to test such setups as a whole. Test each unit individually.

For example

A = () => {
  ...
}

B = () => {
  ...
}

C = () => {
  ...
}

Now we have F, which calls AB & C

F = () => A(B(C))

Or in your case

F = () => A().then(B).then(C)

We can test F, but it will require some setup, and be fragile. Its better to test A, B & C (which give at a proper coverage for F) and ignore F.

Answer 2

Your (partial) controller code

$scope.list = [];

function addNameToList(name) {
  $scope.list.push(name);
}

function doSomething(name){
  addNametoList(name);
}

$scope.functionToTest = function() {
  Name.get($scope.nameId).then(function(name){
    return name;
  }).then(doSomething);
};

Your test

it('should add the name to the list when Name.get() is resolved', function() {
  // Arrange
  var getDeferred = $q.defer();
  spyOn(mockName, 'get').and.returnValue(getDeferred.promise);

  // Act
  $scope.functionToTest();
  getDeferred.resolve('some name that should be in the list');
  $scope.$apply();

  // Assert
  expect($scope.list).toEqual(['some name that should be in the list']);
});

Comment

Note that the following part of your code accomplishes nothing, and can be removed without affecting anything.

.then(function(name){
  return name;
})