Using Jasmine how to check if a local method has been called?

Question

I am trying to test if a local method ( method1 ) is being called from another method( method2 ). I tried something like this but it does not work as the method1() still has the original definition . The code snippet looks like this:

var ClassA = function () {
    var method1 = function () {
        console.log('method1');
    };
    var method2 = function () {
        method1();
    };
    return {method1: method1, method2: method2}
}

Test case:

it("should call method1 when method2 is called", function () {
    var objectA = new ClassA();
    spyOn(objectA, 'method1').andCallThrough;
    objectA.method2();
    expect(objectA, 'method1').toHaveBeenCalled();
});

Tried overriding method1 to without any success:

objectA.method1 = jasmine.createSpy('aSpy').andCallThrough();

Answer 1

When you call new ClassA() you invoke the ClassA function and get a new object with the related prototype. This means that when you spyOn(ClassA, "doSomething") you're not setting up a spy on the object returned by the new call, but on a possible doSomething function on the ClassA function itself. You should be able to do something like:

it("calls method1", function() {
  var originalConstructor = ClassA,
      spiedObj;
  spyOn(window, 'ClassA').and.callFake(function() {
    spiedObj = new originalConstructor();
    spyOn(spiedObj, 'method1');
    return spiedObj;
  });
  method2();
  expect(spiedObj.method1).toHaveBeenCalled();
});

Please let me know if this works if not we can discuss more.