Sinon 监视函数表达式

Sinon spy on function expression

是否可以让 sinon 监视函数表达式?以这段代码为例。

function one() { return 1; }
function two() { return 2; }
function three() { return 3; }

function myMethod() {
  var n1 = one();
  var n2 = two();
  var n3 = three();
  return n1 + n2 + n3;
}


QUnit.module('My test');

QUnit.test('testing functions', (assert) => {
  assert.expect(3);
  
  const spyOne = sinon.spy(one);
  const spyTwo = sinon.spy(two);
  const spyThree = sinon.spy(three);
 myMethod();

  assert.ok(spyOne.called, "called one");
  assert.ok(spyTwo.called, "called two");
  assert.ok(spyThree.called, "called three");
  
  sinon.restore();
});

即使我调用 myMethod() 并且我在 one - two - three 上有间谍,我仍然在 one.called 上得到假(twothree 也是如此)

我在这里错过了什么?

谢谢!

调用 sinon.spy(fn) 不会改变 fn,它只会创建一个 new 函数(间谍),它将调用 fn

为了能够测试 onetwothree,您需要用间谍替换这些函数(或者更确切地说,它们的引用),然后再恢复它们:

// keep references to the original functions
var _one   = one;
var _two   = two;
var _three = three;

// replace the original functions with spies
one   = sinon.spy(one);
two   = sinon.spy(two);
three = sinon.spy(three);

// call our method
myMethod();

// test
assert.ok(one.called,   "called one");
assert.ok(two.called,   "called two");
assert.ok(three.called, "called three");

// restore the original functions
one   = _one;
two   = _two;
three = _three;

但这并不理想,如果可能的话,我可能会将所有功能组合到一个对象中。这样也能让诗乃自己恢复原貌。