[英]Sinon spy on function expression
是否可以使sinon監視函數表達式? 例如看下面的代碼。
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(); });
即使我調用myMethod()
並且對one - two - three
有間諜,但對one - two - three
仍然是假的。被one.called
(對two
和three
相同)
我在這里想念什么?
謝謝!
調用sinon.spy(fn)
不會更改fn
,它只是創建一個將調用fn
的新函數(間諜)。
為了能夠測試one
, two
, three
,您需要用間諜替換這些功能(或更確切地說,它們的引用),然后再將其還原:
// 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;
不過這並不理想,如果可能的話,我可能會將所有功能分組到一個對象中。 那也將使Sinon能夠恢復原件本身。
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