javascript – 用Sinon.js绑定一个类方法

我试图使用sinon.js存根方法但我收到以下错误：

未捕获的TypeError：尝试将未定义的属性sample_pressure包装为函数

我也去了这个问题(Stubbing and/or mocking a class in sinon.js?)并复制并粘贴了代码,但我得到了同样的错误.

这是我的代码：

Sensor = (function() {
  // A simple Sensor class

  // Constructor
  function Sensor(pressure) {
    this.pressure = pressure;
  }

  Sensor.prototype.sample_pressure = function() {
    return this.pressure;
  };

  return Sensor;

})();

// Doesn't work
var stub_sens = sinon.stub(Sensor, "sample_pressure").returns(0);

// Doesn't work
var stub_sens = sinon.stub(Sensor, "sample_pressure", function() {return 0});

// Never gets this far
console.log(stub_sens.sample_pressure());

这是上面代码的jsFiddle(http://jsfiddle.net/pebreo/wyg5f/5/),以及我提到的SO问题的jsFiddle(http://jsfiddle.net/pebreo/9mK5d/1/).

我确保在jsFiddle甚至jQuery 1.9的外部资源中包含sinon.我究竟做错了什么？

解决方法:

您的代码正在尝试在Sensor上存根函数,但您已在Sensor.prototype上定义了该函数.

sinon.stub(Sensor, "sample_pressure", function() {return 0})

基本上与此相同：

Sensor["sample_pressure"] = function() {return 0};

但它很聪明,看不到Sensor [“sample_pressure”]不存在.

那么你想要做的是这样的事情：

// Stub the prototype's function so that there is a spy on any new instance
// of Sensor that is created. Kind of overkill.
sinon.stub(Sensor.prototype, "sample_pressure").returns(0);

var sensor = new Sensor();
console.log(sensor.sample_pressure());

要么

// Stub the function on a single instance of 'Sensor'.
var sensor = new Sensor();
sinon.stub(sensor, "sample_pressure").returns(0);

console.log(sensor.sample_pressure());

要么

// Create a whole fake instance of 'Sensor' with none of the class's logic.
var sensor = sinon.createStubInstance(Sensor);
console.log(sensor.sample_pressure());

标签：javascript,node-js,sinon
来源： https://codeday.me/bug/20190918/1811577.html