如何确定变量是否为JavaScript错误？

Question

我在数组中混合了几个变量，其中一些可能是通用错误或自定义错误（即，由var v = new Error()或var v = new MyCustomError() ）。

是否存在将Error实例与任何其他变量区分开的通用方法？ 谢谢。

编辑：自定义错误的格式为：

function FacebookApiException(res) { this.name = "FacebookApiException"; this.message = JSON.stringify(res || {}); this.response = res; } FacebookApiException.prototype = Error.prototype;

Answer 1

您的FacebookApiException无法从Error正确继承，我建议您执行以下操作：

function FacebookApiException(res) {
    //re use Error constructor
    Error.call(this,JSON.stringify(res || {}));
    this.name = "FacebookApiException";
    this.response = res;
}
//Faceb... is an Error but Error is not Faceb...
//  so you can't set it's prototype to be equal to Error
FacebookApiException.prototype = Object.create(Error.prototype);
//for completeness, not strictly needed but prototype.constructor
// is there automatically and should point to the right constructor
// re assigning the prototype has it pointing to Error but should be
// FacebookApiException
FacebookApiException.prototype.constructor = FacebookApiException;

Answer 2

在大多数情况下，您可以使用instanceof运算符，例如

var err = new Error();
if (err instanceof Error) {
  alert('That was an error!');
}

看到这个jsfiddle 。

Mozilla开发人员网络（MDN）在此处具有有关instanceof运算符的更多详细信息，说明：

instanceof运算符用于测试对象原型链中是否有builder.prototype。

这样，给定以下输入，您的输出将反映在注释中：

function C(){} // defining a constructor
function D(){} // defining another constructor

var o = new C();
o instanceof C; // true, because: Object.getPrototypeOf(o) === C.prototype
o instanceof D; // false, because D.prototype is nowhere in o's prototype chain
o instanceof Object; // true, because:
C.prototype instanceof Object // true

C.prototype = {};
var o2 = new C();
o2 instanceof C; // true
o instanceof C; // false, because C.prototype is nowhere in o's prototype chain anymore

D.prototype = new C(); // use inheritance
var o3 = new D();
o3 instanceof D; // true
o3 instanceof C; // true