Question

我有以下代码示例来获取函数名称，该函数名称在参数中传递以用于回调目的。

var operation = {
    display : function(fun) {
        console.log('operation "'+ fun.toString() +'" performing' );
        //here i am getting
        /*operation "function () { 
                console.log('operation "subtraction" performed' ); }" performing
        */
        // **instead of entire function i would like to get only method name**
        fun.call();
    },
    addition : function() {
        console.log('operation "addition" performed' );
    },
    subtraction : function() {
        console.log('operation "subtraction" performed' );
    },
    multiplication : function() {
        console.log('operation "multiplication" performed' );
    }
}
operation.display(operation.subtraction);
//Output should be
// operation "subtraction" performing
// operation "subtraction" performed
operation.display(operation.multiplication);
//Output should be
// operation "multiplication" performing
// operation "multiplication" performed
operation.display(operation.addition);
//Output should be
// operation "addition" performing
// operation "addition" performed

Answer 1

你的职能是匿名的;即它们是无名函数表达式。在获得他们的名字之前，你必须给他们起名字。

var o = {
    foo: function bar() {} // named function expression, name is "bar"
};
function getName(fn) {
    return fn.name;
}
getName(o.foo); // "bar"

但是，这些功能在您的operation 对象中确实有相关的密钥，但您无法找到对象<的密钥某些东西是肯定的，找到匹配的代码是猜测，因为价值观不一定是唯一的。

function guessKey(val, obj) { var key; for (key in obj) if (obj[key] === val) return key; return null; } guessKey(o.foo, o); // "foo"

Function.prototype.name为IE9 +填充（也许是8;不确定）

if (Function.prototype.name === undefined) { // may be "" so can't use ! Object.defineProperty(Function.prototype, "name", { get: function () { var s = Function.prototype.toString.call(this); return s.slice(9, s.indexOf('(')); } }); }

Answer 2

函数对象包含返回方法名称的“name”属性

<强>更新 AS IE没有实现函数的name属性使用RegEx;）这个小函数可以帮助你：

function getFnName(fn) {
  var f = typeof fn == 'function';
  var s = f && ((fn.name && ['', fn.name]) || fn.toString().match(/function ([^\(]+)/));
  return (!f && 'not a function') || (s && s[1] || 'anonymous');
}

见到这里：https://gist.github.com/dfkaye/6384439

如何使用Javascript将函数名称作为参数传递给回调目的

2 个答案: