無法返回對象方法名稱

Question

我試圖獲取delicious和love方法的名稱，然後將其放入數組quotes。我需要這樣做，因爲報價將顯示在選擇框中。我嘗試使用Function.name。它適用於Say，但是當我在delicious和love上使用它時，我得到的是空字符串。

這裏有什麼問題？

console.log('>>START') 

function Say(name) { 
    this.name = name 
} 

function Food(name) { 
    this.name = name 
} 

Say.delicious = function(food) { 
    if (!food) food = 'food' 
    console.log('This '+food+' is delicious') 
} 
Say.love = function(food) { 
    if (!food) food = 'food' 
    console.log('I love '+food) 
} 

var pizza = new Food('Pizza') 
var donut = new Food('Donut') 
var quotes = [Say.delicious.name, Say.love.name] 

Say.delicious(donut.name) // This Donut is delicious 
Say.love(pizza.name) // I love Pizza 
console.log(quotes) // ["",""] 
console.log(Say.name) // Say 

Answer 1

這是因爲功能delicious實際上是有Say.delicious參考，如果我理解正確的匿名函數。

你可以通過指定它像這樣使它成爲一個命名函數：

Say.delicious = function delicious(food) { 
    if (!food) food = 'food' 
    console.log('This '+food+' is delicious') 
} 

console.log('>>START') 
 

 
function Say(name) { 
 
    this.name = name 
 
} 
 

 
function Food(name) { 
 
    this.name = name 
 
} 
 

 
Say.delicious = function delicious(food) { 
 
    if (!food) food = 'food' 
 
    console.log('This '+food+' is delicious') 
 
} 
 
Say.love = function love(food) { 
 
    if (!food) food = 'food' 
 
    console.log('I love '+food) 
 
} 
 

 
var pizza = new Food('Pizza') 
 
var donut = new Food('Donut') 
 
var quotes = [Say.delicious.name, Say.love.name] 
 

 
Say.delicious(donut.name) // This Donut is delicious 
 
Say.love(pizza.name) // I love Pizza 
 
console.log(quotes) // ["",""] 
 
console.log(Say.name) // Say