將 for 循環變成 foreach 循環不起作用

Question

我正在嘗試將for循環變成forEach循環，但它似乎不起作用......

這是我的代碼：

const townDataURL = "[some link I probably can't disclose...]"
const towns2get = [
    "Preston",
    "Fish Haven",
    "Soda Springs"
]

fetch(townDataURL)
    .then((response) => {
        return response.json()
    })
    .then((jsonData) => {
        const towns = jsonData["towns"].filter((item) => {
            // for (let i = 0; i<towns2get.length; i++) {
            //     if (item.name == towns2get[i]) {
            //         return item
            //     }
            // }
            return (towns2get.forEach(elem => {
                return ( (item.name == elem) ? (item) : "Hello?" )
            }))
        })
        console.log(towns)
    })

當我運行注釋代碼時，它給了我這個：

(3) [{…}, {…}, {…}]
0: {name: "Fish Haven", photo: "fishhaven.jpg", motto: "This is Fish Heaven.", yearFounded: 1864, currentPopulation: 501, …}
1: {name: "Preston", photo: "preston.jpg", motto: "Home of Napoleon Dynamite.", yearFounded: 1866, currentPopulation: 5204, …}
2: {name: "Soda Springs", photo: "sodasprings.jpg", motto: "Historic Oregon Trail Oasis. The Soda is on Us.", yearFounded: 1858, currentPopulation: 2985, …}
length: 3
__proto__: Array(0)

這正是我想要的，但我想簡化我的代碼......我現在擁有的是：

[]
length: 0
__proto__: Array(0)

我做了一些調試，我知道我的條件語句與三元運算符工作正常，它正在返回一個值......但我似乎無法弄清楚為什么它沒有將它返回到filter方法...

這樣不行嗎？ 或者我是否必須以某種方式將forEach與filter放在一起？

感謝您的任何幫助！

Answer 1

這里更好的方法是在過濾方法中使用.includes function

const townDataURL = "[some link I probably can't disclose...]"
const towns2get = [
    "Preston",
    "Fish Haven",
    "Soda Springs"
]

fetch(townDataURL)
    .then((response) => {
        return response.json()
    })
    .then((jsonData) => {
        const towns = jsonData["towns"].filter((item) => {
            if(towns2get.includes(item.name) > -1) return true;
            else return false;
        })
        console.log(towns)
    })

Answer 2

Cris G 給了我答案：

.forEach()不返回任何內容，因此您的.filter()回調為每個元素返回undefined ，這是一個錯誤的值。 使用filter(item => towns2get.includes(item.name)) – Chris G

所以代碼應該是：

const townDataURL = "[some link I probably can't disclose...]"
const towns2get = [
    "Preston",
    "Fish Haven",
    "Soda Springs"
]

fetch(townDataURL)
    .then((response) => {
        return response.json()
    })
    .then((jsonData) => {
        const towns = jsonData["towns"].filter(item => towns2get.includes(item.name))
        console.log(towns)
    })