从特定键中提取键值[重复]

Question

假设我有 json 为：

const details = {
    "something1":1,
    "something2":20,
    "something3":123,
    "something4":[{"a":"apple","b":"ball"},
    "something5":1211
}

为此我尝试过：

let final = {}

for (let keys in details) {
    // console.log(keys)
    if (keys == 'something1'|| 'something2'|| 'something3') {
        final[keys] = details[keys]
    }
}

但它给我的结果与 details

相同

预期的 O/p 应该是 ...

console.log(final) => {"something1":1,"something2":20,"something3":123}

Answer 1

您可以filter()对象的条目，然后使用Object.fromEntries将其转换回对象

const details = {
    "something1":1,
    "something2":20,
    "something3":123,
    "something4":[{"a":"apple","b":"ball"}],
    "something5":1211
}

const removeProps = (obj, props) => (
     Object.fromEntries(Object.entries(obj).filter(([k, _]) => !props.includes(k)))
)

const res = removeProps(details, ['something1', 'something2']);
console.log(res)

Answer 2

问题出在keys === 'something1'|| 'something2'|| 'something3':

const details = {
    "something1":1,
    "something2":20,
    "something3":123,
    "something4":[{"a":"apple","b":"ball"}],
    "something5":1211,
    }
    
let final = {}

for (let keys in details) {
    // console.log(keys)
    if (keys === 'something1'|| keys === 'something2'|| keys==='something3') {
        final[keys] = details[keys]
    }
}

console.log(final)

Answer 3

您可以使用解构函数并获取新对象。

const
    getParts = ({ something1, something2, something3 }) => ({ something1, something2, something3 }),
    details = { something1: 1, something2: 20, something3: 123, something4: [{ a: "apple", b: "ball" }], something5: 1211 },
    final = getParts(details);

console.log(final);

动态方法

const
    details = { something1: 1, something2: 20, something3: 123, something4: [{ a: "apple", b: "ball" }], something5: 1211 },
    wanted = ['something1', 'something2', 'something3'],
    final = Object.fromEntries(wanted.map(k => [k, details[k]]));

console.log(final);

Answer 4

已经有很好的答案可以说明如何解决问题，但它们没有解释为什么您的代码不起作用。

原因是你误解了这个布尔条件的作用：

keys == 'something1'|| 'something2'|| 'something3')

不是的意思是：key 是任何一个。
这确实意味着：

1. 如果keys == 'something1'
2. 或者如果'something2'
3. 或者如果'something3'

您的第二个和第三个 or 表达式只是纯字符串。
并且任何非空字符串的计算结果都是布尔值true。

// non-empty strings are boolean TRUE
console.log(`Boolean: 'something2'`, Boolean('something2'));
// empty strings are boolean FALSE
console.log(`Boolean: ''`, Boolean(''));

因此，您的第二个和第三个 OR 条件始终为 true，因此整个条件为 true。

因此，如果您想坚持显式比较，请使用：

if (keys === 'something1'|| keys === 'something2'|| keys==='something3') {

顺便说一句：您可能想使用=== 而不是==：请参阅Javascript === (triple equals)