TA貢獻1802條經驗 獲得超4個贊
您可以遞歸地構建對象任意數量的嵌套對象。因此，此功能與您的情況無關：
var enrollment = {
user: {
id: 'string',
name: 'string'
},
finished: 'boolean',
path: 'boolean'
}
var enrollment2 = {
user: {
id: 'string',
name: 'string'
},
test: {
test1: {
test2: {
val0:'val0',
test4: { //3rd level nested object for example
val1: 'val1',
val2: 'val2'
}
}
}
},
finished: 'boolean',
path: 'boolean'
}
const flat = (obj, out) => {
Object.keys(obj).forEach(key => {
if (typeof obj[key] == 'object') {
out = flat(obj[key], out) //recursively call for nesteds
} else {
out[key] = obj[key] //direct assign for values
}
})
return out
}
console.log(flat(enrollment, {}))
console.log(flat(enrollment2, {}))
TA貢獻1848條經驗 獲得超2個贊
我需要一些東西來避免重寫原始對象中不同級別的同名鍵。所以我寫了以下內容：
const flattenObject = (obj, parentKey = '') => {
if (parentKey !== '') parentKey += '.';
let flattened = {};
Object.keys(obj).forEach((key) => {
if (typeof obj[key] === 'object' && obj[key] !== null) {
Object.assign(flattened, flattenObject(obj[key], parentKey + key))
} else {
flattened[parentKey + key] = obj[key]
}
})
return flattened;
}
var test = {
foo: 'bar',
some: 'thing',
father: {
son1: 'son1 value',
son2: {
grandchild: 'grandchild value',
duplicatedKey: 'note this is also used in first level',
},
},
duplicatedKey: 'note this is also used inside son2',
}
let flat = flattenObject(test);
console.log(flat);
// how to access the flattened keys:
let a = flat['father.son2.grandchild'];
console.log(a);
還檢查對象是否為空，因為我在使用時遇到了一些問題。
TA貢獻1828條經驗 獲得超13個贊
使用遞歸和歸約。
請注意，如果 value 本身是一個包含對象的數組，您可能需要!Array.isArray(value)根據您的情況添加另一個檢查
function flatObj(obj) {
return Object.entries(obj).reduce(
(flatted, [key, value]) =>
typeof value == "object"
? { ...flatted, ...flatObj(value) }
: { ...flatted, [key]: value },
{}
);
}
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