我環顧四周，但沒有完全找到我要找的東西：我想在不合并的情況下附加到子物件值。
例如：
const ADDRESS = {
address1: {
line1: 'blah 1'
// ...
},
address2: {
line1: 'blah 2'
// ...
},
address3: {
line1: 'blah3'
// ...
},
address4: {
line1: 'blah3'
// ...
}
}
如果我使用擴展運算子，它們會被合并，我只想將它們附加
const addressesappended = {
...ADDRESS.address1,
...ADDRESS.address2
}
期望的結果：
{
address1: {
line1: 'blah 1'
// ...
},
address2: {
line1: 'blah 2'
// ...
}
}
uj5u.com熱心網友回復：
您只能獲取值。
const
ADDRESS = { address1: { line1: 'blah 1' }, address2: { line1: 'blah 2' }, address3: { line1: 'blah3' }, address4: { line1: 'blah3' } };
result = Object.values(ADDRESS);
console.log(result);uj5u.com熱心網友回復：
您可以使用克隆整個物件
const result = Object.assign({}, ADDRESS);
如果匹配“地址”，您可以手動分配子物件
let result = {};
for (const [key, value] of Object.entries(ADDRESS)) {
if(key.includes("address")) result[key] = value;
}
uj5u.com熱心網友回復：
您將需要在目標物件中宣告相關傳播資料的鍵。
const addresses = {
address1: { line1: 'blah 1' },
address2: { line1: 'blah 2' },
address3: { line1: 'blah3' },
address4: { line1: 'blah3' }
};
const addressesAppended = {
address1: { ...addresses.address1 },
address2: { ...addresses.address2 }
};
console.log(addressesAppended);.as-console-wrapper { top: 0; max-height: 100% !important; }轉載請註明出處，本文鏈接：https://www.uj5u.com/caozuo/405743.html
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