4 回答

TA貢獻1811條經驗 獲得超5個贊
forEach是數組的方法,同時你oldObj是對象
首先你必須將它轉換為數組,這里我們可以做的是將對象轉換為鍵值對數組
使用 withmap可以使代碼更短
const oldObj = {
Georgia: {
notes: "lorem ipsum",
lat: "32.1656",
long: "82.9001",
},
Alabama: {
notes: "lorem ipsum",
lat: "32.3182",
long: "86.9023",
},
}
const res = Object.entries(oldObj).map(([name, obj]) => ({ name, ...obj }))
console.log(res)

TA貢獻2003條經驗 獲得超2個贊
const oldObj = {
Georgia : {
notes: "lorem ipsum",
lat: "32.1656",
long: "82.9001"
},
Alabama : {
notes: "lorem ipsum",
lat: "32.3182",
long: "86.9023"
}
}
const desireArray = Object.keys(oldObj).map((key) => ({ name: key, ...oldObj[key] }));
解釋一下
const keys = Object.keys(oldObj);
const desireArray = keys.map((key) => {
return {
name: key,
notes: oldObj[key].notes,
lat: oldObj[key].lat,
long: oldObj[key].long
}
});

TA貢獻1799條經驗 獲得超8個贊
使用查找條目
Object.entries
destructuring
通過將鍵添加到累加器來減少當前對象推入結果數組
const oldObj = {
? ? Georgia : {
? ? ? ? notes: "lorem ipsum",
? ? ? ? lat: "32.1656",
? ? ? ? long: "82.9001"
? ? },
? ? Alabama : {
? ? ? ? notes: "lorem ipsum",
? ? ? ? lat: "32.3182",
? ? ? ? long: "86.9023"
? ? }
};
const result = Object.entries(oldObj).reduce((acc, curr) => {
? ? const [key, val] = curr;
? ??
? ? acc.push({
? ? ? ? name: key,
? ? ? ? ...val
? ? });
? ? return acc;
}, []);
console.log(result);

TA貢獻1810條經驗 獲得超4個贊
const oldObj = {
Georgia : {
notes: "lorem ipsum",
lat: "32.1656",
long: "82.9001"
},
Alabama : {
notes: "lorem ipsum",
lat: "32.3182",
long: "86.9023"
}
};
function convertObjToArr(obj) {
let result = [];
for(let key in obj) {
result.push({name: key, ...obj[key]});
}
return result;
}
console.log(convertObjToArr(oldObj));
或嘗試其他簡單的解決方案
return Object.keys(obj).map(item => ( {name: item, ...obj[item]} ));
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