2 回答

TA貢獻1818條經驗 獲得超7個贊
假設您希望名稱作為鍵,yxou可以構建條目并從中構建對象。
var tab = [{ firstName: 'John', lastName: 'Doe', situation: 'maried', familyMembers: 5 }, { firstName: 'Jack', lastName: 'Daniel', situation: 'single', familyMembers: 6 }],
result = Object.fromEntries(tab.map(({ firstName, lastName, familyMembers }) =>
[firstName, { [lastName]: familyMembers }]
));
console.log(result);

TA貢獻2037條經驗 獲得超6個贊
map()始終返回函數結果的數組。
您可以改用reduce()
var tab = [{
firstName: 'John',
lastName: 'Doe',
situation: 'maried',
familyMembers: 5,
},
{
firstName: 'Jack',
lastName: 'Daniel',
situation: 'single',
familyMembers: 6,
}
];
var result = tab.reduce((obj, item) => {
obj[item.firstName] = {
[item.lastName]: item.familyMembers
};
return obj;
}, {});
console.log(result);
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