Javascript Typescript:合并由Object.defineProperty创建的对象
我有一个这样的数组:Javascript Typescript:合并由Object.defineProperty创建的对象,javascript,Javascript,我有一个这样的数组: let aa = [ Object.defineProperty({alignedType: "t"}, "f1", {value: 1}), Object.defineProperty({alignedType: "t"}, "f2", {value: 2}), Object.defineProperty({alignedType: "t"}, "f3", {value: 3}), Object.defineProperty({aligne
let aa = [
Object.defineProperty({alignedType: "t"}, "f1", {value: 1}),
Object.defineProperty({alignedType: "t"}, "f2", {value: 2}),
Object.defineProperty({alignedType: "t"}, "f3", {value: 3}),
Object.defineProperty({alignedType: "t"}, "f4", {value: 4})
];
我需要将此数组简化为一个简单的对象,如:
let yyy = aa.reduce((acc, e) => {
return {...acc, ...e};
});
我运行的yyy仅包含一个仅具有alignedType:t的对象。它没有更多的属性
这里有调试器:
我期望的对象是:{alignedType:t,f1:1,f2:2,f3:3,f4:4}
有什么想法吗?您需要设置{enumerable:true}来说明此属性在枚举相应对象上的属性时显示:
设aa=[
Object.defineProperty{alignedType:'t'},'f1',{value:1,enumerable:true},
Object.defineProperty{alignedType:'t'},'f2',{value:1,enumerable:true},
Object.defineProperty{alignedType:'t'},'f3',{value:1,enumerable:true},
Object.defineProperty{alignedType:'t'},'f4',{value:1,enumerable:true},
];
设yyy=aa.reduceac,e=>{
返回{…acc,…e};
};
console.logyyy;