Question

我有以下

 {
 bill: [ 
        { satisfy: 'true', comments: '' } 
       ],
 permission_title: [ 
        { satisfy: 'false', comments: '5' } 
       ],
 final_status: [ 
       { satisfy: 'true', comments: '' } 
       ] 
 }

我需要发送：

{
 bill: { satisfy: 'true', comments: '' },
 permission_title: { satisfy: 'false', comments: '5' },
 final_status: { satisfy: 'true', comments: '' } 
 }

获得它的最佳和快捷方式是什么？

Answer 1

使用Object.entries()提取键/值对（条目）。使用Array.map()迭代条目。对于每个条目，返回一个带有键的对象，以及没有包装数组的值。通过spreading合并回Object.assign()对象：

const obj = {"bill":[{"satisfy":"true","comments":""}],"permission_title":[{"satisfy":"false","comments":"5"}],"final_status":[{"satisfy":"true","comments":""}]};

const result = Object.assign(
  ...Object.entries(obj).map(([k, v]) => ({
    [k]: v[0]
  }))
);

console.log(result);

Answer 2

我们假设您的对象名为 obj 。我会这样做：

for(var p in obj) obj[p] = obj[p][0];

＆＃13;

const obj={
 bill: [ 
        { satisfy: 'true', comments: '' } 
       ],
 permission_title: [ 
        { satisfy: 'false', comments: '5' } 
       ],
 final_status: [ 
       { satisfy: 'true', comments: '' } 
       ] 
 }
let obj_new=Object.assign(obj);
for(var p in obj_new) { obj_new[p] = obj_new[p][0] };
console.log(obj_new);

＆＃13;

Answer 3

假设每个值只有一个索引，另一种方法是使用函数reduce和函数Object.keys。

var obj = {  bill: [{    satisfy: 'true',    comments: ''  }],  permission_title: [{    satisfy: 'false',    comments: '5'  }],  final_status: [{    satisfy: 'true',    comments: ''  }]},
    result = Object.keys(obj).reduce((a, k) => (Object.assign(a, {[k]: obj[k][0]})), {});

console.log(result);

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Answer 4

const src = {
 bill: [ 
        { satisfy: 'true', comments: '' } 
       ],
 permission_title: [ 
        { satisfy: 'false', comments: '5' } 
       ],
 final_status: [ 
       { satisfy: 'true', comments: '' } 
       ] 
 };
 
 const dest = Object.keys(src).reduce((prev, key) => {
    prev[key] = src[key][0];
    return prev;
 }, {});
 
 console.log(dest);

Answer 5

我不确定效率，但你也可以使用find功能！

const src = {
 bill: [ 
        { satisfy: 'true', comments: '' } 
       ],
 permission_title: [ 
        { satisfy: 'false', comments: '5' } 
       ],
 final_status: [ 
       { satisfy: 'true', comments: '' } 
       ] 
 };
 
 const dest = Object.keys(src).reduce((prev, key) => {
    if(Array.isArray(src[key])) {
        prev[key] = src[key].find(() => true);
    }
    
    return prev;
 }, {});
 
 console.log(dest);

Javascript数组以对象的最佳方式

5 个答案: