Question

什么是最好的转换方式

const mockResults = [
    [{ user: { firstName: '1', lastName: '1' }, status: 'WRONG' }],
    [{ user: { firstName: '2',lastName: '2' }, status: 'WRONG' }],
    [{ user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }]
];

到

const mockResults = [
    { user: { firstName: '1', lastName: '1' }, status: 'WRONG' },
    { user: { firstName: '2',lastName: '2' }, status: 'WRONG' },
    { user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }
];

整个任务是将mockResults转换为requiredFormat，这就是为什么我需要删除嵌套数组的原因：

const requiredFormat = [
  {
    status: 'WRONG',
    data: [{ user: {firstName: '1', lastName: '1'}}, { user: {firstName: '2', lastName: '2'}}],
  },
  {
    status: 'CORRECT',
    data: [{ user: {firstName: '3', lastName: '3'}}],
  },
];

这是我到目前为止尝试过的： https://jsfiddle.net/9uLje3sg/

谢谢！

Answer 1

您可以使用Array javascript对象中的flat方法。 https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/flat

如果要更改数据源并更改其形状，则使用map和reduce方法可以为您提供帮助。

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

在您的精确示例中，reduce将适合您在按状态属性创建新的对象分组时使用。

const mockResults = [
    [{ user: { firstName: '1', lastName: '1' }, status: 'WRONG' }],
    [{ user: { firstName: '2',lastName: '2' }, status: 'WRONG' }],
    [{ user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }]
];
const flattedAndReduced = mockResults.flat().reduce( (acc, curr)=>{
  const statusIndex = { 'WRONG' : 0, 'CORRECT': 1 };
  acc[statusIndex[curr.status]].data.push({ user: curr.user} );
  return acc;
}, [
    {
      status: 'WRONG',
      data: [],
    },
    {
      status: 'CORRECT',
      data: [],
    }
  ]
  );
console.log(flattedAndReduced);

Answer 2

按如下所示使用函数map，该函数将返回包含所需对象的数组。

let result = mockResults.map(([user]) => user);

该方法假设原始数组中每个数组只有一个索引。

根据requiredFormat
的方法

您可以使用函数reduce进行分组，而使用函数Object.values进行所需的输出。

const mockResults = [
    [{ user: { firstName: '1', lastName: '1' }, status: 'WRONG' }],
    [{ user: { firstName: '2',lastName: '2' }, status: 'WRONG' }],
    [{ user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }]
];

let requiredFormat = Object.values(mockResults.reduce((a, [{user, status}]) => {
  (a[status] || (a[status] = {data: [], status})).data.push(user);
  return a;
}, Object.create(null)));

console.log(requiredFormat);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

只需使用here即可从第一个索引返回对象。

请注意：用Array.prototype.map()声明的变量无法修改，请改用* const。

let mockResults = [
    [{ user: { firstName: '1', lastName: '1' }, status: 'WRONG' }],
    [{ user: { firstName: '2',lastName: '2' }, status: 'WRONG' }],
    [{ user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }]
];

mockResults = mockResults.map(i => i[0]);
console.log(mockResults);

Answer 4

const mockResults = [
    [{ user: { firstName: '1', lastName: '1' }, status: 'WRONG' }],
    [{ user: { firstName: '2',lastName: '2' }, status: 'WRONG' }],
    [{ user: { firstName: '3',lastName: '3' }, status: 'CORRECT' }]
];
const requiredFormat = [
  {status: 'WRONG', data: []},
  {status: 'CORRECT', data: []},
];

for(let [{user,status}] of mockResults) {  
  requiredFormat[ status==="WRONG" ? 0 : 1].data.push({user});
}
console.log(requiredFormat);

从对象中删除嵌套数组

4 个答案: