Question

我正在尝试将具有三个属性的对象数组解构为三个单独的数组，例如

可能像

const {wlAddresses,wlTokens,wlTickets} = Object.map()

或

const [wlAddresses,wlTokens,wlTickets] = Object.map()

Object.map（）返回类似这样的内容

[{ wlAddresses: '23',
    wlTokens: 1,
    wlTickets: 3 },
  { wlAddresses: '24',
    wlTokens: 1,
    wlTickets: 2 },
  { wlAddresses: '25',
    wlTokens: 1,
    wlTickets: 3 }]

我尝试使用该方法，它只返回第一个对象，而不是之后的对象。我知道我可以通过映射对象并将所有内容作为单独的数组返回来解决此问题，但也许我可以使用解构来执行此操作。

注意：它只是一个可以做或不做的问题，我不是强迫回答

Answer 1

当你有一个对象或数组并且需要提取特定项目时，解构是很好的，但这不是这里的情况。您的数据不是简单的对象或数组 - 它是一个对象数组。你不可能通过一个简单的任务来做到这一点。您需要将数据转换为所需的格式。例如：

let arr = [{
    wlAddresses: '23',
    wlTokens: 1,
    wlTickets: 3
  },
  {
    wlAddresses: '24',
    wlTokens: 1,
    wlTickets: 2
  },
  {
    wlAddresses: '25',
    wlTokens: 1,
    wlTickets: 3
  }
]

let r = arr.reduce((acc, curr) => {
    for ([key, value] of Object.entries(curr)) {
        if (! acc[key]) acc[key] = []
        acc[key].push( curr[key])
    }
    return acc
}, {})


const {wlAddresses,wlTokens,wlTickets} = r

console.log(wlAddresses,wlTokens,wlTickets)

Answer 2

假设数组中的每个对象都具有相同的键，您可以只映射它。

＆＃13;

const data = [{wlAddresses:'23',wlTokens:1,wlTickets:3},{wlAddresses:'24',wlTokens:1,wlTickets:2},{wlAddresses:'25',wlTokens:1,wlTickets:3}];
    
const r = Object.keys(data[0]).map((v) => ({ [v]: data.map((c) => c[v]) }));

console.log(JSON.stringify(r));

＆＃13;

Answer 3

您可以使用嵌套循环和Object.entries()

let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];

let entries = Object.entries(o);
    let res = Array.from({length:entries.length}).fill([]);
    for (let [, {wlAddresses,wlTokens,wlTickets}] of entries) {   
      Object.entries({wlAddresses,wlTokens,wlTickets})
      .forEach(([, prop], index) => {
        res[index] = [...res[index], prop]
      })
    }
    
let [wlAddresses,wlTokens,wlTickets] = res;

console.log(wlAddresses,wlTokens,wlTickets);

单独使用解构分配

let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];

let len = o.length;

let [wlAddresses,wlTokens,wlTickets] = Array.from({length:len}, () => []);

let i = 0;

if (i < len) {
  ({wlAddresses:wlAddresses[wlAddresses.length]
  , wlTokens:wlTokens[wlTokens.length]
  , wlTickets:wlTickets[wlTickets.length]} = o[i]);
  console.log(wlAddresses,wlTokens,wlTickets);
  ++i;
} else {
  console.log("done");
  console.log(wlAddresses,wlTokens,wlTickets);
}

if (i < len) {
  ({wlAddresses:wlAddresses[wlAddresses.length]
  , wlTokens:wlTokens[wlTokens.length]
  , wlTickets:wlTickets[wlTickets.length]} = o[i]);
  console.log(wlAddresses,wlTokens,wlTickets);
  ++i;
} else {
  console.log("done");
  console.log(wlAddresses,wlTokens,wlTickets);
}

if (i < len) {
  ({wlAddresses:wlAddresses[wlAddresses.length]
  , wlTokens:wlTokens[wlTokens.length]
  , wlTickets:wlTickets[wlTickets.length]} = o[i]);
  console.log(wlAddresses,wlTokens,wlTickets);
  ++i;
} else {
  console.log("done");
  console.log(wlAddresses,wlTokens,wlTickets);
}

if (i < len) {
  ({wlAddresses:wlAddresses[wlAddresses.length]
  , wlTokens:wlTokens[wlTokens.length]
  , wlTickets:wlTickets[wlTickets.length]} = o[i]);
  console.log(wlAddresses,wlTokens,wlTickets);
  ++i;
} else {
  console.log("done");
  console.log(wlAddresses,wlTokens,wlTickets);
}

在第二个硬编码片段中使用带代码的循环

let o = [{"wlAddresses":"23","wlTokens":1,"wlTickets":3},{"wlAddresses":"24","wlTokens":1,"wlTickets":2},{"wlAddresses":"25","wlTokens":1,"wlTickets":3}];

let len = o.length;

let [wlAddresses,wlTokens,wlTickets] = [[], [], []];

let i = 0;

while (i < len) {
  ({wlAddresses:wlAddresses[wlAddresses.length]
  , wlTokens:wlTokens[wlTokens.length]
  , wlTickets:wlTickets[wlTickets.length]} = o[i]);
  ++i;
}
  
console.log("done");
console.log(wlAddresses,wlTokens,wlTickets);

使用es6解构对象数组

3 个答案: