Question

我想要通过对象中键的值来过滤对象的对象。例如，我的对象看起来像：

const Financials = {
  xxxxx: {
    creditid: "yyyy",
    aggRevs: 2000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2015
  },
  zzzz: {
    creditid: "yyyy",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  },
  aaaa: {
    creditid: "bbbb",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  }
};

我希望能够通过creditid过滤对象。例如，我想返回一个包含所有具有“yyyy”信用的对象的对象。

var { creditid: "yyyy" } = Financials;

结果如下：

{
  xxxxx: {
    creditid: "yyyy",
    aggRevs: 2000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2015
  },
  zzzz: {
    creditid: "yyyy",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  }
}

这是否可以使用解构？

Answer 1

为此，您必须遍历Financials的每个属性，如下所示：

const Financials = {
  xxxxx: {
    creditid: "yyyy",
    aggRevs: 2000,
    aggexpenses: 1000,
    dateOf: '12 / 31 / 2015'
  },
  zzzz: {
    creditid: "yyyy",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: '12 / 31 / 2016'
  },
  aaaa: {
    creditid: "bbbb",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: '12 / 31 / 2016'
  }
};

var resultFinancials = {};

for (var financial in Financials) {
    if (Financials.hasOwnProperty(financial)) {
        if(Financials[financial] && Financials[financial].creditid =='yyyy' ){
            resultFinancials[financial] = Financials[financial];
        }
    }
}

console.log(resultFinancials)

Answer 2

就解构而言，我不知道这可以做到，只是因为解构更像.map()而不是.filter()。但是，您可以使用.reduce()函数轻松地执行此操作，如下所示：

＆＃13;

const Financials = {
  xxxxx: {
    creditid: "yyyy",
    aggRevs: 2000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2015
  },
  zzzz: {
    creditid: "yyyy",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  },
  aaaa: {
    creditid: "bbbb",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  }
};

var filtered = Object.keys(Financials).reduce((res, key) => {
  if (Financials[key].creditid === "yyyy") {
    res[key] = Financials[key]
  }
  return res;
}, {});

console.log(filtered);

＆＃13;

Answer 3

您可以只过滤对象条目，然后将其映射回新对象

＆＃13;

const Financials = {
  xxxxx: {
    creditid: "yyyy",
    aggRevs: 2000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2015
  },
  zzzz: {
    creditid: "yyyy",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  },
  aaaa: {
    creditid: "bbbb",
    aggRevs: 1000,
    aggexpenses: 1000,
    dateOf: 12 / 31 / 2016
  }
};

let arr    = Object.entries(Financials).filter( set => set[1].creditid === "yyyy");
let result = Object.assign(...arr.map(d => ({[d[0]]: d[1]})))

console.log(result)

＆＃13;

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

＆＃13;

带过滤器的JS对象解构

3 个答案: