Question

我有一个具有以下结构的对象

{ cardControlItem: 'Tidak',
  cardDirectUseItem: 'Tidak',
  kbUnspscUuid: '6a564b8e-2976-4fde-8759-7951970d7500',
  substoreUuid: '2f2b04bb-8b80-4b1f-b827-bf20311e31ee',
  cardDetailMin: 654,
  cardDetailMax: 65,
  cardDetailIncrement: 754,
  cardDetailPriceOverall: 4534,
  cardDetailPriceUnit: 0,
  ltMeasurementMinUuid: 'fddca37a-d0a3-40a4-8537-e84375b01601',
  ltMeasurementMaxUuid: '2bc6d7d2-5167-4459-9910-a65839008afd' }

我有一个数组中的键列表

['cardControlItem', 'cardDirectUseItem', 'kbUnspscUuid', 'substoreUuid', 'stockCardGroupUuid', 'stockCardBatchUuid']

如何根据指定的数组减少它

预期产出：

{ cardControlItem: 'Tidak',
  cardDirectUseItem: 'Tidak',
  kbUnspscUuid: '6a564b8e-2976-4fde-8759-7951970d7500',
  substoreUuid: '2f2b04bb-8b80-4b1f-b827-bf20311e31ee' }

P.S：我花了3个小时在StackOverflow上找到类似的问题。这类问题的正确标题是什么：D

Answer 1

 const filter = (obj, arr) => Object.assign(...arr.map(el => ({[el]: obj[el]})));

所以可以做到

const result = filter({a:1, b:2, c:3}, ["a", "b"]);

Answer 2

试试这个解决方案。我过滤了obj（filter部分）中的属性，然后遍历存在的属性（reduce部分）并将属性值分配给新对象（{{ 1}}）。

＆＃13;

acc

＆＃13;

Answer 3

。遍历数组并检查对象是否有此键，如果是，则使用此键创建一个新对象，而不是改变原始对象

＆＃13;

var orgObject = {
  cardControlItem: 'Tidak',
  cardDirectUseItem: 'Tidak',
  kbUnspscUuid: '6a564b8e-2976-4fde-8759-7951970d7500',
  substoreUuid: '2f2b04bb-8b80-4b1f-b827-bf20311e31ee',
  cardDetailMin: 654,
  cardDetailMax: 65,
  cardDetailIncrement: 754,
  cardDetailPriceOverall: 4534,
  cardDetailPriceUnit: 0,
  ltMeasurementMinUuid: 'fddca37a-d0a3-40a4-8537-e84375b01601',
  ltMeasurementMaxUuid: '2bc6d7d2-5167-4459-9910-a65839008afd'
}


var keys = ['cardControlItem', 'cardDirectUseItem', 'kbUnspscUuid', 'substoreUuid', 'stockCardGroupUuid', 'stockCardBatchUuid']

var newObj = {};
keys.forEach(function(item) {
  if (orgObject.hasOwnProperty(item)) {
    newObj[item] = orgObject[item]

  }

})

console.log(newObj)

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Answer 4

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var data = { cardControlItem: 'Tidak',
  cardDirectUseItem: 'Tidak',
  kbUnspscUuid: '6a564b8e-2976-4fde-8759-7951970d7500',
  substoreUuid: '2f2b04bb-8b80-4b1f-b827-bf20311e31ee',
  cardDetailMin: 654,
  cardDetailMax: 65,
  cardDetailIncrement: 754,
  cardDetailPriceOverall: 4534,
  cardDetailPriceUnit: 0,
  ltMeasurementMinUuid: 'fddca37a-d0a3-40a4-8537-e84375b01601',
  ltMeasurementMaxUuid: '2bc6d7d2-5167-4459-9910-a65839008afd' };
  
 var keys = ['cardControlItem', 'cardDirectUseItem', 'kbUnspscUuid', 'substoreUuid', 'stockCardGroupUuid', 'stockCardBatchUuid'];
 
 var new_data = {};
 for(var i = 0; i < keys.length; i++){
  if(data.hasOwnProperty(keys[i])){
    new_data[keys[i]] = data[keys[i]];
  }
 }
 
 console.log(new_data);

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Answer 5

试试这个代码段

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let obj = { cardControlItem: 'Tidak',
  cardDirectUseItem: 'Tidak',
  kbUnspscUuid: '6a564b8e-2976-4fde-8759-7951970d7500',
  substoreUuid: '2f2b04bb-8b80-4b1f-b827-bf20311e31ee',
  cardDetailMin: 654,
  cardDetailMax: 65,
  cardDetailIncrement: 754,
  cardDetailPriceOverall: 4534,
  cardDetailPriceUnit: 0,
  ltMeasurementMinUuid: 'fddca37a-d0a3-40a4-8537-e84375b01601',
  ltMeasurementMaxUuid: '2bc6d7d2-5167-4459-9910-a65839008afd' }
  
let  comparator = ['cardControlItem', 'cardDirectUseItem', 'kbUnspscUuid', 'substoreUuid', 'stockCardGroupUuid', 'stockCardBatchUuid'];
  
  
 Object.keys(obj).map((key)=>{
  if(!comparator.includes(key)){
    delete obj[key]
  }
 })
 
 console.log(obj)

＆＃13;

输出 {cardControlItem：＆＃39; Tidak＆＃39;， cardDirectUseItem：＆＃39; Tidak＆＃39;， kbUnspscUuid：＆＃39; 6a564b8e-2976-4fde-8759-7951970d7500＆＃39;， substoreUuid：＆＃39; 2f2b04bb-8b80-4b1f-b827-bf20311e31ee }

根据指定的数组过滤对象道具

5 个答案: