Question

我有两个数组，一个是原始数组，第二个是原始数组的副本。我将一些新项目放入数组中，并更新其中一些，并希望将其与复制数组进行比较，并希望消除复制数组中的那些项目，并将这些项目保留在Id = null

 var original = [
    {
        "Id": 1,
        "BrandConstruct": 265,
        "YearPlanData": "a"
    },
    {   "Id": 2,
        "BrandConstruct": 236,
        "YearPlanData": "c"
    },
    {   "Id": 3,
        "BrandConstruct": 376,
        "YearPlanData": "b"
    },
    {   "Id": null,
        "BrandConstruct": 476,
        "YearPlanData": "e"
    }, 
    {   "Id": null,
        "BrandConstruct": 576,
        "YearPlanData": "f"
    }

]

    var copy = [
    {
        "Id": 1,
        "BrandConstruct": 165,
        "YearPlanData": "a"
    },
    {   "Id": 2,
        "BrandConstruct": 236,
        "YearPlanData": "c"
    },
    {   "Id": 3,
        "BrandConstruct": 376,
        "YearPlanData": "b"
    }

]

这是两个具有属性ID，BrandConstruct和YearPlanData的数组使用复制数组，但将数据保留在Id = null的地方所以我想在比较后得到这种类型的结果，它必须在IE上可以工作

    var original = [
    {
        "Id": 1,
        "BrandConstruct": 265,
        "YearPlanData": "a"
    },

    {   "Id": null,
        "BrandConstruct": 476,
        "YearPlanData": "e"
    }, 
    {   "Id": null,
        "BrandConstruct": 576,
        "YearPlanData": "f"
    }

]

Answer 1

const newOriginal = original.filter(obj => {
    return !copy.find(copyObj => JSON.stringify(copyObj) === JSON.stringify(obj))
});

这应该可以在Internet Explorer上运行：

const newOriginal = original.filter(function(obj) {
    return !copy.filter(function(copyObj) { return JSON.stringify(copyObj) === JSON.stringify(obj)}).length
});

Answer 2

您可以遍历original数组，并将不属于copy数组的元素推到新数组。

function isInArray(item, array) {
  return JSON.stringify(array).indexOf(JSON.stringify(item)) > -1
}

function getFilteredArray(original, copy) {
  var newOriginal = [];

  for (var i = 0; i < original.length; i++) {
    var item = original[i];
    if (!isInArray(item, copy) || item.id === null) newOriginal.push(item);
  }

  return newOriginal;
}

var original = [{
    "Id": 1,
    "BrandConstruct": 265,
    "YearPlanData": "a"
  },
  {
    "Id": 2,
    "BrandConstruct": 236,
    "YearPlanData": "c"
  },
  {
    "Id": 3,
    "BrandConstruct": 376,
    "YearPlanData": "b"
  },
  {
    "Id": null,
    "BrandConstruct": 476,
    "YearPlanData": "e"
  },
  {
    "Id": null,
    "BrandConstruct": 576,
    "YearPlanData": "f"
  }

]

var copy = [{
    "Id": 1,
    "BrandConstruct": 165,
    "YearPlanData": "a"
  },
  {
    "Id": 2,
    "BrandConstruct": 236,
    "YearPlanData": "c"
  },
  {
    "Id": 3,
    "BrandConstruct": 376,
    "YearPlanData": "b"
  }

]

console.log(getFilteredArray(original, copy));

Answer 3

这是为了处理对象，您可以在找不到要搜索的属性的索引时将它们推入一个空数组，从而创建一个唯一值的新数组：

   for (var key in array_of_objects) {
     var index = empty_array.findIndex(x => x.attribute == array_of_objects.attribute)
        if (index === -1){
            empty_array.push(d);
        }
   }

Answer 4

这可行，但是我认为您应该等待更好的解决方案。

var original = [{
  "Id": 1,
  "BrandConstruct": 265,
  "YearPlanData": "a"
}, {
  "Id": 2,
  "BrandConstruct": 236,
  "YearPlanData": "c"
}, {
  "Id": 3,
  "BrandConstruct": 376,
  "YearPlanData": "b"
}, {
  "Id": null,
  "BrandConstruct": 476,
  "YearPlanData": "e"
}, {
  "Id": null,
  "BrandConstruct": 576,
  "YearPlanData": "f"
}];

var copy = [{
  "Id": 1,
  "BrandConstruct": 165,
  "YearPlanData": "a"
}, {
  "Id": 2,
  "BrandConstruct": 236,
  "YearPlanData": "c"
}, {
  "Id": 3,
  "BrandConstruct": 376,
  "YearPlanData": "b"
}];
copy = copy.map(function(el) {
  return JSON.stringify(el);
});

original = original.map(function(el) {
  return JSON.stringify(el);
}).filter(function(els) {
  return copy.indexOf(els) == -1;
}).map(function(s) {
  return JSON.parse(s);
});
console.log(original);

从数组中删除代表另一个数组的重复数据

4 个答案: