比较两个对象数组和第二个数组中不存在的筛选器元素

Question

假设我有两个对象数组作为：

const array1 = [
{ name: 'detail1', title: 'detail1' },
{ name: 'detail2 ', title: 'detail2 ' },
{ name: 'detail3', title: 'detail3' },
{ name: 'detail4', title: 'detail4' },
{ name: 'detail5', title: 'detail5' },
{ name: 'detail6', title: 'detail6' },
{ name: 'detail7', title: 'detail7' }
]
const array2 = [
{ name: 'detail1', title: 'detail1' },
{ name: 'detail2 ', title: 'detail2 ' },
{ name: 'detail3', title: 'detail3' },
{ name: 'detail4', title: 'detail4' },
]

我想比较两个数组，即array1和array2，得到缺少的元素aaray2。

为此，我尝试作为：

var absent = array2.filter(e=>!array1.includes(e));

但我无法得到数组2中缺少的元素。

我的预期O/p:

[ { name: 'detail5', title: 'detail5' },
{ name: 'detail6', title: 'detail6' },
{ name: 'detail7', title: 'detail7' }]

这些都是数组2中不存在的元素。

我到底做错了什么？

如果有人需要任何进一步的信息，请告诉我。

Answer 1

您可以用键和值构建一个规范化的对象并过滤对象。

const
array1 = [{ name: 'detail1', title: 'detail1' }, { name: 'detail2 ', title: 'detail2 ' }, { name: 'detail3', title: 'detail3' }, { name: 'detail4', title: 'detail4' }, { name: 'detail5', title: 'detail6' }, { name: 'detail7', title: 'detail7' }, { name: 'detail8', title: 'detail8' }],
array2 = [{ name: 'detail1', title: 'detail1' }, { name: 'detail2 ', title: 'detail2 ' }, { name: 'detail3', title: 'detail3' }, { name: 'detail4', title: 'detail4' }],
sortEntriesByKey = ([a], [b]) => a.localeCompare(b),
filter = array2.reduce((r, o) => {
Object
.entries(o)
.sort(sortEntriesByKey)
.reduce((o, [k, v]) => (o[k] ??= {})[v] ??= {}, r);
return r;
}, {});
absent = array1.filter((o) => {
let f = filter;
return !Object
.entries(o)
.sort(sortEntriesByKey)
.every(([k, v]) => f = f[k]?.[v]);
});
console.log(absent);

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

Answer 2

编辑：您希望对象在A中而不是在B中。它是理想的循环通过A，发现元素是否存在于B中。如果是，则不包括它。

在javascript中，当您执行"=="或"==="或其他阵列搜索方法。

{} == {}将返回false。您可以在开发控制台中进行检查。

在您的情况下，您必须检查特定的属性。

var absent = array1.filter(e=>{
let findInd = array2.findIndex((a)=>{
return (a.name == e.name && a.title == e.title);});
return (findInd == -1); });

在内部findIndex中，我正在查找基于条件的索引。在filter方法中，只有当索引为-1(未找到(时，我才会返回true。

Answer 3

这对我有效：

let cart = [{
"id": "6304a51af5726921c0dadd64",
"qty": 1
},
{
"id": "8704a51af5726921c0dadd64",
"qty": 1
},
{
"id": "4704a51af5726921c0dadd64",
"qty": 1
}
]
let cartList = [{
"id": "6304a51af5726921c0dadd64",
"qty": 1
},
{
"id": "9704a51af5726921c0dadd64",
"qty": 1
}
]

let test = cart.some((element) =>
cartList.some((e) => element.id === e.id)
);
console.log(" if any single object matched:", test);
let test1 = cart.filter((element) =>
cartList.some((e) => element.id === e.id)
);
console.log("display matching objects :", test1);

Answer 4

const array1 = [
{ name: 'detail1', title: 'detail1' },
{ name: 'detail2 ', title: 'detail2 ' },
{ name: 'detail3', title: 'detail3' },
{ name: 'detail4', title: 'detail4' },
{ name: 'detail5', title: 'detail5' },
{ name: 'detail6', title: 'detail6' },
{ name: 'detail7', title: 'detail7' }
]
const array2 = [
{ name: 'detail1', title: 'detail1' },
{ name: 'detail2 ', title: 'detail2 ' },
{ name: 'detail3', title: 'detail3' },
{ name: 'detail4', title: 'detail4' },
]
const thirdArray = array1.filter((elem) => {
return !array2.some((ele) => {
return elem.name === ele.name
});
});