在JavaScript中如何檢查物件陣列中是否存在陣列值而不重復

我有一個陣列

let names = ['Devid', 'Aries','James'];

和一個物件陣列

let details = [
{"Name":"John"},
{"Name":"Devid"},
{"Name":"Aries"},
{"Name":"Aries"}
];

我想詳細檢查所有名稱是否存在，它們在 1 個回圈中也不是按名稱重復的

 let names = ['Devid', 'Aries','James'];

 let details = [
    {"Name":"John"},
    {"Name":"Devid"},
    {"Name":"Aries"},
    {"Name":"Aries"}
    ];

let count = 0;

for (var eachDetail of details) {
     // checking here Name exist
     let isNameFound = names.some(el => eachDetail.Name.includes(el));
            if (isNameFound) count  ;
        }
console.log(count);

// console.log(3) expected output 2

輸出應為 2（如 Devid 和 Aries）存在

uj5u.com熱心網友回復：

您可以簡單地遍歷names陣列并執行details.find(...)操作。如果您在 中找到該物件details，請將其添加到結果串列中。

let names = ['Devid', 'Aries','James'];

let details = [
  {"Name":"John"},
  {"Name":"Devid"},
  {"Name":"Aries"},
  {"Name":"Aries"}
];

let result = [];
names.forEach(name => {
  const object = details.find(detail => detail.Name === name);
  if (object) {
    result.push(object);
  }
});

console.log(result.length   ' entries found:');
console.log(JSON.stringify(result, null, 2));

uj5u.com熱心網友回復：

您可以使用reduce來計算陣列中可以在物件值中找到的元素。您可以使用 aSet洗掉重復項。

let names = ['Devid', 'Aries','James'];

let details = [
  {"Name":"John"},
  {"Name":"Devid"},
  {"Name":"Aries"},
  {"Name":"Aries"}
];

const detailsNames = [...new Set(details.map(d => d.Name))];
const count = [...new Set(names)].reduce((acc, el) => acc   detailsNames.includes(el), 0);

console.log(count);

uj5u.com熱心網友回復：

您應該能夠使用.filter()、.some()和.inlcude()內部驗證非重復項的方法來過濾陣列。

請在此處查看有關 JS 陣列方法的檔案：https : //developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array

let names = ['Devid', 'Aries','James'];
let tempNames = [];

let details = [
  {"Name":"John"},
  {"Name":"Devid"},
  {"Name":"Aries"},
  {"Name":"Aries"}
];

var newDetails = details.filter(x => {
    let includeName = names.some(j => j == x.Name) && !tempNames.includes(x.Name);
    tempNames.push(x.Name);
    return includeName;
});

console.log(newDetails);

uj5u.com熱心網友回復：

您可以原型陣列，然后遍歷名稱陣列

Array.prototype._indexOfName = function(name){
  let n = 0;
  for(; n < this.length; n  )
    if(this[n].Name == name)
      return n
  return -1
};

let names = ['Devid', 'Aries','James'];
let details = [
  {"Name":"John","Age":20},
  {"Name":"Devid","Age":23},
  {"Name":"Aries","Age":34},
  {"Name":"Aries","Age":50}
];
let n = 0;
let n2 = 0;
for(; n < names.length; n  )
  if(details._indexOfName(names[n]) > -1)
    n2   
alert('Found '   n2   ' matches')

uj5u.com熱心網友回復：

這里有兩個解決方案filter()的names基礎上在存在陣列details。然后檢索過濾后陣列的長度。

檢查名稱是否存在的一種details方法是使用some(). 回傳true只要回呼的回傳值是truthy。

另一種方法是將 的所有名稱details放在Set第一個中。這允許您將過濾器回呼簡化為name => detailNames.has(name).

const names = ["Devid", "Aries", "James"];
const details = [
  { "Name": "John"  },
  { "Name": "Devid" },
  { "Name": "Aries" },
  { "Name": "Aries" },
];

// solution #1
const count1 = names.filter((name) => (
  details.some(detail => detail.Name == name)
)).length;
console.log(count1);

// solution #2
const detailNames = new Set(details.map(detail => detail.Name));
const count2 = names.filter(name => detailNames.has(name)).length;
console.log(count2);

兩種解決方案都假定names不包含重復名稱。

如果details小，第一個解決方案很可能更快。第二種解決方案有更多的開銷，但如果details很大，則速度更快。這是因為some()從開始到結束搜索專案開始迭代，同時has()使用查找演算法。

標籤：javascript 节点.js 数组 目的

上一篇：如何創建嵌套物件目錄名，直到找到檔案

下一篇：如何正確使用JMeter性能測驗？緊扣面試實際要求