reactjs 比较两个不同的对象数组，并过滤掉一个数组中具有相同ID的项

gblwokeq 于 2022-12-22 发布在 React

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我尝试比较并过滤出两个不同的对象数组，我需要检查两个对象数组中的所有元素，并需要过滤出具有相同ID的项，这意味着我只需要从一个数组中过滤出相同的项，另一个数组应该具有该项。（id），但我只需要从第二个数组中过滤掉。

// Example :

// Lets's say I am having following array of objects:

const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];

// I am trying to get something like below:

const res = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

基本上我需要检查每一个元素，我需要忽略数组b中相同的元素，我只需要得到数组a中的元素，我也不想从const b中删除那个元素，我需要检查两个数组，如果匹配，我只需要使用A，同时B中的元素不应该被删除。
我尝试下面的解决方案给我从两个列表相同的项目。任何人可以帮助我如何才能实现这一点吗？提前感谢'

const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];

let result = a.filter(o => !b.some(v => v.id === o.id));

console.log(result);

reactjs

来源：https://stackoverflow.com/questions/74887590/compare-two-different-array-of-objects-and-filter-out-the-item-which-is-having-s

4条答案

按热度按时间

svujldwt1#

你可以试试这个：

let arr = []

    let res = a.filter(o => !b.some(v => v.id === o.id))
    let res2 = a.filter(o => b.some(x => x.id == o.id))
    arr.push(...res2 , ...res)

赞(0）回复(0）举报 2022-12-22

643ylb082#

您可以在Set上进行闭包，并过滤所有项的数组。

const
    a = [{ id: '1234', description: 'PrdOne', version: '3', categories: '--' }, { id: '12345', description: 'PrdTwo', version: '2', categories: '--' }, { id: '123456', description: 'PrdThree', version: '2', categories: '--' }],
    b = [{ id: '1234', description: 'PrdOne', version: '3', categories: '--' }, { id: '12345', description: 'PrdTwo', version: '2', categories: '--' }],
    result = [...a, ...b].filter(
        (s => ({ id }) => !s.has(id) && s.add(id))
        (new Set)
    );

console.log(result);

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

赞(0）回复(0）举报 2022-12-22

b1uwtaje3#

const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];
function some(el){
    const objectKeys=Object.keys(el);
    for(let elemnets of b){
        if(objectKeys.length!=Object.keys(elemnets).length)continue;
        let found=1;
        for(let key of objectKeys){
            if(el[key]!=elemnets[key]){
                found=0;
                break;
            }
        }
        if(found)return true;
    }
    return false
}
let result = a.filter((el)=>{
    return some(el);
});

console.log(result);

试试这个

赞(0）回复(0）举报 2022-12-22

lkaoscv74#

const a = [
  { id: '1234', description: 'PrdOne', version: '3' },
  { id: '12345', description: 'PrdTwo', version: '2' },
  { id: '123456', description: 'PrdThree', version: '2' },
];
const b = [
  { id: '1234', description: 'PrdOne', version: '3' },
  { id: '12345', description: 'PrdTwo', version: '2' },
];

const concatArr = a.concat(b); // will contain all the entries from both array

const ids = concatArr.map(o => o.id); // ['1234', '12345', '123456', '1234', '12345']

const uniqueIds = [...new Set(ids)]; // ['1234', '12345', '123456']

const filtered = concatArr.filter((obj) => {
    if(uniqueIds.includes(obj['id'])){
        uniqueIds.splice(uniqueIds.indexOf(obj['id']), 1);
        return true;
    }
    return false;
});

console.log(filtered); // [{"id":"1234","description":"PrdOne","version":"3"},{"id":"12345","description":"PrdTwo","version":"2"},{"id":"12345","description":"PrdTwo","version":"2"}]

赞(0）回复(0）举报 2022-12-22

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reactjs 比较两个不同的对象数组，并过滤掉一个数组中具有相同ID的项

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