考虑以下数组:
[
{ url: "https://url.com/file1", md5: "fbbbabcc19264ce7b376ce4c726b9b85" },
{ url: "https://url.com/file2", md5: "d920d140432b961f07695ec34bd2a8ad" },
{ url: "https://url.com/file3", md5: "fbbbabcc19264ce7b376ce4c726b9b85" },
{ url: "https://url.com/file4", md5: "bf80655dbe90123324f88a778efa39f7" },
{ url: "https://url.com/file5", md5: "fbbbabcc19264ce7b376ce4c726b9b85" }
];
文件“ file1”,“ file3”和“ file5”具有相同的内容,因此具有相同的md5。我只想保留具有不同md5(文件1,文件2,文件4)的文件。
现代ES6有哪些可能的方法来实现这一目标?
您可以reduce
用来解决问题:
const list = [
{ url: 'https://url.com/file1', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },
{ url: 'https://url.com/file2', md5: 'd920d140432b961f07695ec34bd2a8ad' },
{ url: 'https://url.com/file3', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },
{ url: 'https://url.com/file4', md5: 'bf80655dbe90123324f88a778efa39f7' },
{ url: 'https://url.com/file5', md5: 'fbbbabcc19264ce7b376ce4c726b9b85' },
];
const removeDup = (arr, key) => {
return arr.reduce((acc, cur) => {
if (acc.some(a => a[key] === cur[key])) return acc;
return acc.concat(cur)
}, [])
};
console.log(removeDup(list, 'md5'))