Как отфильтровать массив в соответствии с массивом

Question

У меня есть список разрешений:

const all_permissions = [
    {
        permission_id: 1,
        permission_name: "Getting User List",
        permission_method: "GET",
        permission_url: /^\/panel\/user$/
    },
    {
        permission_id: 2,
        permission_name: "Getting User Info",
        permission_method: "GET",
        permission_url: /^\/panel\/user\/[0-9]+$/
    },
    {
        permission_id: 3,
        permission_name: "Testing",
        permission_method: "GET",
        permission_url: /^\/panel\/test\/[a-z]+$/
    }
]

const my_permissions = [1,2]

Как отфильтровать all_permissions, чтобы соответствовать только идентификатору внутри my_permissions?Спасибо

Answer 1

Вы можете использовать фильтр и включает в себя

• Перебирать массив all_permission с использованием фильтра
• проверка внутри фильтра для permission_id в my_permission массиве, если он там есть, возвращать true, иначе возвращать false

const all_permissions = [{permission_id: 1,permission_name: "Getting User List",permission_method: "GET",permission_url: /^\/panel\/user$/},{permission_id: 2,permission_name: "Getting User Info",permission_method: "GET",permission_url: /^\/panel\/user\/[0-9]+$/},{permission_id: 3,permission_name: "Testing",permission_method: "GET",permission_url: /^\/panel\/test\/[a-z]+$/}]
const my_permissions = [1,2]

let op = all_permissions.filter(({permission_id})=> my_permissions.includes(permission_id))

console.log(op)

Answer 2

Вы можете попробовать с Array​.prototype​.filter() и Array​.prototype​.includes()

const all_permissions = [
    {
        permission_id: 1,
        permission_name: "Getting User List",
        permission_method: "GET",
        permission_url: /^\/panel\/user$/
    },
    {
        permission_id: 2,
        permission_name: "Getting User Info",
        permission_method: "GET",
        permission_url: /^\/panel\/user\/[0-9]+$/
    },
    {
        permission_id: 3,
        permission_name: "Testing",
        permission_method: "GET",
        permission_url: /^\/panel\/test\/[a-z]+$/
    }
]

const my_permissions = [1,2];
const match = all_permissions.filter(p => my_permissions.includes(p.permission_id));
console.log(match);

Answer 3

Используйте filter и includes, например, так:

const all_permissions = [{permission_id:1,permission_name:"Getting User List",permission_method:"GET",permission_url:/^\/panel\/user$/},{permission_id:2,permission_name:"Getting User Info",permission_method:"GET",permission_url:/^\/panel\/user\/[0-9]+$/},{permission_id:3,permission_name:"Testing",permission_method:"GET",permission_url:/^\/panel\/test\/[a-z]+$/}];
const my_permissions = [1,2];

const res = all_permissions.filter(({ permission_id }) => my_permissions.includes(permission_id));

console.log(res);

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Синтаксис ES5:

var all_permissions = [{permission_id:1,permission_name:"Getting User List",permission_method:"GET",permission_url:/^\/panel\/user$/},{permission_id:2,permission_name:"Getting User Info",permission_method:"GET",permission_url:/^\/panel\/user\/[0-9]+$/},{permission_id:3,permission_name:"Testing",permission_method:"GET",permission_url:/^\/panel\/test\/[a-z]+$/}];
var my_permissions = [1,2];

var res = all_permissions.filter(function(item) {
  return my_permissions.indexOf(item.permission_id) > -1;
});

console.log(res);

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