javascript - 我应该如何测试两个嵌套的不可变 js 数据结构的深度相等性？

Question

假设我有以下嵌套的不可变数据结构:

import { Record, List } from 'immutable'

const foo = List([
        Record({
           id: 1,
           gameState: Record({})
         })
       ])

我如何测试两个嵌套的不可变 js 数据结构彼此相等？

test.js

import jest from 'jest'
import { Record, List } from 'immutable'

describe('test', () => {
    it('that foo equals expected', () => {
        const expected = List([
              Record({
                 id: 1,
                 gameState: Record({})
               })
             ])
        expect(foo).toEqual(expected)
    })

Answer 1

对于大多数 ImmutableJS 类型，您进行的测试都可以正常工作。问题在于您使用记录的方式。与 Map、List 等不同，constructorRecord(...) 不返回新对象，它返回一个构造函数，然后可以使用该构造函数创建新对象。来自文档(https://facebook.github.io/immutable-js/docs/#/Record):

[Record] Creates a new Class which produces Record instances. A record is similar to a JS object, but enforce a specific set of allowed string keys, and have default values.

如果你想记录用户，那么你可以这样写你的测试:

import { Record, List, is } from 'immutable'

describe('test', () => {
    it('that foo equals expected', () => {
        const gameState = Record({});
        const gameRecord = Record({
            id: 1,
            gameState: new gameState()
        });

        const foo = List([
            new gameRecord()
        ]);

        const expected = List([
            new gameRecord()
        ]);
        expect(foo.equals(expected)).toBe(true);
    })
});

将此与使用 map 而不是记录进行比较。下面使用 jest 的 toEqual 函数的测试按预期通过:

import { Map, List, is } from 'immutable'

describe('test', () => {
    it('that foo equals expected', () => {

        const foo = List([
            Map({
                id: 1,
                gameState: Map({})
            })
        ]);

        const expected = List([
            Map({
                id: 1,
                gameState: Map({})
            })
        ])
        expect(foo).toEqual(expected);
    })
});