我正在嘗試模擬 bcrypt 哈希方法實作,但出現以下錯誤:
Error: thrown: "Exceeded timeout of 5000 ms for a test.
Use jest.setTimeout(newTimeout) to increase the timeout value, if this is a long-running test."
我嘗試將超時增加到 30000。我還嘗試模擬整個 bcrypt 模塊,例如 jest.mock('bcrypt')。我是測驗新手,可能存在一些邏輯錯誤或不良做法。如果您指出它們,我將不勝感激。
我的代碼:
import { UserService } from '../user.service';
import { Test, TestingModule } from '@nestjs/testing';
import { Repository } from 'typeorm';
import { getRepositoryToken } from '@nestjs/typeorm';
import * as bcrypt from 'bcrypt';
import { UserEntity } from '../user.entity';
import { UserRepository } from '../user.repository';
import { CreateUserDto } from '../dto/create-user.dto';
import { userStub } from './stubs/user.stub';
describe('UserService', () => {
let userService: UserService;
let userRepository: Repository<UserEntity>;
beforeAll(async () => {
const module: TestingModule = await Test.createTestingModule({
providers: [
UserService,
{
provide: getRepositoryToken(UserRepository),
useClass: Repository,
},
],
}).compile();
userService = module.get<UserService>(UserService);
userRepository = module.get<Repository<UserEntity>>(
getRepositoryToken(UserRepository),
);
});
it('should define UserService', () => {
expect(userService).toBeDefined();
});
it('should define userRepository', () => {
expect(userRepository).toBeDefined();
});
describe('createUser method', () => {
it('has called with valid data', async () => {
const createUserDto: CreateUserDto = {
email: userStub().email,
firstName: userStub().firstName,
lastName: userStub().lastName,
password: userStub().password,
};
const user: UserEntity = userStub();
const spiedBcryptHashMethod = jest
.spyOn(bcrypt, 'hash')
.mockImplementation(() => Promise.resolve(''));
const spiedRepositoryCreateMethod = jest
.spyOn(userRepository, 'create')
.mockReturnValue(user);
const spiedRepositorySaveMethod = jest
.spyOn(userRepository, 'save')
.mockResolvedValue(user);
const createUserResult = await userService.createUser(createUserDto);
expect(spiedBcryptHashMethod).toHaveBeenCalled();
expect(spiedRepositoryCreateMethod).toHaveBeenCalled();
expect(spiedRepositorySaveMethod).toHaveBeenCalledWith(user);
expect(createUserResult).toEqual(user);
});
});
});
這里出現錯誤:
const spiedBcryptHashMethod = jest
.spyOn(bcrypt, 'hash')
.mockImplementation(() => Promise.resolve(''));
我的用戶服務代碼:
import { Injectable } from '@nestjs/common';
import { InjectRepository } from '@nestjs/typeorm';
import * as bcrypt from 'bcrypt';
import { UserRepository } from './user.repository';
import { CreateUserDto } from './dto/create-user.dto';
import { UserEntity } from './user.entity';
@Injectable()
export class UserService {
constructor(
@InjectRepository(UserRepository) private userRepository: UserRepository,
) {}
public async createUser(createUserDto: CreateUserDto): Promise<UserEntity> {
return await this.userRepository.save(
this.userRepository.create({
...createUserDto,
password: await new Promise((resolve, reject) => {
bcrypt.hash(createUserDto.password, 10, (err, encrypted) => {
if (err) {
reject(err);
}
resolve(encrypted);
});
}).then((onFilled: string) => onFilled),
}),
);
}
}
uj5u.com熱心網友回復:
好的,所以關于您的服務代碼有很多話要說...
您遇到問題的直接問題是因為您正在模擬 bcrypt 的hash方法以回傳一個承諾,但使用該方法回傳一個回呼。如果您想繼續使用與承諾方法混合的回呼,您需要執行類似的操作
jest.spyOn(bcrypt, 'hash').mockImplementation((pass, salt, cb) => cb(null, ''))
這將與Promise.resolve('').
但是我不建議這樣做。Bcrypt 內置了 Promise 支持,所以不用你自己的 Promise 包裝回呼,你可以這樣做await bcrypt.hash(pass, salt),你會得到散列的密碼,然后你Promise.resolve('')也會按照你的意愿作業。我認為兩者都沒有必要then((onFullfilled: string) => onFullfilled),但是如果您無論如何都洗掉自定義承諾,那將會消失。
一般來說,盡量堅持使用單一的異步方法。所有回呼(不再建議),所有承諾(更好)或所有異步/等待(現在幾乎是標準)。
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