【发布时间】:2019-02-11 16:50:39
【问题描述】:
我正在测试一个快递应用,我有以下文件:
app.js
const express = require('express');
const path = require('path');
const apiRouter = require('./api');
const app = express();
app.use('/api',apiRouter);
module.exports = app;
api.js
const Router = require( 'express').Router;
const CompanyController =require( './CompanyController');
const router = Router();
router.route('/company/registration').post(CompanyController.register);
module.exports=router;
CompanyController.js
class CompanyRegistration {
constructor() {}
static async register(req, res) {
//some implementations
//if successfull returns status code 200 or 422 if cant process
}
registration.js
import Request from 'supertest';
import app from './app';
import CompanyController from './CompanyController';
const mockComp = jest.fn();
CompanyController.mockImplementation(()=>{
return {
register:mockComp
}
});
it("should work",()=>{
return Request(app).post("/api/company/registration").then(response => {
expect(response.statusCode).toBe(200)
});
});
我试图做的是模拟 CompanyController 中的 register 方法并更改实现,使其返回状态代码 200,但我收到错误:
TypeError: _CompanyController.default.mockImplementation 不是 功能
如何以正确的方式实现这一点?
【问题讨论】:
标签: node.js unit-testing express jestjs