惩戒ES6类方法

我测试的明确应用程序，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)
    });
}); 

什么我试图做的是模拟在CompanyController寄存器方法和改变实现，使其返回200状态码，但我得到的错误：

类型错误：_CompanyController.default.mockImplementation不是一个函数

如何实现这一正确的方法是什么？

如果你想使用mockImplementation你需要告诉玩笑说CompanyController需要被嘲笑（如描述here）：

import CompanyController from './CompanyController';
jest.mock('./CompanyController'); // you miss this line
const mockComp = jest.fn();
CompanyController.mockImplementation(()=>{
    return {
        register:mockComp
    }
});

惩戒ES6类方法

我测试的明确应用程序，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)
    });
}); 

什么我试图做的是模拟在CompanyController寄存器方法和改变实现，使其返回200状态码，但我得到的错误：

类型错误：_CompanyController.default.mockImplementation不是一个函数

如何实现这一正确的方法是什么？

如果你想使用mockImplementation你需要告诉玩笑说CompanyController需要被嘲笑（如描述here）：

import CompanyController from './CompanyController';
jest.mock('./CompanyController'); // you miss this line
const mockComp = jest.fn();
CompanyController.mockImplementation(()=>{
    return {
        register:mockComp
    }
});