Jest, Enzyme: Invariant Violation: 你不应该使用 <Route>或 withRouter() 在 <Router&g

本文介绍了Jest, Enzyme: Invariant Violation: 你不应该使用 <Route>或 withRouter() 在 <Router> 之外的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个 输出一个 组件和由 呈现的联系人列表;.

问题是，在对 的测试中，当我尝试挂载它时，测试输出错误 Invariant Violation: You should not use 或 withRouter() 在 <Router>

之外

withRouter() 用于 组件.

如何在没有路由器的情况下模拟 ContactsComponent 来测试父组件?

我发现了一些类似的问题 www.bountysource/issues/49297944-invariant-violation-you-should-not-use-route-or-withrouter-outside-a-router但它只描述了组件被 withRouter() 本身覆盖的情况，而不是子组件.

UserList.test.jsx

const mockResp = {数:2，项目: [{_id: 1,名称:'用户1'，电子邮件:'email1@gmail'，电话:'+123456'，在线:假的},{_id: 2,名称:'用户2'，电子邮件:'email2@gmail'，电话:'+789123'，在线:假的},{_id: 3,名称:'用户3'，电子邮件:'email3@gmail'，电话:'+258369147'，在线:假的}],下一个: 空}描述('用户列表'，()=> {beforeEach(() => {fetch.resetMocks()});test('应该输出用户列表', () => {fetch.mockResponseOnce(JSON.stringify(mockResp));const wrapper = mount(<UserListComponent user={mockResp.items[2]}/>);期望(wrapper.find('.contact_small')).to.have.length(3);});})

UserList.jsx

export class UserListComponent extends PureComponent {使成为() {const { 用户，错误 } = this.state;返回 (<React.Fragment><联系方式用户名={this.props.user.name}内容={this.props.user.phone}/>{错误 ?<p>{error.message}</p>: <Contacts type="contactList" user={this.props.user} contacts={users}/>}</React.Fragment>);}}

Contacts.jsx

class ContactsComponent 扩展组件 {构造函数(){极好的();this.state = {错误:空，};}使成为() {返回 (<React.Fragment><SectionTitle title="联系人"/><div className="联系人">//联系人

</React.Fragment>);}}export const Contacts = withRouter(ContactsComponent);

解决方案

要测试包含 和 withRouter 的组件(使用 Jest)，您需要在您的测试中导入路由器，而不是在您的组件中

import { BrowserRouter as Router } from 'react-router-dom';

并像这样使用它

app = 浅(<路由器><应用程序/></路由器>);

I have a <UserListComponent /> which outputs one <Contact /> component and list of contacts presentated by <Contacts />.

The problem is that in the test for <UserListComponent /> when I try to mount it, test outputs an error Invariant Violation: You should not use <Route> or withRouter() outside a <Router>

withRouter() is used in <Contacts /> component.

How can I mock ContactsComponent without router in test of parent component?

I found some similar issue www.bountysource/issues/49297944-invariant-violation-you-should-not-use-route-or-withrouter-outside-a-router but it only describes situation when component is cover by withRouter() itself, not children.

UserList.test.jsx

const mockResp = { count: 2, items: [ { _id: 1, name: 'User1', email: 'email1@gmail', phone: '+123456', online: false }, { _id: 2, name: 'User2', email: 'email2@gmail', phone: '+789123', online: false }, { _id: 3, name: 'User3', email: 'email3@gmail', phone: '+258369147', online: false } ], next: null } describe('UserList', () => { beforeEach(() => { fetch.resetMocks() }); test('should output list of users', () => { fetch.mockResponseOnce(JSON.stringify(mockResp)); const wrapper = mount(<UserListComponent user={mockResp.items[2]} />); expect(wrapper.find('.contact_small')).to.have.length(3); }); })

UserList.jsx

export class UserListComponent extends PureComponent { render() { const { users, error } = this.state; return ( <React.Fragment> <Contact userName={this.props.user.name} content={this.props.user.phone} /> {error ? <p>{error.message}</p> : <Contacts type="contactList" user={this.props.user} contacts={users} />} </React.Fragment> ); } }

Contacts.jsx

class ContactsComponent extends Component { constructor() { super(); this.state = { error: null, }; } render() { return ( <React.Fragment> <SectionTitle title="Contacts" /> <div className="contacts"> //contacts </div> </React.Fragment> ); } } export const Contacts = withRouter(ContactsComponent);

解决方案

To test a component (with Jest) that contains <Route> and withRouter you need to import Router in you test, not in your component

import { BrowserRouter as Router } from 'react-router-dom';

and use it like this

app = shallow( <Router> <App /> </Router>);