Javascript React,Jest:模拟提交表单单元测试
我一直坚持使用简单的测试用例,无法找出它不起作用的原因。 有一个我想测试的组件:Javascript React,Jest:模拟提交表单单元测试,javascript,reactjs,jestjs,enzyme,Javascript,Reactjs,Jestjs,Enzyme,我一直坚持使用简单的测试用例,无法找出它不起作用的原因。 有一个我想测试的组件: export class SignInForm extends Component { onSubmit = (user) => { console.log('THERE----', this.props.signin); // ---> i saw it during testing this.props.signin(); }; render()
export class SignInForm extends Component {
onSubmit = (user) => {
console.log('THERE----', this.props.signin); // ---> i saw it during testing
this.props.signin();
};
render() {
return (
<Form onSubmit={this.onSubmit}>
{formApi => (
<form onSubmit={formApi.submitForm}>
<InputFieldGroup field="identifier" label="Username/Email"/>
<InputFieldGroup field="password" label="Password" type="password"/>
<button type="submit">Submit</button>
</form>
)}
</Form>
);
}
}
这是我的测试用例
it('should render in proper way', () => {
const signinMock = jest.fn();
const props = {
signin: signinMock
};
const component = mount(<SignInForm {...props}/>);
component.find('form').simulate('submit');
console.log(signinMock.mock.calls); // ---> saw empty array []
});
在本例中,我希望立即调用signinMock。但我看到了调用的空数组,但在启动测试期间,我看到了组件方法的控制台日志:console.log'THERE--',this.props.sign;我看到了模拟函数本身。console.log在事件处理程序之前执行,因此您会看到一个空数组。下面的代码将允许执行事件处理程序,然后您可以执行断言
setTimeout(()=> {
expect(signinMock.mock.calls.length).to.equal(1);
}, 0);
谢谢这节省了我的时间。