【发布时间】:2019-12-24 12:43:55
【问题描述】:
我有以下 react-native 测试代码。
import { shallow } from 'enzyme';
import React from 'react';
import {
BorderlessButton,
InputBox,
ProgressBar,
} from 'components';
import Name from '../name.component';
describe('Name component', () => {
let wrapper: any;
const mockOnPress = jest.fn();
const mockSaveStep = jest.fn();
const mockProps: any = {
errors: null,
values: [{ givenName: 'givenName', familyName: 'familyName' }],
};
beforeEach(() => {
wrapper = shallow(<Name signUpForm={mockProps} saveStep={mockSaveStep} />);
});
it('should render Name component', () => {
expect(wrapper).toMatchSnapshot();
});
it('should render 2 <InputBox />', () => {
expect(wrapper.find(InputBox)).toHaveLength(2);
});
it('should render a <ProgressBar />', () => {
expect(wrapper.find(ProgressBar)).toHaveLength(1);
});
it('should render a <BorderlessButton /> with the text NEXT', () => {
expect(wrapper.find(BorderlessButton)).toHaveLength(1);
expect(wrapper.find(BorderlessButton).props().text).toEqual('NEXT');
});
it('should press the NEXT button', () => {
wrapper.find(BorderlessButton).simulate('click');
expect(mockOnPress).toHaveBeenCalled();
});
});
但是最后一个测试不能正常工作。如何模拟此按钮单击?这给了我一个错误提示
期望(jest.fn()).toHaveBeenCalled()。 预期的模拟函数已被调用,但未被调用。
这是组件。
class NameComponent extends Component {
componentDidMount() {
const { saveStep } = this.props;
saveStep(1, 'Name');
}
disableButton = () => {
const {
signUpForm: {
errors, values,
},
} = this.props;
if (errors && values && errors.givenName && errors.familyName) {
if (errors.givenName.length > 0 || values.givenName === '') return true;
if (errors.familyName.length > 0 || values.familyName === '') return true;
}
}
handleNext = () => {
navigationService.navigate('PreferredName');
}
resetForm = () => {
const { resetForm } = this.props;
resetForm(SIGN_UP_FORM);
navigationService.navigate('LoginMain');
}
render() {
const { name, required } = ValidationTypes;
const { step } = this.props;
return (
<SafeAreaView style={{ flex: 1 }}>
<KeyboardAvoidingView style={{ flex: 1 }}
behavior={Platform.OS === 'ios' ? 'padding' : null}
enabled>
<ScreenContainer
navType={ScreenContainer.Types.LEVEL_THREE}
levelThreeOnPress={this.resetForm}>
<View style={styles.container}>
<View style={{ flex: 1 }}>
<SinglifeText
type={SinglifeText.Types.H1}
label='Let’s start with your legal name'
style={styles.textLabel}
/>
<View style={styles.names}>
<InputBox
name='givenName'
form={SIGN_UP_FORM}
maxLength={22}
placeholder='Given name'
containerStyle={styles.givenNameContainer}
inputContainerStyle={styles.inputContainer}
errorStyles={styles.inputError}
keyboardType={KeyBoardTypes.default}
validations={[required, name]}
/>
<InputBox
name='familyName'
form={SIGN_UP_FORM}
maxLength={22}
placeholder='Family name'
inputContainerStyle={styles.inputContainer}
errorStyles={styles.inputError}
keyboardType={KeyBoardTypes.default}
validations={[required, name]}
/>
</View>
<SinglifeText
type={SinglifeText.Types.HINT}
label='Please use the same name you use with your bank'
style={styles.hint}
/>
</View>
</View>
</ScreenContainer>
<ProgressBar presentage={(step / MANUAL_SIGNUP_STEP_COUNT) * 100} />
<View style={styles.bottomButtonContainer}>
<BorderlessButton
text='NEXT'
disabled={this.disableButton()}
onPress={this.handleNext}
/>
</View>
</KeyboardAvoidingView>
</SafeAreaView>
);
}
}
我该如何解决这个问题??
【问题讨论】:
-
当您说“无法正常工作”时,您是什么意思?错误是什么?
-
我要测试的方法是,您还必须模拟单击按钮时将触发的函数并更改为 expect(function).toHaveBeencall();
-
你能添加整个测试代码吗?这将帮助我们了解真正的问题出在哪里。
-
用完整的测试代码更新了帖子。
-
您调用的实际点击功能是什么?我猜 mockpress 只是一个模拟名称。您也可以发布组件代码吗?只是想看看你在 onclick 事件上调用了哪个函数。
标签: javascript reactjs react-native jestjs enzyme