Fab*_*zzi 7 javascript reactjs jestjs enzyme formik
我已经整理了一个很好的基本联系表格.但是我现在需要开始编写我的单元测试,并且遇到了大量问题(比如我到目前为止只能设法获得快照测试).
首先,如果您没有填写所有必需的部分,那么当您单击"提交"按钮时,我正在尝试测试表单应该呈现我的验证消息.
我以为我可以通过调用handleSubmit()函数来实现这个目的:
componentRender.find('Formik').instance().props.handleSubmit(badFormValues, { resetForm });
但是,当我运行时componentRender.debug(),我的验证消息没有被渲染.这就像没有调用validationSchema函数?
有什么特别需要做的吗?我觉得这个mapPropsToValues()函数正在工作,从查看状态对象,它正在填充我传递表单的值.我只是不明白为什么验证似乎被跳过了?
我已经在这2天了,并且通过谷歌找不到任何好的例子(可能是我的错)所以任何帮助都会受到大力赞赏.
这里是参考测试文件到目前为止:
import React from 'react';
import { shallow, mount } from 'enzyme';
import { BrowserRouter as Router } from 'react-router-dom';
import PartnerRegistrationForm from 'Components/partner-registration-form/PartnerRegistrationForm';
describe('PartnerRegistrationForm component', () => {
const formValues = {
companyName: 'some company',
countryCode: 'GB +44',
telNumber: 12345678,
selectCountry: 'United Kingdom',
postcode: 'ABC1 234',
addressSelect: '123 street',
siteName: 'blah',
siteURL: 'https://www.blah.com',
contactName: 'Me',
email: 'me@me.com',
};
const componentShallow = shallow(<PartnerRegistrationForm {...formValues} />);
describe('Component Snapshot', () => {
it('should match stored snapshot', () => {
expect(componentShallow).toMatchSnapshot();
});
});
describe('Component functionality', () => {
it('should not submit if required fields are empty', () => {
const badFormValues = {
companyName: 'some company',
countryCode: 'GB +44',
telNumber: 12345678,
};
const resetForm = jest.fn();
const componentRender = mount(
<Router>
<PartnerRegistrationForm {...badFormValues} />
</Router>,
);
componentRender.find('Formik').instance().props.handleSubmit(badFormValues, { resetForm });
// console.log(componentRender.update().find('.validation-error'));
// console.log(componentRender.find('Formik').instance());
// expect(componentRender.find('.validation-error').text()).toEqual('Company Name is required');
});
});
});Run Code Online (Sandbox Code Playgroud)
这是我的withFormik()功能:
const WrappedFormWithFormik = withFormik({
mapPropsToValues({
companyName,
countryCode,
telNumber,
selectCountry,
postcode,
addressSelect,
siteName,
siteURL,
contactName,
email,
}) {
return {
companyName: companyName || '',
countryCode: countryCode || '',
telNumber: telNumber || '',
selectCountry: selectCountry || '',
postcode: postcode || '',
addressSelect: addressSelect || '',
siteName: siteName || '',
siteURL: siteURL || '',
contactName: contactName || '',
email: email || '',
};
},
validationSchema, // This is a standard Yup.object(), just importing it from a separate file
handleSubmit: (values, { resetForm }) => {
console.log('submitting');
const {
companyName,
countryCode,
telNumber,
selectCountry,
postcode,
addressSelect,
siteName,
siteURL,
contactName,
email,
} = values;
const emailBody = `Name: ${contactName},`
+ `Email: ${email},`
+ `Company Name: ${companyName},`
+ `Country Code: ${countryCode},`
+ `Telephone Number: ${telNumber},`
+ `Country: ${selectCountry},`
+ `Postcode: ${postcode},`
+ `Address: ${addressSelect},`
+ `Website Name: ${siteName},`
+ `Website URL: ${siteURL}`;
// TODO set up actual contact submit logic
window.location.href = `mailto:test@test.com?subject=New partner request&body=${emailBody}`;
resetForm();
},
})(PartnerRegistrationForm);Run Code Online (Sandbox Code Playgroud)
如果您尝试通过单击带有的按钮来提交表单,则它可能不起作用 type="submit"
我发现让它提交(并因此运行验证)的唯一方法是直接模拟它:
const form = wrapper.find('form');
form.simulate('submit', { preventDefault: () => {} });
Run Code Online (Sandbox Code Playgroud)
...此外,在 formik 的异步验证和状态更改后,您可能需要使用类似以下内容来更新包装器:
setTimeout(() => {
wrapper.update();
}, 0);
Run Code Online (Sandbox Code Playgroud)
不要忘记使用done()或异步等待,这样测试就不会提前终止。