reactjs - React Enzyme test Child Component if is rendered

Question

Hi I have problem with Enzyme how to make test on parent where child component will show only if status = 'visible'.

Here is Parent Component with Child:

function TableInfo(): ReactElement {
const [status, setStatus] = useState('')
const [data, setData] = useState()
 return (
 <div className="d-flex justify-content-end mx-5 my-3">
   <h1>All Users</h1>
    {status !== 'None' ? (
    'No Data!') : (
   <Table data-testid="table-info" data={data}></Table>
  )}
 </div>
)

}

and in test of TableInfo:

import React from 'react'
import { render, screen, act } from '@testing-library/react'
import { findByTestDataId } from '../../../../../testUtil/getById'
import { shallow } from 'enzyme';

const setup = (props = {}) => {
  return shallow(<TableInfo {...props} />)

 describe('table information', () => {

    it('render table', () => {
      const wrapper = setup({status: 'test'})
      const component = findByTestDataId(wrapper, 'table-info')
      expect(component.length).toBe(1)
    })
}
 })

Answer 1

等待大神答复