TextInput onChangeText 設置 useState 值在 React Native Testing with Enzyme 中不起作用

Question

如果 TextInput 等於“”，我想測試用戶不應該導航到任何其他屏幕。

我試圖讓控制台日志更正確地理解，但現在的問題是，在input.simulate("changeText", "any@email.com");之后發生正確，控制台日志顯示來自測試的傳入值是“any@email.com”但在setValue(text); 發生console.log 說useState 值是“”。

為什么 state 沒有變化？ 我需要用.update()更新包裝器嗎？

這是我的 React Native 代碼：

import React, { useState } from "react";
import { View, Button, TextInput } from "react-native";

export const LoginContainer = (props) => {
  const [value, setValue] = useState("");

  const handleClick = () => {
    if (value !== "") {
      props.navigation.navigate("any other screen");
    }
  }

  return (
    <View>
      <TextInput
        test-id="login-input"
        onChangeText={(text) => {
          console.log("INPUT_TEXT:", text);
          setValue(text);
          console.log("STATE_TEXT:", value);
        }}
        value={value}
      />
      <Button test-id="login-button" onPress={() => handleClick()} />
    </View>
  );
}

控制台日志：

  console.log
    INPUT_TEXT: any@email.com

      at onChangeText (....)

  console.log
    STATE_TEXT: 

      at onChangeText (....)

  expect(jest.fn()).toHaveBeenCalledTimes(expected)

  Expected number of calls: 1
  Received number of calls: 0

測試代碼：

import "react-native";
import React from "react";
import { shallow } from 'enzyme';
import { LoginContainer } from "...";
import { findByTestAttr } from '...';

const navigation = {
  navigate: jest.fn()
}

describe('correct login action', () => {
    const wrapper = shallow(<LoginContainer navigation={navigation} />);
    let input = findByTestAttr(wrapper, "login-input");
    let button = findByTestAttr(wrapper, "login-button");

    test('should not navigate to login mail screen if email adress is not entered', () => {
      input.simulate("changeText", "any@email.com");
      button.simulate("press");

      expect(navigation.navigate).toHaveBeenCalledTimes(1);

      //input.simulate("changeText", "");
      //button.simulate("press");
      //expect(navigation.navigate).toHaveBeenCalledTimes(0);
    });
});

Answer 1

你好 setValue 是一個異步的 function 所以它不會立即更新 state，你可以看到下一個值直到下一次渲染

添加一個console.log，如下所示

import React, { useState } from "react";
import { View, Button, TextInput } from "react-native";

export const LoginContainer = (props) => {
  const [value, setValue] = useState("");

  const handleClick = () => {
    if (value !== "") {
      props.navigation.navigate("any other screen");
    }
  }
 // try to add console.log here and you will see the new value after click
 console.log("STATE_TEXT:", value);

  return (
    <View>
      <TextInput
        test-id="login-input"
        onChangeText={(text) => {
          console.log("INPUT_TEXT:", text);
          setValue(text);
        }} 
      // if you want the setValue to be executed immediately you can call it inside promise like below, however the value will still the same even after re-rendering
        //onChangeText={(text) => {
          //console.log("INPUT_TEXT:", text);
          //Promise.resolve().then(() => {
            //setValue(text);
            //console.log("STATE_TEXT after calling it inside promise:", value);
          //});
          //console.log("STATE_TEXT:", value);
        //});
        value={value}
      />
      <Button test-id="login-button" onPress={() => handleClick()} />
    </View>
  );
}

默認情況下，反應在反應事件處理程序結束時重新渲染組件。 有關如何反應更新 state 的更多信息，請在此處查看此博客

Answer 2

我用這里的解決方案解決了這個問題。

test('should not navigate to login mail screen if email adress is not entered', () => {
    const wrapper = shallow(<LoginContainer navigation={navigation} />);
    input = findByTestAttr(wrapper, "login-input");

    input.simulate("changeText", "any@email.com");
    wrapper.update(); // update after changing the state
   
    // After performing update, should find button element
    button = findByTestAttr(wrapper, "login-button");
    button.simulate("press");
    expect(navigation.navigate).toHaveBeenCalledTimes(1);
});