1

如果 TextInput 等于“”，我想测试用户不应该导航到任何其他屏幕。

我试图让控制台日志更正确地理解，但现在的问题是，在input.simulate("changeText", "any@email.com");正确发生之后，控制台日志说来自测试的传入值是“any@email.com”，但是在setValue(text);发生之后，console.log 说 useState 值是“ ”。

为什么状态没有改变？我需要用 更新包装器.update()吗？

这是我的 React Native 代码：

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

测试代码：

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);
    });
});
4

2 回答 2

0

你好 setValue 是一个异步函数所以它不会立即更新状态，你可以看到下一个值直到下一次渲染

添加一个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>
  );
}

默认情况下，反应在反应事件处理程序结束时重新渲染组件。有关如何反应更新状态的更多信息，请在此处查看此博客

于 2021-02-14T19:39:17.333 回答
0

我用这里的解决方案解决了这个问题。

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);
});
于 2021-02-14T20:53:13.877 回答