javascript - 未定义不是对象(评估this.props.navigation.navigate)

我正在尝试第一次学习本机反应。我试图介绍一个stacknavigator，但是行const {navigate} = this.props.navigation;导致了错误：


  未定义不是对象（评估
  'this.props.navigation.navigate'）


这是我的代码：

import React, { Component } from "react";
import { StackNavigator } from "react-navigation";
import { AppRegistry, Text, View } from "react-native";

export default class App extends Component {
  static navigationOptions = {
    title: "Login",
    headerStyle: {
      backgroundColor: "#000000"
    },
    headerTitleStyle: {
      color: "#fff"
    }
  };

  constructor(props) {
    super(props);
  }
  render() {
    console.log(this.props);

    const { navigate } = this.props.navigation;
    return (
      <View>
        <Text>Hello</Text>
      </View>
    );
  }
}

const myscreens = StackNavigator({
  Home: { screen: App }
});
AppRegistry.registerComponent("app", () => myscreens);

我在做什么错，我该如何解决？

我还应该提到props在render函数和构造函数中为空。

这是我的package.json以防万一

{
  "name": "myapp",
  "version": "0.0.1",
  "private": true,
  "scripts": {
    "start": "node node_modules/react-native/local-cli/cli.js start",
    "test": "jest"
  },
  "dependencies": {
    "react": "16.0.0-alpha.12",
    "react-native": "0.48.3",
    "react-navigation": "git+https://github.com/react-community/react-navigation.git"
  },
  "devDependencies": {
    "babel-jest": "21.0.2",
    "babel-preset-react-native": "4.0.0",
    "jest": "21.1.0",
    "react-test-renderer": "16.0.0-alpha.12"
  },
  "jest": {
    "preset": "react-native"
  }
}

这是我的index.android.js

import React, { Component } from 'react';
import App from './components/app';
import {
  AppRegistry,
  View
} from 'react-native';

export default class myapp extends Component {
  render() {
    return (
        <App />
    );
  }
}


AppRegistry.registerComponent('myapp', () => myapp);

最佳答案

两个错误：


您不应该这样打两次AppRegistry。您可以将其从app.js文件中删除。
您误解了react-navigation路由器的工作方式。 StackNavigator（以及Tab和Drawer）是需要直接渲染（例如，传递给AppRegistry [A]的内容）或在应用程序中某些时候[B]渲染的组件。


因此，要解决此问题，您将要导出myscreens而不是App。为了说明这两种方式：

答：您可以看到此in the docs的示例，但是由于将代码分成两个文件，因此如下所示：

./components/app.js

import React, { Component } from 'react';
import { StackNavigator } from 'react-navigation';
import { Text, View } from 'react-native';

class App extends Component {
  static navigationOptions = {
    title: 'Login',
    headerStyle: {
      backgroundColor: '#000000',
    },
    headerTitleStyle: {
      color: '#fff',
    },
  };

  constructor(props) {
    super(props);
  }
  render() {
    console.log(this.props);

    const { navigate } = this.props.navigation;
    return (
      <View>
        <Text>Hello</Text>
      </View>
    );
  }
}

const myscreens = StackNavigator({
  Home: { screen: App },
});

export default myscreens;

index.android.js

import { AppRegistry } from 'react-native';
import myscreens from './components/app';

AppRegistry.registerComponent('myapp', () => myscreens);

B.您将在另一个类中渲染StackNavigator，然后导出呈现它的类。这更适合您已经在做的事情，并且从长远来看更加灵活（例如：如果您在某些时候使用redux，则需要这样做），因此您应该这样做：

./components/app.js-注意将myscreens更改为MyScreens的大小写。这是必需的，因为React Native将complain about this称为lowercase JSX tags are considered to be HTML。直接使用AppRegistry不会触发此错误，因为您没有在JSX中使用myscreens。

import React, { Component } from 'react';
import { StackNavigator } from 'react-navigation';
import { Text, View } from 'react-native';

class App extends Component {
  static navigationOptions = {
    title: 'Login',
    headerStyle: {
      backgroundColor: '#000000',
    },
    headerTitleStyle: {
      color: '#fff',
    },
  };

  constructor(props) {
    super(props);
  }
  render() {
    console.log(this.props);

    const { navigate } = this.props.navigation;
    return (
      <View>
        <Text>Hello</Text>
      </View>
    );
  }
}

const MyScreens = StackNavigator({
  Home: { screen: App },
});

export default MyScreens;

index.android.js

import React, { Component } from 'react';
import { AppRegistry } from 'react-native';
import MyScreens from './components/app';

export default class myapp extends Component {
  render() {
    return <MyScreens />;
  }
}

AppRegistry.registerComponent('myapp', () => myapp);