onClick未呈现不同的react组件

Question

我已经创建了一个SocialShare组件，当单击任何其他组件上的Share按钮时，我希望呈现该组件。

import React, {Component} from 'react';
import {View, Share, Button} from 'react-native';

export class SocialShare extends Component {

  onShare = async () => {
    try {
      const result = await Share.share({
        message:
          'React Native | A framework for building native apps using React',
      });

      if (result.action === Share.sharedAction) {
        if (result.activityType) {
          // shared with activity type of result.activityType
        } else {
          // shared
        }
      } else if (result.action === Share.dismissedAction) {
        // dismissed
      }
    } catch (error) {
      alert(error.message);
    }
  };

  render() {
    return (

      this.onShare()

      );
  }
}

这就是我从另一个组件调用这个组件的方式：

<View>
    <Button onClick={() => this.onShareButtonClick()}>Button</Button>
         {this.state.showShareComponent ?
             <SocialShare /> :
             null
         }
</View>

onShareButtonClick函数：

onShareButtonClick(){        
        this.setState({
            showShareComponent: !this.state.showShareComponent,
        })
    }

单击按钮时，这是我得到的错误：

Invariant Violation: Objects are not valid as a React child (found: object with keys {_40, _65, _55, _72}). If you meant to render a collection of children, use an array instead.
    in SocialShare 

我的代码出了什么问题？

编辑:根据建议，将我的SocialShare类修改为：

import React, {Component} from 'react';
import {View, Share, Button} from 'react-native';

export class SocialShare extends Component {
  constructor(props) {
    super(props);
    this.state = {
        asyncCompleted: false,
    };
  }
  onShare = async () => {
    try {
      const result = await Share.share({
        message:
          'React Native | A framework for building native apps using React',
      }).then(
        this.setState({asyncCompleted: true})
      );



      if (result.action === Share.sharedAction) {

        if (result.activityType) {
          // shared with activity type of result.activityType
        } else {
          // shared
        }
      } else if (result.action === Share.dismissedAction) {
        // dismissed
      }
    } catch (error) {
      alert(error.message);
    }
  };


  render() {
    return (

      <View>
      {this.state.asyncCompleted ? this.onShare() : null}
      </View>
      );
  }
}

现在，在我的另一个类中单击按钮时没有任何反应。

Answer 1

在我看来，主要问题是您的render方法试图直接呈现promise，而异步onShare()方法将返回该promise。相反，您应该让异步代码更新组件的状态，这样就可以触发基于该状态值的呈现，而不是基于onShare()的直接输出