如何在 tabNavigator (React Native) 的 if 语句中调用驻留在父屏幕中的函数

Question

所以我的问题是：父组件是选项卡导航器中的一个屏幕，其中嵌套了另一个选项卡导航器。我可以通过 screenProps 将道具向下传递给子选项卡。但是，我无法从子选项卡导航器中的任何屏幕调用父屏幕中的函数。我究竟做错了什么？

ParentScreen.js //One of the screens of a parent tab navigator


import React, { Component } from 'react';
import { ChildTabs } from '../navigatation/router.js';

export default class ParentScreen extends Component {
    state = { name: '' }

    componentWillMount() {
        this.parentFxn();
    }

    parentFxn() {
        this.setState({ name: 'Staker' });
    }

    render() {
        screenProps = { parentFxn: this.parentFxn.bind(this) };
        return(
            <ChildTabs  // this is a tab Navigator
                screenProps={screenProps}
            />
        );
    }
}

现在子屏幕之一：

ChildScreen.js
//childScreen is one of the screens in the tab navigator

import React, { Component } from 'react';
import {Button} from 'react-native-elements';
import { ChildTabs } from '../navigatation/router.js';

export default class ChildScreen extends Component {

    childFxn() {
        axios.get(someRemoteUrl)
        .then((response) => {
            if (response.status == 200) {
                return this.props.screenProps.parentFxn; //this never gets called
            }
        });
    }

    render() {
        return(
            <Button
                onPress={this.childFxn.bind(this)}
            />
        );
    }
}

Answer 1

函数没有被调用，因为你没有调用它。应该这样称呼

  if (response.status == 200) {
     return this.props.screenProps.parentFxn(); //this always gets called
  }

如何在 tabNavigator (React Native) 的 if 语句中调用驻留在父屏幕中的函数

How to call a function residing in a parent screen within an if statement in a tabNavigator (React Native)

javascript

reactjs

react-native

react-navigation