Javascript 在另一个屏幕导航V5-NATIVE中从屏幕获取参数

我试图从屏幕B中的屏幕A获取参数，但总是抛出错误，我就是这样做的：

go = (songs) => {
        this.props?.navigation.navigate("Artists_Song", {hola: "hola"});
 }

 <TouchableOpacity onPress = {() => this.go(songs)}>
                                    <View style = {{
                                        backgroundColor: "yellow", 
                                        width: Dimensions.get("window").width * 0.48, 
                                        marginBottom: 20, 
                                        marginLeft: 5, 
                                        alignItems: "center"}}
                                        key = {artist}>
                                        <Image source = {require("../assets/default_image.jpg")} style={{ width: Dimensions.get("window").width * 0.48, height: 100}}></Image>
                                        <Text style = {{fontSize: 20, textAlign: "center"}}>
                                            {
                                                (artist.length > 15) ?  
                                                artist.substring(0,13).padEnd(15,".") 
                                                : artist
                                            }
                                        </Text>
                                    </View>
                                </TouchableOpacity>

---

你能检查一下

this.props.navigation.state.params吗？

它告诉你：undefined不是一个评估

this.props.navigation.state.params的对象。你正在使用的react导航的版本是什么？@jogesh_pi navigation v5
class Artists_Song extends React.Component {
render() {
    return(
        <>
            <Artist_Song {...this.props}></Artist_Song>
        </>
    )
}

 componentDidMount() {
    const songs = this.props.route.params;
    console.log(songs);
}

render() {
    return (
        <>
            <Text>Hola</Text>
        </>
    )
}

this.props.route.params.hola

this.props.navigation.route.params