抽屉导航器(reactnavigation v2)是否可以不更改屏幕而是改为redux状态?
示例:
api = [{id: 1, name: 'Pippo'}, {id: 2, name: 'Pluto'}, {id: 3, name: 'Paperino'}]
抽屉
Menu title 1
Menu title 2
Menu title 3
屏幕
Welcome **state.name**
单击元素1-> actionChangeName(element_1)->状态名称='pippo'
Screen -> Welcome Pippo
单击元素2-> actionChangeName(element_2)->状态名称='pluto'
Screen -> Welcome Pluto
单击元素3-> actionChangeName(element_3)->状态名称='paperino'
Screen -> Welcome Paperino
谢谢
答案 0 :(得分:0)
我解决了我的问题:
我创建了一个只包含一个包含“ Welcome ** state.name **”的屏幕和一个名为 DrawerMenu
createDrawerNavigator({ main: WelcomeScreen }, { contentComponent: DrawerMenu })
DrawerMenu(示例):
class DrawerMenu extends Component {
render() {
return (
<View style={styles.container}>
<View style={styles.titleContainer}>
<Text style={styles.title}> Drawer Menu </Text>
</View>
<ScrollView>
{this._renderList()}
</ScrollView>
</View>
);
}
_renderList = () => this.props.list.map(element => (
<TouchableOpacity key={element.id} onPress={this._updateElementSelected(element.id)}>
<Text>Menu title {element.id}</Text>
</TouchableOpacity>
);
_updateElementSelected = (elementId) => () => {
this.props.actionSetElementSelected(elementId, this.props.list);
this.props.navigation.toggleDrawer();
};
}