And*_*ini 8 react-native react-navigation
在官方react-navigationgithub 上打开了很多问题:https://github.com/react-community/react-navigation/issues但是没有人提供真正的解决方案,包括一篇帖子说"嘿伙计们,看看这个问题,很多问题":https://github.com/react-community/react-navigation/issues/691没有回答.
基本上:我有DrawerNavigator一些嵌套的StackNavigator.
export const MainNavigator = DrawerNavigator(
{
OrderNew: {
screen: OrderNew,
navigationOptions: {
drawerLabel: DrawerLabel('New Order')
}
},
Orders: {
screen: OrderNavigator, //a StackNavigator
navigationOptions: {
drawerLabel: DrawerLabel('Orders')
}
},
MenuNavigator: {
screen: MenuNavigator, //a StackNavigator
navigationOptions: {
drawerLabel: DrawerLabel('Menu')
}
},
//more screens...
{
initialRouteName: 'Orders',
drawerPosition: 'right',
contentComponent: props => <ScrollView style={{backgroundColor: 'rgba(227, 100, 29, .95)'}}><DrawerMenu state={props} /></ScrollView>,
}
);
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所以,我正在contentComponent使用我自己的菜单组件.我的想法是覆盖组件的onItemPress事件DrawerItems并将我的逻辑放在这里.
我想要的是:MenuNavigator每当用户按下该项目时,我想重置(和其他堆栈屏幕).我只是无法做到这一点.正如我所说,我尝试了很多代码.实际上,它是这样的:
onItemPress={
router => {
const navigateAction = NavigationActions.navigate({
routeName: router.route.routeName,
});
this.props.state.navigation.dispatch(navigateAction);
}
}
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上面的代码工作,我可以浏览抽屉项目.现在,在我的脑海中,我应该将子动作重置为导航器上的第一个项目.所以我尝试过:
onItemPress={
router => {
const navigateAction = NavigationActions.navigate({
routeName: router.route.routeName,
action: NavigationActions.reset({
index: 0,
actions: [
NavigationActions.navigate({routeName: router.route.routes[0].routeName}),
]
})
});
this.props.state.navigation.dispatch(navigateAction);
}
}
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并爆炸出错:
没有为键菜单定义路径.必须是以下之一:'订单','OrderClient'
总结:StackNavigator当我导航到另一个DrawerNavigator项目时,我怎么能重置?
Jay*_*tel -2
您缺少密钥:null。
导航动作应该是这样的。
action: NavigationActions.reset({
index: 0,
key: null,
actions: [
NavigationActions.navigate({routeName: router.route.routes[0].routeName}),
]
})
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