【发布时间】:2017-07-12 17:30:13
【问题描述】:
我有这个结构,我有一个MyComponent 的列表:
class MyComponent extends Component
{
props: { navigation: Object, data: Object };
ShowScreenB(data: Object){
this.props.navigation.navigate('ScreenB', {data});
}
render()
{
return (
<Menu>
<MenuTrigger> <Text> Menu </Text> </MenuTrigger>
<MenuOptions>
<MenuOption onSelect={() => this.ShowScreenB.bind(this, this.props.data)} text='Show Screen B' />
</MenuOptions>
</Menu>
);
}
}
class MyScreen extends Component
{
render()
{
let renderRow = (row) => { return (<MyComponent data= {row} navigation= {this.props.navigation} /> );}
return (
<View >
<ListView dataSource={this.state.DataSource} renderRow={renderRow.bind(this)}/>
</View>
);
}
}
但是ShowScreenB() 没有转到另一个屏幕。
我还尝试在MyScreen 类中准备导航器,然后将其作为函数传递给MyComponent。但也不行:
class MyComponent extends Component
{
props: { OnPress: Function, data: Object };
render()
{
return (
<Menu>
<MenuTrigger> <Text> Menu </Text> </MenuTrigger>
<MenuOptions>
<MenuOption onSelect={() => this.OnPress.bind(this)} text='Show Screen B' />
</MenuOptions>
</Menu>
);
}
}
class MyScreen extends Component
{
ShowScreenB(data: Object){
this.props.navigation.navigate('ScreenB', {data});
}
render()
{
let renderRow = (row) => { return (<MyComponent data= {row} OnPress= {this.ShowScreenB.bind(this, row)} /> );}
return (
<View >
<ListView dataSource={this.state.DataSource} renderRow={renderRow.bind(this)}/>
</View>
);
}
}
可能是什么问题?
编辑:Menu 是 PopUp Menu。
【问题讨论】:
标签: javascript reactjs react-native react-native-ios react-navigation