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react-native - react 导航 v5 : Pass function as parameter to be used in headerRight button

转载 作者:行者123 更新时间:2023-12-01 21:38:49 25 4
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我想调用 headerRight 按钮点击屏幕上的函数。我在 useEffect 中将 func 作为参数传递,如下所示。

useEffect(() => {

navigation.setParams({ _confirmClick: confirmNotification })

}, [navigation])
useLayoutEffect(() => {
navigation.setOptions({
headerRight: () => (
<TouchableOpacity
onPress={() => params._confirmClick('New') }
style={[theme.marginRight15]}>
<View style={[styles.sendNotificationButton,
{
backgroundColor: notification ? theme.colors.notificationDeleteButtonColor :
theme.colors.sendbuttonColor,
borderColor: notification ? theme.colors.notificationDeleteButtonColor :
theme.colors.sendbuttonColor,
}]}>
<Ionicons name="ios-send" size={15}
style={theme.padding3}
color={theme.colors.whiteColor} />
</View>
</TouchableOpacity>
),
});
}, [navigation]);



function confirmNotification(status) {
...
}

当我点击按钮时,它说:TypeError: Cannot read property '_confirmClick' of undefined

最佳答案

您不需要在标题上设置参数。您可以直接将该方法传递给 headerRight :

function yourScreenName({ navigation }) {
useLayoutEffect(() => {
navigation.setOptions({
headerRight: () => (
<TouchableOpacity
onPress={() => confirmNotification('New')}
style={[theme.marginRight15]}>
<View style={[styles.sendNotificationButton,
{
backgroundColor: notification ? theme.colors.notificationDeleteButtonColor :
theme.colors.sendbuttonColor,
borderColor: notification ? theme.colors.notificationDeleteButtonColor :
theme.colors.sendbuttonColor,
}]}>
<Ionicons name="ios-send" size={15}
style={theme.padding3}
color={theme.colors.whiteColor} />
</View>
</TouchableOpacity>
),
});
}, [navigation, confirmNotification]); // pass method directly here
}

Here是一些文档。

关于react-native - react 导航 v5 : Pass function as parameter to be used in headerRight button,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61606346/

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