[英]how do I select an Item in list view with item.description
here is the app, I want to create diferent screens with diferent catergories in this case I have Dermatologista and Hospital, how can I select just one description这是应用程序,我想创建具有不同类别的不同屏幕,在这种情况下,我有皮肤科和医院,我如何只选择一个描述
const [state, setState] = useState({
places: [
{
id: 1,
title: 'Clinica da pele',
description: 'Dermatologista',
latitude:-2.42206406,
longitude:-54.71947789,
},
{
id: 2 ,
title:'Unimed',
description:'Hospital',
latitude:-2.42501721,
longitude:-54.71146077,
},
{
id: 3,
title: 'Dra. Josimar',
description:'Dermatologista',
latitude: -2.4288346,
longitude:-54.7290553,
}
]
});
return(
I just want to select the items with the description == dermatologista how can I do this ?我只想选择带有描述 == dermatologista 的项目我该怎么做?
<SafeAreaView>
<FlatList
styles = {styles.PlaceContainer}
showsVerticalScrollIndicator
data={state.places}
keyExtractor={item => item.id}
renderItem={({ item }) => {
return(
<View key={item.id} style={styles.place} >
<Text>{item.title}</Text>
<Text>{item.description}</Text>
</View>
)
}
}
/>
</SafeAreaView>
) } ) }
You can use array.filter
:您可以使用array.filter
:
const filteredPlaces = state.places.filter( place => place.description === "Dermatologista" )
and pass filteredPlaces
instead of the entire object to the child component.并将filteredPlaces
而不是整个对象传递给子组件。
Try this尝试这个
<SafeAreaView>
<FlatList
styles = {styles.PlaceContainer}
showsVerticalScrollIndicator
data={state.places}
keyExtractor={item => item.id}
renderItem={({ item }) => {
item.description == "dermatologista" ? (
<View key={item.id} style={styles.place} >
<Text>{item.title}</Text>
<Text>{item.description}</Text>
</View>
):""
}
}
/>
</SafeAreaView>
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