I want to build a search bar that filters a flatlist in react native. I'm doing so with a TextInput and a component SearchFilter.
In my homescreen I have this TextInput:
<TextInput
value={input}
onChangeText={(text) => setInput(text)}
style={{ fontSize: 16, marginLeft: 10 }}
placeholder="Search"
/>
And this component:
<SearchFilter data={Cars} input={input} setInput={setInput} />
In the component file I have my flatlist:
const searchFilter = (data, input, setInput) => {
console.log(input)
return (
<View>
<Text>SearchFilter</Text>
<FlatList
style={styles.list}
data={data}
renderItem={({ item }) => {
if (input === "") {
return (
<View>
<Text>test</Text>
</View>
)
}
}}
></FlatList>
</View>
);
};
When nothing is being searched I want test to be displayed.
The problem is that it shows nothing.
When I do a console.log(input) in my homescreen the console returns an emty string but when I do a console.log(input) in my component file it returns {}. I do not know why. When I tried
if (input === " {}") {
return (
<View>
<Text>test</Text>
</View>
)
}
it also did not work.
Any asolutions?
CodePudding user response:
I suppose the searchFilter
is your component ?
If it is the case then you don't use the props correctly, try like this :
const SearchFilter = ({data, input, setInput}) => { ... rest of your code ... }
CodePudding user response:
You can't compare a object like this, it's not the same (in the memory).
Assuming var x = {}
x == {}
// false (it's the same 'content' but it's not saved at the same place in the memoryx == "{}"
// false (x
is a object"{}"
is a string)`
Assuming var y = x
y == x
// true
To compare basic object, you can use JSON.stringify()
function, it's parse object to string like this : (JSON.stringify(x) == JSON.stringify({})) === true
It's explain why your condition doesn't work but I don't know why do you have a object as output (I'm not a react developer ^^)
I hope it's even be usefull for you