how to apply useEffect and map function without repeat more than one a time

Question

I have a component in react native that gets data from mongodb

const Num = () => {
    console.log("how many times do you see this message?")
    const [data, setData] = useState([])
    const [loading,setloading] = useState(true)

    const url = "http://localhost:3100/cuentas"

    
    useEffect(()=>{
        fetch(url)
        .then((response)=>response.json())
        .then((json)=>setData(json))
        .catch((error)=>console.error(error))
        .finally(()=>setloading(false))
    },[])
    

  
        return(
            
            <View style={styles.sectionContainer}>
                {loading ? ( <Text>Loading....</Text>) : ( 
                    
                data.map((post,xid)=>(
                
               <TouchableOpacity  key={xid} style={styles.button}><Text style={styles.text}>#{post._id.slice(-2)}</Text></TouchableOpacity>
    
                         ))
                    )}
            </View>
        )

}

Then i put the component into another component to set the multiple items into a scrollView

const Notas = () => {

    
    return(
    
        <ScrollView Style={styles.container}>
            <View style={styles.sectionContainer}>
            
                <Num></Num>
            </View>
        </ScrollView>
    
    )
}

But the problem here is that it doesn't apply the map one time,it does it 3 times and you can see it by the console.log that is set first at the Num component

I've been debugging and i find that if i remove the .then's it doesn't repeat the map function 3 times, i tried to put the whole funcion inside the useEffect but i get an error that said "useEffect must not return anything besides a function"

If some one have a tip, it could be useful.

Answer 1

Try using async await:

useEffect( async ()=>{
    await fetch(url)
    .then((response)=>response.json())
    .then((json)=>setData(json))
    .catch((error)=>console.error(error))
    .finally(()=>setloading(false))
},[])