Question 34547 in React native Question posted in
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how to call in return inside of if statement's variable outside of statement – React native

if (condition works){
  let x = 20;
  let y= 30;
  console.log(x);
  console.log(y);
}

using react native:
I don’t want to declare the variable outside the if condition so I use the let inside the if statement. Now, I want to call this outside of the if statement in return (),

How can I call the x, y in return and get the values?

suppose:

return (
  <View style={styles.container}>
    <Text style={styles.text}>
       X:{x} & Y:{y} 
    </Text>
  </View>
);
Tags:

2

Answers


  1. 0

    Akash Ghosh, this is not the right approach. You should save your x and y in useState, and show them if your condition is met, smt like this:

    const [x,setX]=useState(null);
const [y,setY]=useState(null);

    And in you function you just set the value:

        if(conditionMet){
     setX(20);
     setY(30);
    }

    And in your rendering if you don’t want to show them if they value didn’t change you do this:

    return (
  <View style={styles.container}>
  { x && y ? <Text style={styles.text}>
    X:{x} & Y:{y} 
  </Text> :null }

  </View>
);

enter code here
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  2. 0 
     you can define like it  


function myfunction (){
    if (condition works){
                let x = 20;
                let y= 30;
                let data={x,y}
                return data
             }
}

    and can use like it

    return (
  <View style={styles.container}>
  { x && y ? <Text style={styles.text}>
    X:{myfunction().x} & Y:{myfunction().y} 
  </Text> :null }

  </View>
);
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