How do I use a variable outside a function in react native? - react-native

Question

I need to use that variable "let result" outside of this function. I need to use that variable inside of another function. How can I get it?
My code looks like this:
method: "GET",
headers: {
'Accept': 'application/json',
'Content-Type': 'application/json',
'Cache-Control': 'no-cache',
'Authorization': 'Bearer ' + this.state.clientToken,
},
})
.then((response) => response.json())
.then((responseJson) => {
let result= JSON.parse((responseJson.lstsurveyoncode))
let answer = JSON.parse(responseJson.lstsurveyoncode)[0].qoptions
console.log("responsejson",result[1].QUESTIONCODE)
console.log("answerrr",answer);
console.log("data length",result.length);
this.setState({
isLoading:false,
dataresponse:result,
// count:Object.keys(dataresponse).length
},function(){
});

Answer 1

You can create a global variable, and then update it inside your first function and in the second function you check if the variable is defined
let result;
function firstFunction(){
result = 'Your Json Result'
}
function secondFunction(){
if (!result) return;
// Here you can use your variable result
}

Answer 2

You should read about async-await and state managment in react-native.
For quick solution, you should mark your function with async modifier and await for result. Sample code below:
async function yourRequest() {
return new Promise((resolver, rejected) => {
// your async request
})
.then(response => response.json())
.then(responseJson => {
const result = JSON.parse(responseJson.lstsurveyoncode)
const answer = JSON.parse(responseJson.lstsurveyoncode)[0].qoptions
// ...
return {
result: result,
answer: answer,
}
})
}
async function handleResponse() {
const response = await yourRequest()
// and now, you can access to variables declared at `makeRequest` function
console.log(response.result)
console.log(response.answer)
}