how create future function in flutter? - flutter

Question

i trying to create post function in flutter
here is my code:
Future<String> makePostRequest() async {
String requestResult;
var body = this.toMapRequest();
String requestUrl = RequestZarinpal.PAYMENT_REQUEST_URL;
String gateWayUrl;
String jsonBody = json.encode(body);
final encoding = Encoding.getByName('utf-8');
Response response = await post(
requestUrl,
headers: headers,
body: jsonBody,
encoding: encoding,
).timeout(const Duration(seconds: 10), onTimeout: () {
throw TimeoutException('The connection has timed out, Please try again!');
});
responseBody = response.body;
var parsedJson = json.decode(responseBody);
var data = parsedJson['data'];
var error = parsedJson['errors'];
if (error.toString() != "[]") {
var errorcode = parsedJson['errors']['code'];
print("$body va $requestUrl va $parsedJson");
requestResult = " شما ارور زیر را دریافت کرده اید \n$error";
} else if (data.toString() != "[]") {
var authority = parsedJson['data']['authority'];
requestResult = "اتوریتی شما با موفقیت ساخته شد و به درگاه متصل می شود";
_request.setAuthority(authority);
print(parsedJson);
String gateWay = RequestZarinpal.PAYMENT_GATEWAY_URL;
gateWayUrl = "$gateWay$authority";
if (await canLaunch(gateWayUrl)) {
await launch(
gateWayUrl,
forceSafariVC: false,
forceWebView: false,
headers: headers,
);
} else {
throw 'Could not launch $requestUrl';
}
print(requestResult);
return requestResult;
}
}
but i got this error:
The body might complete normally, causing 'null' to be returned, but the return type is a potentially non-nullable type.
Try adding either a return or a throw statement at the end.dart(body_might_complete_normally)
what should i do?

Answer 1

With the line if (error.toString() != "[]") you divided your function in 2 possibile outcome.
The positive one, doesn't have a return.
Maybe you should move return requestResult; after the curly bracket, so that the return gets fired regardless.

Answer 2

You are missing the return statement inside the if condition
if (error.toString() != "[]") {
var errorcode = parsedJson['errors']['code'];
print("$body va $requestUrl va $parsedJson");
requestResult = " شما ارور زیر را دریافت کرده اید \n$error";
return requestResult;
}