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JavaScript Build a Simple Dynamic Site with Node.js Creating a Basic Template Engine in Node.js Making Our Views DRY

Posted by Laxmikanta Nayak
.a{fill-rule:evenodd;}techdegree
Laxmikanta Nayak
Front End Web Development Techdegree Student 554 Points

I cant able to return data from a method

my server.js file:

var http=require('http');
var router= require('./router.js');
var weatherDetails=require('./weatherDetails.js');

var prettyJson=weatherDetails.getWeatherReport('USA');
console.dir(prettyJson);

var server=http.createServer(function(request,response){

    response.writeHead(200,{'Content-Type':'text/plain'});
    //console.log(request.url);
    if(request.url==='/'){
    router.home(request,response);
    }
    else{
        router.searchResult(request,response);
    }
    //response.write(weatherDetails.getWeatherReport('USA'));
    response.end("/nend");
}).listen(1337);
server.on("error",function(error){
console.error("Sorry we have this error :: "+error.message);
});

route.js file

function home(request,response){
    response.write('SEARCHING');
}
function searchResult(request,response){

    response.write('SEARCH RESULT');
}
module.exports.home=home;
module.exports.searchResult=searchResult;

weatherdetails.js

var http=require('http');
var body='';
var prettyJson='';
function getWeatherReport(country){
    var request=http.get("http://api.apixu.com/v1/current.json?key=5fccf98036ed460abb8195912162406&q="+country,function(response){
        response.on("data",function(chunk){
            body+=chunk;
        });
        response.on("end",function(){
            prettyJson=JSON.parse(body);
            while(prettyJson.location.name!='undefined'){
                console.log('here');
                return prettyJson.location.name; <-- on this link i cant able to get the value reutrned to my server.js file
            }
        });

    });

}
module.exports.getWeatherReport=getWeatherReport;

2 Answers

Steven Parker
Steven Parker
229,708 Points
Steven Parker
Steven Parker
229,708 Points

The line you indicated is inside an asynchronous callback function. Where do you expect the return value to go?

The bottom line is your own code doesn't directly call that function, so anything it tries to return will be lost. But you could assign it to some global variable, or call another function to do something with it.

Laxmikanta Nayak
.a{fill-rule:evenodd;}techdegree
Laxmikanta Nayak
Front End Web Development Techdegree Student 554 Points

Actually i want to take that value and use as field in my output , like location name , i have passed that to another function also assigned to global variable but still the same result is 'undefined'.

i want to access the locationname from getWeatherReoport Method from weatherdeails.js file in server.js to show to the use the result

Farid Wilhelm Zimmermann
Farid Wilhelm Zimmermann
16,753 Points
Farid Wilhelm Zimmermann
Farid Wilhelm Zimmermann
16,753 Points

I´d also recommend just putting in another callback function that runs once the data is received, using the received data as parameter.

I had a similiar problem to yours, when I was trying to make an API Call, that relied on JSON from a previous API Call, and got around by just declaring the second API Call inside a function, that got triggered .on("end") inside the callback function of the first API Call.