Simple weather only in farenheit ? #226

vimorozov opened this Issue Mar 24, 2016 · 8 comments


None yet

7 participants


Hi, any idea why simple weather only display the weather in farenheit ?


Yahoo has made some changes to their API . So I guess that is temporary solution.

biapar commented Mar 25, 2016


KibGim commented Mar 25, 2016


biapar commented Mar 25, 2016

The yahoo api work now on public url.

biapar commented Mar 25, 2016

Simple workaround

temp = F temperature
function getMyAltTemp(unit, temp) { if(unit === 'c') { return Math.round((5.0/9.0)*(temp-32.0)); } else { return Math.round((9.0/5.0)*temp+32.0); } }


Celsius also not displaying for my WOEID


As the bug is only displaying Farenheit, I came to this solution:

// Based on @biapar answer (y)
var getCelsius = function (temp) {
    return Math.round((5.0/9.0)*(temp-32.0));

var isFarenheitBug = weather.units.temp === "F";
if (isFarenheitBug) {
    weather.temp = getCelsius(parseInt(weather.temp));
    weather.units.temp = "C";

Hope its useful :)


Celsius seems to be working with the exception of "wind chill" ( feels like ). I really hate asking but I don't suppose someone could point methe right direction for a work around for this? Been banging my head for a few days now on this one with no luck .

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment