JQuery 3.x Object doesn't support property or method 'success'

Question

I am using the Yahoo Weather API. It works fine with JQuery 1.x. The issue is with JQuery 3.x. I am getting this error: Object doesn't support property or method 'success'. What can I use instead of .success?

I tried .done based on the documentation but it does not show any data.

https://api.jquery.com/deferred.done/

$(document).ready(function(){
  var city = "Erie, PA";
  var searchtext = "select * from weather.forecast where woeid in (select woeid from geo.places(1) where text='" + city + "') and u='f'"
  $.getJSON("https://query.yahooapis.com/v1/public/yql?q=" + searchtext + "&format=json").success(function(data){
      $('#weather-temp').html(data.query.results.channel.item.condition.temp + "°F");
      $("#weather-title").text(data.query.results.channel.title);
      $("#weather-text").text(data.query.results.channel.item.condition.text);
      $("#weather-speed").text("Wind: " + data.query.results.channel.wind.speed + " mph");
      $("#weather-sunset").text("Sunset: " + data.query.results.channel.astronomy.sunset);
      var iconUrl = 'https://s.yimg.com/zz/combo?a/i/us/we/52/'
      var weatherCode = data.query.results.channel.item.condition.code;
      $(".weather_icon").attr('src', iconUrl + weatherCode + '.gif');
  });
});

`success` is an ajax option on the properties of the request, not a function on the deferred. — Taplar, Mar 15 '19 at 16:01

score 2 · Accepted Answer · answered Mar 15 '19 at 16:03

2

For getJSON the success callback can be passed in as the second argument.

Ref. http://api.jquery.com/jQuery.getJSON/

$.getJSON("https://query.yahooapis.com/v1/public/yql?q=" + searchtext + "&format=json", function(data){
      $('#weather-temp').html(data.query.results.channel.item.condition.temp + "°F");
      $("#weather-title").text(data.query.results.channel.title);
      $("#weather-text").text(data.query.results.channel.item.condition.text);
      $("#weather-speed").text("Wind: " + data.query.results.channel.wind.speed + " mph");
      $("#weather-sunset").text("Sunset: " + data.query.results.channel.astronomy.sunset);
      var iconUrl = 'https://s.yimg.com/zz/combo?a/i/us/we/52/'
      var weatherCode = data.query.results.channel.item.condition.code;
      $(".weather_icon").attr('src', iconUrl + weatherCode + '.gif');
  });

answered Mar 15 '19 at 16:03

Taplar

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This means that the callback shall be fired in case the request failed also so validation has to be done in the callback. – Michel Hanna Mar 15 '19 at 16:05
That's not the case at all. The callback is only fired here when the request is successful – Rory McCrossan Mar 15 '19 at 16:06
@RoryMcCrossan so what if i need to fire some logic in case of failure, i should handle the promise? – Michel Hanna Mar 15 '19 at 16:07
Call `fail()` on the deferred object returned from `$.getJSON()` https://api.jquery.com/category/deferred-object/ – Rory McCrossan Mar 15 '19 at 16:14
Yeah, you'd have to attach additional logic for the failure case – Taplar Mar 15 '19 at 16:15

Onix · Answer 2 · 2019-03-15T16:02:42.427

0

Look here: http://api.jquery.com/jQuery.ajax/

.done(function() {
  //do stuff
});

edited Mar 15 '19 at 16:02

answered Mar 15 '19 at 16:01

Onix

2,129
2
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Ok, now the link is not relevant to your updated example. I think you meant https://api.jquery.com/deferred.done instead – Rory McCrossan Mar 15 '19 at 16:05
@RoryMcCrossan Yes you are correct. I made a few windows open and copied the wrong link. – Dan Nick Mar 15 '19 at 16:08

JQuery 3.x Object doesn't support property or method 'success'

2 Answers2