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mpatel-211-Visitor
1-Visitor
February 22, 2016
Solved

Problem in yahoo weather

  • February 22, 2016
  • 2 replies
  • 10886 views

Hello Expert ,

I am getting this error on get yahoo weather :

If you know the solution for this error please suggest me.@

Execution error in service script [WeatherThing GetWeather] : Wrapped org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 429; The content of elements must consist of well-formed character data or markup. Cause: The content of elements must consist of well-formed character data or markup.

Code :

var stringLocation=location+" ";

var arrayLocations=stringLocation.split(",");

var lat=arrayLocations[0];

var long=arrayLocations[1];

//Modify the open weather url here if you are using your own API key

var params = {

url: "http://api.openweathermap.org/data/2.5/forecast?lat="+lat+"&lon="+long+ "&units=imperial" +"&type=accurate&mode=xml&APPID=4abc8a3c9f101ccf9c7cfe7cbf3dfed7" /* STRING */, 

    timeout: 60 /* NUMBER */

};

// result: XML

var xmlPage = Resources["ContentLoaderFunctions"].LoadXML(params);   // Error Generate on This line

var params = {

    infoTableName : "InfoTable",

    dataShapeName : "OpenWeatherFeed"

};

// CreateInfoTableFromDataShape(infoTableName:STRING("InfoTable"), dataShapeName:STRING):INFOTABLE(YahooWeatherFeed)

var result = Resources["InfoTableFunctions"].CreateInfoTableFromDataShape(params);

var from_today_value=xmlPage.forecast.time[0].@from.substring(0,10);

logger.warn("from_today_value= "+from_today_value);

for each (var item in xmlPage.forecast.time) {

    var row = new Object();

    var from_current_value=item.@from.substring(0,10);

    if(from_current_value!=from_today_value){

        row.WindDirection = item.windDirection.@name;

        row.WindSpeed=item.windSpeed.@mps;

        row.WeatherTemp=parseFloat(item.temperature.@value);

        row.WeatherHumidity=parseFloat(item.humidity.@value);

        row.From=item.@from;

        row.To=item.@to;

        row.Title=xmlPage.location.name+" "+xmlPage.location.country;

        row.Precipitation=(item.precipitation.@type+" "+item.precipitation.@value)==" "?"No Precipitation":item.precipitation.@type;

        result.AddRow(row);

    }

}

Thanks ,

Mayank

    Best answer by vmihai

    Hi Mayank,

    Yahoo Weather API is no longer in service. Please use OpenWeatherMap as a replacement or Wunderground API.

    Below is an implementation for a  OpenWeatherMap API service call:

    var stringLocation=location+" ";

    var arrayLocations=stringLocation.split(",");

    var lat=arrayLocations[0];

    var long=arrayLocations[1];

    var params = {

    url: "http://api.openweathermap.org/data/2.5/forecast?lat="+lat+"&lon="+long+"&units=metric&type=accurate&mode=xml&APPID=4abc8a3c9f101ccf9c7cfe7cbf3dfed7" /* STRING */,

    timeout: 160 /* NUMBER */

    };

    // result: XML

    var xmlPage = Resources["ContentLoaderFunctions"].LoadXML(params);

    var params = {

    infoTableName : "InfoTable",

    dataShapeName : "OpenWeatherFeed"

    };

    var result = Resources["InfoTableFunctions"].CreateInfoTableFromDataShape(params);

    var from_today_value=xmlPage.forecast.time[0].@from.substring(0,10);

    1. logger.warn("from_today_value= "+from_today_value);

    for each (var item in xmlPage.forecast.time) {

    var row = new Object();

    var from_current_value=item.@from.substring(0,10);

    row.WindDirection = item.windDirection.@name;

        row.WindSpeed=item.windSpeed.@mps;

    row.WeatherTemp=parseInt(item.temperature.@value);

    row.WeatherHumidity=parseFloat(item.humidity.@value);

    row.From =  item.@from;

    row.To =  item.@to;

    row.Title=xmlPage.location.name+" "+xmlPage.location.country;

    row.Precipitation=(item.precipitation.@type+" "+item.precipitation.@value)==" "?"NoPrecipitation":item.precipitation.@type;

    result.AddRow(row);

    }


    Here are the inputs and outputs of the service

    Please let me know if you are in need of anything else.

    Thank you,

    Veronica

    2 replies

    A
    Aanjan5-Regular Member
    5-Regular Member
    February 22, 2016

    Mayank, you might've already checked this, but just in case - are you getting a valid xml from that url?

    M
    mpatel-211-VisitorAuthor
    1-Visitor
    February 23, 2016

    Yes Ravi , i got valid xml and Json also , but problem in parsing this xml and as well json in Thingworx service, I tested this online parser and no any error shown me on online parser. Would you provide me any url for weather. so i can test it.

    Thanks ,

    V
    vmihai1-VisitorAnswer
    1-Visitor
    February 23, 2016

    Hi Mayank,

    Yahoo Weather API is no longer in service. Please use OpenWeatherMap as a replacement or Wunderground API.

    Below is an implementation for a  OpenWeatherMap API service call:

    var stringLocation=location+" ";

    var arrayLocations=stringLocation.split(",");

    var lat=arrayLocations[0];

    var long=arrayLocations[1];

    var params = {

    url: "http://api.openweathermap.org/data/2.5/forecast?lat="+lat+"&lon="+long+"&units=metric&type=accurate&mode=xml&APPID=4abc8a3c9f101ccf9c7cfe7cbf3dfed7" /* STRING */,

    timeout: 160 /* NUMBER */

    };

    // result: XML

    var xmlPage = Resources["ContentLoaderFunctions"].LoadXML(params);

    var params = {

    infoTableName : "InfoTable",

    dataShapeName : "OpenWeatherFeed"

    };

    var result = Resources["InfoTableFunctions"].CreateInfoTableFromDataShape(params);

    var from_today_value=xmlPage.forecast.time[0].@from.substring(0,10);

    1. logger.warn("from_today_value= "+from_today_value);

    for each (var item in xmlPage.forecast.time) {

    var row = new Object();

    var from_current_value=item.@from.substring(0,10);

    row.WindDirection = item.windDirection.@name;

        row.WindSpeed=item.windSpeed.@mps;

    row.WeatherTemp=parseInt(item.temperature.@value);

    row.WeatherHumidity=parseFloat(item.humidity.@value);

    row.From =  item.@from;

    row.To =  item.@to;

    row.Title=xmlPage.location.name+" "+xmlPage.location.country;

    row.Precipitation=(item.precipitation.@type+" "+item.precipitation.@value)==" "?"NoPrecipitation":item.precipitation.@type;

    result.AddRow(row);

    }


    Here are the inputs and outputs of the service

    Please let me know if you are in need of anything else.

    Thank you,

    Veronica

    R
    rzalman1-Visitor
    1-Visitor
    February 23, 2016

    Hi Veronica,

    this code is for forecast in weather application, is possible to get code for actual weather in same application? If I am wrong please correct me.

    http://openweathermap.org/forecast5  Here you use this, it's for forecast in 3 hours interval

    http://openweathermap.org/current

    And i would like to get data from this page if it is possible.

    So i would like exchange yahoo weather for this one:YahooWeatherFeed.

    Thank you for any reply,

    Robert

    V
    vmihai1-Visitor
    1-Visitor
    February 24, 2016

    Hi Robert,

    Indeed, the code I posted is for retrieving forecast data. You can use the  Rest call described in the tutorial link you mentioned.

    The parsing of the data once you get it in your thingworx service will be different from the one did for the forecast call. Let me know if you are experiencing problems with parsing it.

    Thank you,

    Veronica

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