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Problem in yahoo weather

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Level 8

Problem in yahoo weather

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

1 ACCEPTED SOLUTION

Accepted Solutions

Re: Problem in yahoo weather

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

14 REPLIES 14

Re: Problem in yahoo weather

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

Re: Problem in yahoo weather

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 ,

Re: Problem in yahoo weather

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

Re: Problem in yahoo weather

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

Re: Problem in yahoo weather

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

Re: Problem in yahoo weather

I am working on parsing of the data but it still doesn't work. If you can help me with that i will be glad.

Thank you,

Robert

Re: Problem in yahoo weather

Hi Robert,

I would be glad to help. Please post the code your are experiencing trouble with.

Thank you,

Veronica

Re: Problem in yahoo weather

I have solved it but thank you for your interest.

Robert.

Re: Problem in yahoo weather

You're welcome Robert. Please don't hesitate to post your questions on the forum if you have trouble with anything else.

Thank you,

Veronica