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Problem in yahoo weather

SOLVED

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

Re: Problem in yahoo weather

Hi,

I'm still having an error message of

  • org.xml.sax.SAXParseException; lineNumber: 1; columnNumber: 1; Content is not allowed in prolog Cause: Content is not allowed in prolog

I guess it's related to this issue

( Mark S. Kolich - Java: Resolving org.xml.sax.SAXParseException: Content is not allowed in prolog )

But I have no idea how to modify the code

Here is the code

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=ff07c7c18d3a5d86f13a93134f1d94a9" /* 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);

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);

}

Thank you.

Re: Problem in yahoo weather

Hello Lee ,

Take input parameter as a Location ,

var test=Location+" ";

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

var lat=arrayLocations[0];

var long=arrayLocations[1];

var prm = {

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

    timeout: 160 /* NUMBER */  

};

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

// result: XML

var params = {

    infoTableName : "InfoTable",

    dataShapeName : "OpenWeatherFeed"

};

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);

    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);

}

Thanks ,

Mayank

Re: Problem in yahoo weather

Hello Im have a diferent error mi code is

// Need to make a free wunderground.com account to get this key

var apiKey = "381603bc1002b5ce";

var temperature;

var humidity;

  //Inputted location needs to be valid lat, long: 40.7127, 74.0059

if(location != null){

    var test=location+" ";

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

var lat=arrayLocations[0];

var long=arrayLocations[1];

var prm = {

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

    timeout: 160 /* NUMBER */ 

};

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

// result: XML

var params = {

    infoTableName : "InfoTable",

    dataShapeName : "OpenWeatherFeed"

};

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 O

atbject();

    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);

}

    }

attach image error.png

Highlighted

Re: Problem in yahoo weather

Have you create a data shape for store value ?