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JSON format from the API

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Mert
Mert on 24 May 2024 at 22:46
Edited: Rik on 25 May 2024 at 19:44
Hello, I'm going to fetch data from the OpenWeatherMap website as an app designer. The API will be in JSON format, but I keep getting errors. Can you help?
  3 Comments
Mert
Mert on 25 May 2024 at 13:30
Edited: Mert on 25 May 2024 at 13:30
I could not solve this problem
Please help me
Write a MATLAB function (fetchFiveDayForecast) that retrieves city-based weather data from the OpenWeatherMap API. This function should pull data in JSON format from the API and process this data to convert it into a format that can be utilized within MATLAB.
Rik
Rik on 25 May 2024 at 14:01
Sounds like homework to me. What code did you try?

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Accepted Answer

Abhishek Kumar Singh
Abhishek Kumar Singh on 25 May 2024 at 14:43
Hi @Mert,
I generated an API key from https://home.openweathermap.org/api_keys and tried a simple MATLAB function, from which I was able to retrieve data without any errors.
Refer to the MATLAB function code below:
function getWeatherForecast(cityName, apiKey)
% Construct the API request URL
url = ['http://api.openweathermap.org/data/2.5/forecast?q=', cityName, '&appid=', apiKey, '&units=metric'];
try
% Send the request and get the response
responseData = webread(url);
% Process and display the temp for the first time point
firstTemp = responseData.list{1}.main.temp;
disp(['First temperature forecast for ', cityName, ' is: ', num2str(firstTemp), ' Celsius']);
catch ME
% Error handling
disp('Failed to retrieve or process weather data.');
disp(['Error: ', ME.message]);
end
end
You can call the function as:
% cityName - Name of the city (string)
% apiKey - Your OpenWeatherMap API key (string)
getWeatherForecast('London', 'YOUR_API_KEY')
The API gives you a lot of information organized in lists. You'll need to think about how you want to use this information and might need to add more steps to work with it properly. Here's a quick look at what kind of data you get from the API:
K>> responseData.list{1}
ans =
struct with fields:
dt: 1.7166e+09
main: [1×1 struct]
weather: [1×1 struct]
clouds: [1×1 struct]
wind: [1×1 struct]
visibility: 10000
pop: 0.3100
sys: [1×1 struct]
dt_txt: '2024-05-25 15:00:00'
K>> responseData.list{1}.main
ans =
struct with fields:
temp: 20.0400
feels_like: 19.4800
temp_min: 19.8100
temp_max: 20.0400
pressure: 1015
sea_level: 1015
grnd_level: 1012
humidity: 53
temp_kf: 0.2300
K>> responseData.list{1}.main.temp
ans =
20.0400
I hope this helps you get preliminary idea of how to get started. You can further process and visualize the forecast data in various formats within MATLAB.
  2 Comments
Mert
Mert on 25 May 2024 at 16:24
Edited: Rik on 25 May 2024 at 19:44
I tried something but I have an error, This is my code and I gave an error.Can you check ?
error is : Undefined function 'fetchFiveDayForecast' for input arguments of type 'char'.
methods (Access = private)
function weatherData = fetchFiveDayForecast(cityName,apiKey)
baseURL = 'http://api.openweathermap.org/data/2.5/forecast';
query = sprintf('%s?q=%s&units=metric&appid=%s', baseURL, cityName, apiKey);
data = webread(query);
temperature = data.list.main.temp;
humidity = data.list.main.humidity;
windSpeed = data.list.wind.speed;
end
end
function GetInformationButtonPushed(app, event)
cityName = app.SelectCityDropDown.Value;
apiKey = 'myAPİ';
weatherData = fetchFiveDayForecast(cityName, apiKey);
plot(app.TemperatureAxes, weatherData.temperature);
title(app.TemperatureAxes, sprintf('Temperature Forecast for %s', cityName));
ylabel(app.TemperatureAxes, 'Temperature (°C)');
plot(app.HumidityAxes,weatherData.humidity);
title(app.HumidityAxes, sprintf('Humidity Forecast for %s', cityName));
ylabel(app.HumidityAxes, 'Humidity (%)');
plot(app.WindSpeedAxes, weatherData.windSpeed);
title(app.WindSpeedAxes, sprintf('Wind Speed Forecast for %s', cityName));
ylabel(app.WindSpeedAxes, 'Wind Speed (m/s)');
currentWeather = weatherData(1, :);
app.TemperatureLabel.Text = sprintf('Temperature: %.1f °C', currentWeather.temperature);
app.HumidityLabel.Text = sprintf('Humidity: %.0f%%', currentWeather.humidity);
app.WindSpeedLabel.Text = sprintf('Wind Speed: %.2f m/s', currentWeather.windSpeed);
end
Abhishek Kumar Singh
Abhishek Kumar Singh on 25 May 2024 at 17:46
Can you replace the line:
weatherData = fetchFiveDayForecast(cityName, apiKey);
with:
weatherData = app.fetchFiveDayForecast(cityName, apiKey);
and try again?
Let me know how it goes.

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