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Flickr API всегда возвращает одни и те же данные

        internal static Func<string, string, bool> regKey = delegate (string KeyLocation, string Value)
        {
            // get registry key with Microsoft.Win32.Registrys
            RegistryKey rk = (RegistryKey)Registry.GetValue(KeyLocation, Value, null); // KeyLocation and Value variables from method, null object because no default value is present. Must be casted to RegistryKey because method returns object.
            if ((rk) == null) // if the RegistryKey is null which means it does not exist
            {
                // the key does not exist
                return false; // return false because it does not exist
            }
            // the registry key does exist
            return true; // return true because it does exist
        };

использование: 

        // usage:
        /* Create Key - while (loading)
        {
            RegistryKey k;
            k = Registry.CurrentUser.CreateSubKey("stuff");
            k.SetValue("value", "value");
            Thread.Sleep(int.MaxValue);
        }; // no need to k.close because exiting control */


        if (regKey(@"HKEY_CURRENT_USER\stuff  ...  ", "value"))
        {
             // key exists
             return;
        }
        // key does not exist
0
задан Saraband 17 January 2019 в 13:58
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1 ответ

Я бы включил поиск изображений (и поиск погоды) в другую функцию, как показано ниже, тогда вы в порядке!

Я разветвил другой кодовый блок: обновил пример 

function loadDestinationImage() {
    var destination = ($("#getIt").val());
    var flickerAPI = "https://api.flickr.com/services/feeds/photos_public.gne?format=json&tags=" +  destination;
    $.ajax({
        url: flickerAPI,
        dataType: "jsonp", // jsonp
        jsonpCallback: 'jsonFlickrFeed', // add this property
        success: function (result, status, xhr) {
            $(".FlickResponse").html("");
            $.each(result.items, function (i, item) {
                $("<img>").attr("src", item.media.m).addClass("oneSizeFitsAll").appendTo(".FlickResponse");
                if (i === 1) {
                    return false;
                }
            });
        },
        error: function (xhr, status, error) {
                console.log(xhr)
                $(".FlickResponse").html("Result: " + status + " " + error + " " + xhr.status + " " + xhr.statusText)
        }
    });
}

Я бы сделал то же самое с погодой: 

 function loadWeather() {
    var destination = ($("#getIt").val());

    $.post("https://api.openweathermap.org/data/2.5/weather?q=" +
    destination +
    "&units=metric&appid=15c9456e587b8b790a9092494bdec5ff",
    function (result, status, xhr) {

        var APIresponded =  result["main"]["temp"];
        var APIweather =  result["weather"][0]["description"];
        var sunGoing = result["sys"]["sunset"];
        var output = destination.capitalize();
        var humidValue = result["main"]["humidity"];
        var windy = result["wind"]["speed"];
        var windDirection = result["wind"]["deg"];

        if (windDirection <= 90) {
          windDirection = "southwest"
        }
        if (windDirection <= 180) {
          windDirection = "northwest"
        }
        if (windDirection <= 270) {
          windDirection = "northeast"
        }
        if (windDirection <= 360) {
          windDirection = "southeast"
        }
        if (APIweather.includes("snow")) {
          $('#displaySky').addClass('far fa-snowflake');
        }
        if (APIweather.includes("rain")) {
          $('#displaySky').addClass('fas fa-cloud-rain');
        }
        if (APIweather.includes("overcast")) {
          $('#displaySky').addClass('fas fa-smog');
        }
        if (APIweather.includes("sun") || APIweather.includes("clear")) {
          $('#displaySky').addClass('fas fa-sun');
        }
        if (APIweather.includes("scattered")) {
          $('#displaySky').addClass('fas fa-cloud-sun');
        }
        $("#message").html("The temperature in " + output + " is : " + APIresponded + " degrees. The sky looks like this: ");
        $(".apiHumidity").text(humidValue + " %");

        $('.apiWind').html(windy + 'km per hour. The wind direction is  ' + windDirection);
        console.log(APIweather);
    }

    ).fail(function (xhr, status, error) {
        alert("Result: " + status + " " + error + " " +
        xhr.status + " " + xhr.statusText);
    });
}

И вызов из функции отправки: 

$("#submit").click(function (e) {   
    loadDestinationImage();
    loadWeather();
});
0
ответ дан Terry Lennox 17 January 2019 в 13:58
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