Question

我有这个普通的对象，我已经从weather API返回了代码，所以根据代码我想显示不同的HTML。代码应该与codeArrays中的一个匹配，并且根据匹配我应该具有相应的html

var WeatherIcons = {
    sunShower: {
        html: '<div class="icon sun-shower"></div>',
        codes: [1009, 1150, 1153, 1240, 1249]
    },
    thunderStorm: {
        html: '<div class="icon thunder-storm"></div>',
        codes: [1087, 1117, 1147, 1273, 1276, 1279]
    },
    cloudy: {
        html: '<div class="icon cloudy"></div>',
        codes: [1006, 1030, 1135]
    },
    snowing: {
        html: '<div class="icon snowing"></div>',
        codes: [1066, 1114, 1210, 1213, 1219, 1222, 1225, 1255, 1258, 1261, 1264, 1282]
    },
    sunny: {
        html: '<div class="icon sunny"></div>',
        codes: [1000, 1003, 1216]
    },
    rainy: {
        html: '<div class="icon rainy"></div>',
        codes: [1063, 1069, 1072, 1168, 1171, 1180, 1183, 1186, 1189, 1192, 1195, 1198, 1201, 1204, 1237, 1243, 1246, 1252]
    }
};

Answer 1

这是我得到的地方......

var resolveWeatherIcon = function (Condition) {
    var iconHtml = '';

    $j.each(WeatherIcons, function (key, val) {

        if (val.codes.indexOf(Condition.code) >= 0) {
            iconHtml = WeatherIcons[key].html;
            return false;
        };
    })

    return iconHtml;

}

它有效...很抱歉打扰并感谢在Plain Object和JSON之间明确

在jquery对象中查找另一个属性值

1 个答案: