ajax与请求openWeatherMap

Question

所以我需要在地图上画一个图标，当状态正常的时候。但是当我将请求改为ajax时，我得到了

位置0处JSON中的意外令牌u

当我回去要求的时候一切正常..。但是我需要异常处理，所以这就是Ajax的原因。

JS

    function displayXML()
{
    if (this.readyState == 4)
    { 
        proccessResults();

        if (this.status == 200)
        {
            console.log(this.responseXML);

        } else {
            console.log(this.status);
            alert('Something went wrong')
        }
    }
}
var getWeather = function(northLat, eastLng, southLat, westLng) {
    gettingData = true;
    var requestString = "http://api.openweathermap.org/data/2.5/box/city?bbox="
        + westLng + "," + northLat + ","
        + eastLng + "," + southLat + ","
        + map.getZoom()
        + "&cluster=yes&format=json"
        + "&APPID=" + openWeatherMapKey;
    var ajax = new XMLHttpRequest();
    ajax.open('GET', requestString, true);
    ajax.onreadystatechange = displayXML;
    ajax.send();
    // request = new XMLHttpRequest();
    // request.onload = proccessResults;
    // request.open("get", requestString, true);
    // request.send();
};

var proccessResults = function() {

    var results = JSON.parse(this.responseText);
    if (results.list.length > 0) {
        resetData();
        for (var i = 0; i < results.list.length; i++) {
            geoJSON.features.push(jsonToGeoJson(results.list[i]));
        }
        drawIcons(geoJSON);
    }
};

答复：

{"cod":"200","calctime":0.0007,"cnt":2,"list":[{"id":3082473,"name":"Wejherowo","coord":{"lon":18.23559,"lat":54.605679},"main":{"temp":15.33,"pressure":1012,"humidity":93,"temp_min":15,"temp_max":15.56},"dt":1468456804,"wind":{"speed":0.5,"deg":0},"clouds":{"all":20},"weather":[{"id":801,"main":"Clouds","description":"few clouds","icon":"02n"}]},{"id":3099424,"name":"Gdynia","coord":{"lon":18.531879,"lat":54.51889},"main":{"temp":15.3,"pressure":1012,"humidity":93,"temp_min":15,"temp_max":15.56},"dt":1468456816,"wind":{"speed":0.5,"deg":0},"clouds":{"all":20},"weather":[{"id":801,"main":"Clouds","description":"few clouds","icon":"02n"}]}]}

Answer 1

不使用任何参数调用proccessResults函数。它不知道this (或this.responseText)是什么。您正在尝试将字符串"undefined“解析为JSON。这就是你的“意外标记u”错误的来源。

尝试将processResults函数更改为接受一个参数

var proccessResults = function(jsonString) {    
    var results = JSON.parse(jsonString);
    if (results.list.length > 0) {
        resetData();
        for (var i = 0; i < results.list.length; i++) {
            geoJSON.features.push(jsonToGeoJson(results.list[i]));
        }
        drawIcons(geoJSON);
    }
}

然后当你调用它时，把this.responseText传递给它

function displayXML(){
    if (this.readyState == 4){ 
        proccessResults(this.responseText);

        if (this.status == 200){
            console.log(this.responseXML);

        }
        else {
            console.log(this.status);
            alert('Something went wrong')
        }
    }
}