如何在javascript中从回调函数中获取整个数据

我编写了以下函数，该函数从url获取json数据。

function getWeatherDataForCities(cityArray){ var arrAllrecords = []; var toDaysTimestamp = Math.round((new Date()).getTime() / 1000) - (24*60*60); for(var i in cityArray){ for(var j=1; j<=2; j++){ var jsonurl = "http://api.openweathermap.org/data/2.5/history/city?q="+cityArray[i]+"&dt="+toDaysTimestamp; $.ajax({ url: jsonurl, dataType: "jsonp", mimeType: "textPlain", crossDomain: true, contentType: "application/json; charset=utf-8", success: function(data){ arrAllrecords[j]["cityName"] = data.list[0].city.name; arrAllrecords[j]["weather"] = data.list[0].weather[0].description; } }); toDaysTimestamp = toDaysTimestamp - (24*60*60); } } return arrAllrecords; //returning arrAllrecords } $(document ).ready(function() { var cityArray = new Array(); cityArray[0] = "pune"; var arrAllRecords = getWeatherDataForCities(cityArray); //Print All records returned by getWeatherDataForCities(cityArray); }); 

我在上面的代码中写了一些注释。我调用了getWeatherDataForCities function ，它返回url所有记录。我在函数中声明了getWeatherDataForCities array 。我想在该array中添加所有返回的记录。我已经尝试过如上所述但是没有任何进展array 。

在控制台中显示j和arrAllrecords未定义。

如何从回调函数中获取数组中的所有记录？

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Answer 1

你的getWeatherDataForCities函数不会返回任何内容，因为ajax操作是异步的。 你需要使用回调。

修改您的函数以接受回调函数：

 function getWeatherDataForCities(cityArray, callback){ var arrAllrecords = []; var toDaysTimestamp = Math.round((new Date()).getTime() / 1000) - (24*60*60); for(var i in cityArray){ for(var j=1; j<=2; j++){ var jsonurl = "http://api.openweathermap.org/data/2.5/history/city?q="+cityArray[i]+"&dt="+toDaysTimestamp; $.ajax({ url: jsonurl, dataType: "jsonp", mimeType: "textPlain", crossDomain: true, contentType: "application/json; charset=utf-8", success: function(data){ arrAllrecords[j]["cityName"] = data.list[0].city.name; arrAllrecords[j]["weather"] = data.list[0].weather[0].description; // call the callback here callback(arrAllrecords); } }); toDaysTimestamp = toDaysTimestamp - (24*60*60); } } } 

并像这样使用它：

 $(document ).ready(function() { var cityArray = new Array(); cityArray[0] = "pune"; getWeatherDataForCities(cityArray, function(arrAllrecords) { // Do something with your data }); }); 

Answer 2

您正在尝试使用空数组。 获取值时，它将始终返回undefined。

 var arrAllrecords = []; arrAllrecords[2]; //undefined arrAllrecords[2]["cityname"]; //undefined 

你应该更好地使用对象数组。

我不知道你为什么使用变量j。 以下代码适用于我。

 var arrAllrecords = []; function getWeatherDataForCities(cityArray){ var toDaysTimestamp = Math.round((new Date()).getTime() / 1000) - (24*60*60); for(var i in cityArray){ var jsonurl = "http://api.openweathermap.org/data/2.5/history/city?q="+cityArray[i]+"&dt="+toDaysTimestamp; $.ajax({ url: jsonurl, dataType: "jsonp", mimeType: "textPlain", crossDomain: true, contentType: "application/json; charset=utf-8", success: function(data){ arrAllrecords.push({ "cityName" : data.list[0].city.name, "weather" : data.list[0].weather[0].description }); if(arrAllrecords.length === cityArray.length) { callback(arrAllrecords); } } }); } } function callback(arrAllrecords) { console.log(arrAllrecords); } $(document).ready(function() { var cityArray = new Array(); cityArray[0] = "pune"; cityArray[1] = "mumbai"; cityArray[2] = "delhi"; getWeatherDataForCities(cityArray); });