我已经成功地在Parse.com中存储了一些geopoints,现在在另一个页面中,我想将它们全部记录下来,这样我就可以将它们放入一些变量中,然后在google中放置一个标记。所以我试着用这个代码来得到它们,但是我确实错过了一些东西,我需要你的建议。
Parse.initialize("APPID", "JSKEY");
var PhotoObject = Parse.Object.extend('magazia');
var photoObject = new PhotoObject();
var query = new Parse.Query(PhotoObject);
query.select('latlon');
query.find({
success: function(locationList) {
alert("Successfully retrieved " + locationList.length + " locations.");
for (var i = 0; i < locationList.length; i++) {
var locationsBlock = {};
locationsBlock = JSON.parse(JSON.stringify(locationList[i]));
var location = {};
location = JSON.parse(JSON.stringify(locationsBlock.geolocation));
alert(location.latitude);
};
},
error: function(error) {
alert("Error: " + error.code + " " + error.message);
}
});所以我有一个叫做"magazia“的类,在这个类里面有一个列,叫做"latlon”,它是Geopoint。例如,该列的内容是48.29124, 28.52015浮点数。警报显示"magazia“类中正确的行数。
有人知道为什么我不能从上面的代码中得到结果吗?提前谢谢。
发布于 2016-01-21 10:30:24
好吧,那是个愚蠢的错误
var PhotoObject = Parse.Object.extend('magazia');
var photoObject = new PhotoObject();
var query = new Parse.Query(PhotoObject);
query.select('latlon');
query.find({
success: function(locationList) {
console.log("Successfully retrieved " + locationList.length + " locations.");
for (var i = 0; i < locationList.length; i++) {
var locationsBlock = {};
locationsBlock = JSON.parse(JSON.stringify(locationList[1]));
var location = {};
location = JSON.parse(JSON.stringify(locationsBlock.latlon));
var lat = location.latitude;
var lon = location.longitude;
console.log(lat, lon);
};
},
error: function(error) {
alert("Error: " + error.code + " " + error.message);
}
});而不是locationsBlock.geolocation,应该有locationsBlock.latlon
https://stackoverflow.com/questions/34919661
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