我有一个通过AJAX从PHP文件调用GPS坐标的功能，可以在地图上显示它们，这一切都很好。
该函数检查它获取的标记是否已经在地图上，如果它是否更新了它的位置，如果它不在地图上，它也将它添加到地图中。
我现在需要的东西似乎无法正常工作，我需要删除以前使用过的不再更新的标记。
这是我的代码：
setInterval(
$.ajax({
type: "POST",
url: 'markers.php',
success: function(data){
var json_obj = jQuery.parseJSON(JSON.stringify(data));
for (var i = 0; i < json_obj.length; i++) {
newcoordinate = new google.maps.LatLng(json_obj[i].lat, json_obj[i].lng);
if (markersArray[json_obj[i].name] && markersArray[json_obj[i].name].setPosition){
markersArray[json_obj[i].name].setPosition(newcoordinate);
if(json_obj[i].type == 'onderweg') {
markersArray[json_obj[i].name].setIcon(onderweg);
var label = markersArray[json_obj[i].name].getLabel();
label.color="black";
markersArray[json_obj[i].name].setLabel(label);
}
if(json_obj[i].type == 'leeg') {
markersArray[json_obj[i].name].setIcon(leeg);
var label = markersArray[json_obj[i].name].getLabel();
label.color="black";
markersArray[json_obj[i].name].setLabel(label);
}
if(json_obj[i].type == 'bezet') {
markersArray[json_obj[i].name].setIcon(bezet);
var label = markersArray[json_obj[i].name].getLabel();
label.color="white";
markersArray[json_obj[i].name].setLabel(label);
}
if(json_obj[i].type == 'afwachten') {
markersArray[json_obj[i].name].setIcon(afwachten);
var label = markersArray[json_obj[i].name].getLabel();
label.color="black";
markersArray[json_obj[i].name].setLabel(label);
}
if(json_obj[i].type == 'pauze') {
markersArray[json_obj[i].name].setIcon(pauze);
var label = markersArray[json_obj[i].name].getLabel();
label.color="black";
markersArray[json_obj[i].name].setLabel(label);
}
if(json_obj[i].type == 'geendienst') {
markersArray[json_obj[i].name].setIcon(geendienst);
var label = markersArray[json_obj[i].name].getLabel();
label.color="black";
markersArray[json_obj[i].name].setLabel(label);
}
} else {
addMarker(json_obj[i].lat, json_obj[i].lng, json_obj[i].name, json_obj[i].afkorting, json_obj[i].type);
}
}
},
dataType: "json"//set to JSON
})
}
,3000);
有人可以指出我正确的方向来解决这个问题吗？
Json数据：
[{"id":"1505108","0":"1505108","name":"Benny","1":"Benny","afkorting":"Be","2":"Be","address":"","3":"","speed":"0","4":"0","lat":"53.198181","5":"53.198181","lng":"6.563437","6":"6.563437","type":"onderweg","7":"onderweg","ts":"2017-04-17 18:26:34","8":"2017-04-17 18:26:34","ritid":"483648","9":"483648","bedrijf":"1","10":"1","ipadres":"188.207.111.80","11":"188.207.111.80","versie":"10","12":"10","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"},{"id":"1505110","0":"1505110","name":"Rene","1":"Rene","afkorting":"Re","2":"Re","address":"","3":"","speed":"0","4":"0","lat":"53.182274","5":"53.182274","lng":"6.577085","6":"6.577085","type":"leeg","7":"leeg","ts":"2017-04-17 18:26:37","8":"2017-04-17 18:26:37","ritid":"0","9":"0","bedrijf":"1","10":"1","ipadres":"188.207.126.119","11":"188.207.126.119","versie":"10","12":"10","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"},{"id":"1505112","0":"1505112","name":"Peter","1":"Peter","afkorting":"Pr","2":"Pr","address":"","3":"","speed":"0","4":"0","lat":"53.196129","5":"53.196129","lng":"6.581492","6":"6.581492","type":"bezet","7":"bezet","ts":"2017-04-17 18:26:39","8":"2017-04-17 18:26:39","ritid":"483650","9":"483650","bedrijf":"1","10":"1","ipadres":"92.69.203.230","11":"92.69.203.230","versie":"10","12":"10","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"},{"id":"1505113","0":"1505113","name":"Ivar","1":"Ivar","afkorting":"Iv","2":"Iv","address":"","3":"","speed":"0","4":"0","lat":"53.360897","5":"53.360897","lng":"6.414318","6":"6.414318","type":"onderweg","7":"onderweg","ts":"2017-04-17 18:26:39","8":"2017-04-17 18:26:39","ritid":"483649","9":"483649","bedrijf":"1","10":"1","ipadres":"188.207.122.118","11":"188.207.122.118","versie":"10","12":"10","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"},{"id":"1505114","0":"1505114","name":"Niels","1":"Niels","afkorting":"Ni","2":"Ni","address":"","3":"","speed":"0","4":"0","lat":"53.205700","5":"53.205700","lng":"6.586080","6":"6.586080","type":"bezet","7":"bezet","ts":"2017-04-17 18:26:40","8":"2017-04-17 18:26:40","ritid":"483633","9":"483633","bedrijf":"1","10":"1","ipadres":"92.69.223.185","11":"92.69.223.185","versie":"9","12":"9","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"},{"id":"1505109","0":"1505109","name":"John","1":"John","afkorting":"Jo","2":"Jo","address":"","3":"","speed":"0","4":"0","lat":"53.200897","5":"53.200897","lng":"6.596990","6":"6.596990","type":"leeg","7":"leeg","ts":"2017-04-17 18:26:36","8":"2017-04-17 18:26:36","ritid":"0","9":"0","bedrijf":"1","10":"1","ipadres":"92.69.230.150","11":"92.69.230.150","versie":"10","12":"10","inactief":"0","13":"0","beta":"0","14":"0","onzichtbaar":"0","15":"0"}]
数组回声：
[_.Ne, _.Ne, _.Ne, _.Ne, _.Ne, _.Ne, Benny: _.Ne, Rene: _.Ne, Peter: _.Ne, Ivar: _.Ne, Niels: _.Ne…]
这就是我得到的......但是每个对象都有自己的值，脚本确实有效..
答案 0 :(得分：0)
我没有机会对此进行全面测试，但你可以尝试一下。我会说这不是很有效率。循环遍历两个数组，但至少循环不是嵌套的。
首先稍微更改您的成功处理程序（添加空行以突出显示）。
success: function(data){
var idArray = [];
var json_obj = jQuery.parseJSON(JSON.stringify(data));
for (var i = 0; i < json_obj.length; i++) {
idArray.push(json_obj[i].id);
newcoordinate = new google.maps.LatLng(json_obj[i].lat, json_obj[i].lng);
...
然后循环遍历标记数组（此处称为数组，使用数组名称）并拼接新idArray中不存在的标记数组。
for (var i = array.length - 1; i >= 0; i--)
{
if (idArray.indexOf(array[i].id) < 0) array.splice(i, -1);
}
通过反向工作，你不会搞砸早期数组成员的索引。
我在地图应用程序中有一系列点，我必须动态添加和删除点。在那里我们替换数组，然后重新渲染地图。
有关拼接的更多信息 - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/splice