批量地理编码 [英] Batch Geocoding
本文介绍了批量地理编码的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我需要扭转地理codeA十万纬度列表/经度坐标。我能做到这一点的一组坐标没有问题，但是当它涉及到许多人来说，只会做5的时间。我需要弄清楚如何做到这一步，时间的循环，所以它不会被谷歌拒绝。
我在别处被告知，我需要利用回调函数，但我不能确定如何做到这一点。
下面是一些演示code
＆LT; HEAD＆GT;
＆LT; META HTTP-当量=内容类型内容=text / html的;字符集= UTF-8/＆GT;
＆LT;脚本类型=文/ JavaScript的SRC =JS / jQuery的-1.7.2.min.js＆GT;＆LT; / SCRIPT＆GT;
＆LT;脚本类型=文/ JavaScript的SRC =http://maps.google.com/maps/api/js?sensor=false＆GT;＆LT; / SCRIPT＆GT;
＆LT;脚本类型=文/ JavaScript的＆GT;
VAR地理codeR =新google.maps.Geo codeR（）;
变种C = [];
C [0] =43.5878，-79.6777;
C [1] =49.2828，-123.1414;
C [2] =49.2895，-123.127;
C [3] =49.2784，-123.1364;
C [4] =49.0327，-122.2529;
C [5] =49.2696，-123.0587;
C [6] =49.269，-122.9954;
C [7] =49.235，-122.882;
C [8] =49.1908，-122.7508;
C [9] =49.085，-122.4111;
变种一个= [];
函数codeLatLng（输入，I）{
变种latlngStr = input.split（，，2）;
变种纬度= parseFloat（latlngStr [0]）;
变种LNG = parseFloat（latlngStr [1]）;
VAR经纬度=新google.maps.LatLng（纬度，经度）;
地理coder.geo code（{
经纬度：经纬度
}，功能（结果状态）{
A [1] =结果[0] .formatted_address;
$（＃考）追加（A [1] +＆LT; BR /＆gt;中）。
}）;
}
函数此Showmethemoney（）{
对于（VAR I = 0; I＆LT; = c.length - 1;我++）{
$（＃考）追加（I +）+ C（I）+＆LT; BR /＆gt;中）;
codeLatLng（C [I]，I）;
};
}
＆LT; / SCRIPT＆GT;
＆LT; /头＆GT;
＆LT;身体GT;
＆LT; DIV ID =测试＆GT;＆LT; / DIV＆GT;
＆LT; BR /＆GT;
＆LT;按钮的onclick =此Showmethemoney（）;＆GT;给我钱＆LT;！/按钮＆GT;
＆LT; /身体GT;
我需要的结果是内嵌在自己的坐标。
任何帮助将是很大的AP preciated！
感谢。
编辑：
获取得远一点：
VAR地理codeR =新google.maps.Geo codeR（）;
变种C = [];
C [0] =43.5878，-79.6777;
C [1] =49.2828，-123.1414;
C [2] =49.2895，-123.127;
C [3] =49.2784，-123.1364;
C [4] =49.0327，-122.2529;
C [5] =49.2696，-123.0587;
C [6] =49.269，-122.9954;
C [7] =49.235，-122.882;
C [8] =49.1908，-122.7508;
C [9] =49.085，-122.4111;
变种一个= [];
函数codeLatLng（输入，I）{
变种latlngStr = input.split（，，2）;
变种纬度= parseFloat（latlngStr [0]）;
变种LNG = parseFloat（latlngStr [1]）;
VAR经纬度=新google.maps.LatLng（纬度，经度）;
地理coder.geo code（{
经纬度：经纬度
}，功能（结果状态）{
//$("#test").append(results[0].formatted_address +＆LT; BR /＆gt;中）;
A [1] =结果[0] .formatted_address;
$（＃考）追加（A [1] +＆LT; BR /＆gt;中）。
}）;
}
变种I = 0;
功能此Showmethemoney（起点，终点）{
对于（i =启动; I＆LT; = c.length - 1安培;＆安培; I＆LT;结束;我++）{
$（＃考）追加（I +）+ C [I] + - ）;
codeLatLng（C [I]，I）;
}
如果（ⅰ＆GT = c.length - 1）{
返回;
}
的setTimeout（函数（）{
此Showmethemoney（I，I + 1）;
}，1500）;
}
＆LT; / SCRIPT＆GT;
＆LT; /头＆GT;
＆LT;身体GT;
＆LT; DIV ID =测试＆GT;＆LT; / DIV＆GT;
＆LT; BR /＆GT;
＆LT;按钮的onclick =此Showmethemoney（0,1）;＆GT;给我钱＆LT;！/按钮＆GT;
＆LT; /身体GT;
解决方案
修改功能此Showmethemoney 。通过使用的setTimeout 的jsfiddle 。您可以从10更改间隔任何你想要的数字。
变种i = 0;
功能此Showmethemoney（起点，终点）{
对于（i =启动; I＆LT; = c.length - 1安培;＆安培; I＆LT;结束;我++）{
$（＃考）追加（I +）+ C（I）+＆LT; BR /＆gt;中）;
codeLatLng（C [I]，I）;
}
如果（ⅰ＆GT = c.length - 1）{
返回;
}
的setTimeout（函数（）{
此Showmethemoney（I，I + 5）;
}，10）;
}
此Showmethemoney（I，I + 5）;
I need to reverse geocode a list of thousands of latitude/longitude coordinates. I can do this for one set of coordinates no problem, but when it comes to many it will only do 5 at a time. I need to figure out how to do this in one step and time the loop so it will not get denied by google.
I have been told elsewhere that I need to utilize callback functions but I am unsure how to do this.
Here is some demo code
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
<script type="text/javascript" src="js/jquery-1.7.2.min.js" ></script>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
var geocoder = new google.maps.Geocoder();
var c = [];
c[0]="43.5878,-79.6777";
c[1]="49.2828,-123.1414";
c[2]="49.2895,-123.127";
c[3]="49.2784,-123.1364";
c[4]="49.0327,-122.2529";
c[5]="49.2696,-123.0587";
c[6]="49.269,-122.9954";
c[7]="49.235,-122.882";
c[8]="49.1908,-122.7508";
c[9]="49.085,-122.4111";
var a = [];
function codeLatLng(input, i) {
var latlngStr = input.split(",", 2);
var lat = parseFloat(latlngStr[0]);
var lng = parseFloat(latlngStr[1]);
var latlng = new google.maps.LatLng(lat, lng);
geocoder.geocode({
'latLng' : latlng
}, function(results, status) {
a[i] = results[0].formatted_address;
$("#test").append(a[i] + "<br />");
});
}
function showmethemoney(){
for (var i=0; i <= c.length - 1; i++) {
$("#test").append(i + ") " + c[i] + "<br/>");
codeLatLng(c[i], i);
};
}
</script>
</head>
<body>
<div id="test"></div>
<br />
<button onclick="showmethemoney();">Show me the money!</button>
</body>
I need my results to be inline with their coordinates.
Any help would be greatly appreciated!
Thanks.
Edit:
Getting a little bit further:
var geocoder = new google.maps.Geocoder();
var c = [];
c[0]="43.5878,-79.6777";
c[1]="49.2828,-123.1414";
c[2]="49.2895,-123.127";
c[3]="49.2784,-123.1364";
c[4]="49.0327,-122.2529";
c[5]="49.2696,-123.0587";
c[6]="49.269,-122.9954";
c[7]="49.235,-122.882";
c[8]="49.1908,-122.7508";
c[9]="49.085,-122.4111";
var a = [];
function codeLatLng(input, i) {
var latlngStr = input.split(",", 2);
var lat = parseFloat(latlngStr[0]);
var lng = parseFloat(latlngStr[1]);
var latlng = new google.maps.LatLng(lat, lng);
geocoder.geocode({
'latLng' : latlng
}, function(results, status) {
//$("#test").append(results[0].formatted_address + "<br />");
a[i] = results[0].formatted_address;
$("#test").append(a[i] + "<br />");
});
}
var i = 0;
function showmethemoney(start, end) {
for (i = start; i <= c.length - 1 && i < end; i++) {
$("#test").append(i + ") " + c[i] + " - ");
codeLatLng(c[i], i);
}
if( i >= c.length - 1){
return;
}
setTimeout(function(){
showmethemoney(i, i+1);
}, 1500);
}
</script>
</head>
<body>
<div id="test"></div>
<br />
<button onclick="showmethemoney(0, 1);">Show me the money!</button>
</body>
解决方案
Modified function showmethemoney. by using setTimeout jsfiddle. You can change interval from 10 to any number you want.
var i = 0;
function showmethemoney(start, end ) {
for (i = start; i <= c.length - 1 && i < end; i++) {
$("#test").append(i + ") " + c[i] + "<br/>");
codeLatLng(c[i], i);
}
if( i >= c.length - 1){
return;
}
setTimeout(function(){
showmethemoney(i, i+5);
}, 10);
}
showmethemoney(i, i+5);
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