避免具有多个标记的自定义谷歌地图的地理编码限制

发布于 2024-12-08 09:36:27 字数 233 浏览 0 评论 0原文

我创建了一个包含大约 150 个标记的自定义地图。在达到地理编码限制之前，它仅绘制大约 20 个左右。

我可以在代码中加入哪些内容以避免达到限制？

更新： 如何为每个请求添加一个时间延迟，比如 0.25 秒（或者我能做到的最低时间）？

我尝试了一些不同的建议，但我似乎无法让它们与我的代码一起使用，因此请任何示例都可以使用我的代码。

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一片旧的回忆 2024-12-15 09:36:27

因此，我认为与其在 for 循环中几乎立即发送所有请求，不如在地理编码返回结果时发送下一个结果。如果返回的结果是 OVER_QUERY_LIMIT，请重新发送请求并增加延迟。我还没有找到产生最少 OVER_QUERY_LIMIT 的最佳超时时间，但至少它会创建所有标记（对于有效地址）。在我的机器上大约需要 45 秒左右才能完成。

<!DOCTYPE html>
<html>
<head>
<link href="http://code.google.com/apis/maps/documentation/javascript/examples/default.css"
rel="stylesheet" type="text/css" />
<script type="text/javascript" src="http://maps.googleapis.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
var geocoder;
var map;
var infowindow = new google.maps.InfoWindow();

var places = [];
var popup_content = [ /* all your popup_content */];
var address = [/* all of your addresses */];
var address_position = 0;

var timeout = 600;

function initialize() {
    geocoder = new google.maps.Geocoder();
    var latlng = new google.maps.LatLng(52.40, -3.61);
    var myOptions = {
      zoom: 8,
      center: latlng,
      mapTypeId: 'roadmap'
    }
    map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
    addMarker(address_position);
}

function addMarker(position)
{
    geocoder.geocode({'address': address[position]}, function(results, status)
    {
        if (status == google.maps.GeocoderStatus.OK) {
            places[position] = results[0].geometry.location;

            var marker = new google.maps.Marker({
                position: places[position],
                map: map
            });

            google.maps.event.addListener(marker, 'click', function() {
                if (!infowindow) {
                    infowindow = new google.maps.InfoWindow();
                }
                infowindow.setContent(popup_content[position]);
                infowindow.open(map, marker);
            });
        }
        else
        {
            if (status == google.maps.GeocoderStatus.OVER_QUERY_LIMIT)
            {
                setTimeout(function() { addMarker(position); }, (timeout * 3));
            }
        }
        address_position++;
        if (address_position < address.length)
        {
            setTimeout(function() { addMarker(address_position); }, (timeout));
        }
    });
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="height: 80%; top:10px; border: 1px solid black;"></div>
</body>
</html>

So instead of sending all the requests almost instantly in a for-loop, I thought it might be better that when a geocode returns a result, to send the next one. If the result coming back is a OVER_QUERY_LIMIT, resend the request with an increase on the delay. I haven't found the sweet spot timeout that produces the fewest OVER_QUERY_LIMIT's, but at least it will create all the markers (for valid addresses). It takes about 45 seconds or so to finish on my machine.

<!DOCTYPE html>
<html>
<head>
<link href="http://code.google.com/apis/maps/documentation/javascript/examples/default.css"
rel="stylesheet" type="text/css" />
<script type="text/javascript" src="http://maps.googleapis.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
var geocoder;
var map;
var infowindow = new google.maps.InfoWindow();

var places = [];
var popup_content = [ /* all your popup_content */];
var address = [/* all of your addresses */];
var address_position = 0;

var timeout = 600;

function initialize() {
    geocoder = new google.maps.Geocoder();
    var latlng = new google.maps.LatLng(52.40, -3.61);
    var myOptions = {
      zoom: 8,
      center: latlng,
      mapTypeId: 'roadmap'
    }
    map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
    addMarker(address_position);
}

function addMarker(position)
{
    geocoder.geocode({'address': address[position]}, function(results, status)
    {
        if (status == google.maps.GeocoderStatus.OK) {
            places[position] = results[0].geometry.location;

            var marker = new google.maps.Marker({
                position: places[position],
                map: map
            });

            google.maps.event.addListener(marker, 'click', function() {
                if (!infowindow) {
                    infowindow = new google.maps.InfoWindow();
                }
                infowindow.setContent(popup_content[position]);
                infowindow.open(map, marker);
            });
        }
        else
        {
            if (status == google.maps.GeocoderStatus.OVER_QUERY_LIMIT)
            {
                setTimeout(function() { addMarker(position); }, (timeout * 3));
            }
        }
        address_position++;
        if (address_position < address.length)
        {
            setTimeout(function() { addMarker(address_position); }, (timeout));
        }
    });
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="height: 80%; top:10px; border: 1px solid black;"></div>
</body>
</html>

回复收藏 0 原文

德意的啸 2024-12-15 09:36:27

我当然不是代码大师，但我没有使用超时，而是继续从函数内部调用该函数（我确信资源占用）。如果成功，我会迭代计数器，否则如果不成功，我会使用当前计数器值调用我的函数。按照这个速度，我可能会为我的 37 条折线（或标记，如果这就是您的绘图）循环 1000 次，但我相信它会尽快完成。这是一个例子 - 我首先将我的绘图函数 onclick 称为plot(0)：

function plot(k)
{
...
if (status == google.maps.GeocoderStatus.OK)
        {
            draw(r.routes[0]);
            if(k<origArray.length) plot(k++);
        }
else plot(k)
...
}

i'm certainly no codemaster, but instead of using a timeout i just keep calling the function from within the function (resource hog i'm sure). i iterate through my counter if i'm successful, otherwise i call my function with the current counter value if unsuccessful. at this rate, i may loop 1000 times for my 37 polylines (or markers if that's what your drawing), but i believe it will finish at the earliest time possible. here's an example - i first call my plot function onclick as plot(0):

function plot(k)
{
...
if (status == google.maps.GeocoderStatus.OK)
        {
            draw(r.routes[0]);
            if(k<origArray.length) plot(k++);
        }
else plot(k)
...
}

回复收藏 0 原文

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