这是我用来显示一个有3个针/标记的地图:
<script>
function initialize() {
var locations = [
['DESCRIPTION', 41.926979, 12.517385, 3],
['DESCRIPTION', 41.914873, 12.506486, 2],
['DESCRIPTION', 41.918574, 12.507201, 1]
];
var map = new google.maps.Map(document.getElementById('map'), {
zoom: 15,
center: new google.maps.LatLng(41.923, 12.513),
mapTypeId: google.maps.MapTypeId.ROADMAP
});
var infowindow = new google.maps.InfoWindow();
var marker, i;
for (i = 0; i < locations.length; i++) {
marker = new google.maps.Marker({
position: new google.maps.LatLng(locations[i][1], locations[i][2]),
map: map
});
google.maps.event.addListener(marker, 'click', (function(marker, i) {
return function() {
infowindow.setContent(locations[i][0]);
infowindow.open(map, marker);
}
})(marker, i));
}
}
function loadScript() {
var script = document.createElement('script');
script.type = 'text/javascript';
script.src = 'https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false&' + 'callback=initialize';
document.body.appendChild(script);
}
window.onload = loadScript;
</script>
<div id="map" style="width: 900px; height: 700px;"></div>
我正在寻找的是一种避免必须“手动”找到地图中心的方法:new google.maps.LatLng(41.923, 12.513)。是否有一种方法可以自动将地图置于三个坐标的中心?
以下是我的看法,以防有人看到这个帖子:
这有助于防止非数值数据破坏确定纬度和液化天然气的最终变量。
它的工作原理是获取所有坐标,将它们解析为数组中单独的latat和lng元素,然后确定每个元素的平均值。平均值应该是中心(在我的测试用例中已经证明了这一点)。
var coords = "50.0160001,3.2840073|50.014458,3.2778274|50.0169713,3.2750587|50.0180745,3.276742|50.0204038,3.2733474|50.0217796,3.2781737|50.0293064,3.2712542|50.0319918,3.2580816|50.0243287,3.2582281|50.0281447,3.2451177|50.0307925,3.2443178|50.0278165,3.2343882|50.0326574,3.2289809|50.0288569,3.2237612|50.0260081,3.2230589|50.0269495,3.2210104|50.0212645,3.2133541|50.0165868,3.1977592|50.0150515,3.1977341|50.0147901,3.1965286|50.0171915,3.1961636|50.0130074,3.1845098|50.0113267,3.1729483|50.0177206,3.1705726|50.0210692,3.1670394|50.0182166,3.158297|50.0207314,3.150927|50.0179787,3.1485753|50.0184944,3.1470782|50.0273077,3.149845|50.024227,3.1340514|50.0244172,3.1236235|50.0270676,3.1244474|50.0260853,3.1184879|50.0344525,3.113806";
var filteredtextCoordinatesArray = coords.split('|');
centerLatArray = [];
centerLngArray = [];
for (i=0 ; i < filteredtextCoordinatesArray.length ; i++) {
var centerCoords = filteredtextCoordinatesArray[i];
var centerCoordsArray = centerCoords.split(',');
if (isNaN(Number(centerCoordsArray[0]))) {
} else {
centerLatArray.push(Number(centerCoordsArray[0]));
}
if (isNaN(Number(centerCoordsArray[1]))) {
} else {
centerLngArray.push(Number(centerCoordsArray[1]));
}
}
var centerLatSum = centerLatArray.reduce(function(a, b) { return a + b; });
var centerLngSum = centerLngArray.reduce(function(a, b) { return a + b; });
var centerLat = centerLatSum / filteredtextCoordinatesArray.length ;
var centerLng = centerLngSum / filteredtextCoordinatesArray.length ;
console.log(centerLat);
console.log(centerLng);
var mapOpt = {
zoom:8,
center: {lat: centerLat, lng: centerLng}
};
有一个更简单的方法,通过扩展一个空的LatLngBounds,而不是显式地从两个点创建一个。(详见这个问题)
应该是这样的,添加到你的代码:
//create empty LatLngBounds object
var bounds = new google.maps.LatLngBounds();
var infowindow = new google.maps.InfoWindow();
for (i = 0; i < locations.length; i++) {
var marker = new google.maps.Marker({
position: new google.maps.LatLng(locations[i][1], locations[i][2]),
map: map
});
//extend the bounds to include each marker's position
bounds.extend(marker.position);
google.maps.event.addListener(marker, 'click', (function(marker, i) {
return function() {
infowindow.setContent(locations[i][0]);
infowindow.open(map, marker);
}
})(marker, i));
}
//now fit the map to the newly inclusive bounds
map.fitBounds(bounds);
//(optional) restore the zoom level after the map is done scaling
var listener = google.maps.event.addListener(map, "idle", function () {
map.setZoom(3);
google.maps.event.removeListener(listener);
});
这样,您可以使用任意数量的点,并且不需要事先知道顺序。
演示jsFiddle在这里:http://jsfiddle.net/x5R63/
以下是我的看法,以防有人看到这个帖子:
这有助于防止非数值数据破坏确定纬度和液化天然气的最终变量。
它的工作原理是获取所有坐标,将它们解析为数组中单独的latat和lng元素,然后确定每个元素的平均值。平均值应该是中心(在我的测试用例中已经证明了这一点)。
var coords = "50.0160001,3.2840073|50.014458,3.2778274|50.0169713,3.2750587|50.0180745,3.276742|50.0204038,3.2733474|50.0217796,3.2781737|50.0293064,3.2712542|50.0319918,3.2580816|50.0243287,3.2582281|50.0281447,3.2451177|50.0307925,3.2443178|50.0278165,3.2343882|50.0326574,3.2289809|50.0288569,3.2237612|50.0260081,3.2230589|50.0269495,3.2210104|50.0212645,3.2133541|50.0165868,3.1977592|50.0150515,3.1977341|50.0147901,3.1965286|50.0171915,3.1961636|50.0130074,3.1845098|50.0113267,3.1729483|50.0177206,3.1705726|50.0210692,3.1670394|50.0182166,3.158297|50.0207314,3.150927|50.0179787,3.1485753|50.0184944,3.1470782|50.0273077,3.149845|50.024227,3.1340514|50.0244172,3.1236235|50.0270676,3.1244474|50.0260853,3.1184879|50.0344525,3.113806";
var filteredtextCoordinatesArray = coords.split('|');
centerLatArray = [];
centerLngArray = [];
for (i=0 ; i < filteredtextCoordinatesArray.length ; i++) {
var centerCoords = filteredtextCoordinatesArray[i];
var centerCoordsArray = centerCoords.split(',');
if (isNaN(Number(centerCoordsArray[0]))) {
} else {
centerLatArray.push(Number(centerCoordsArray[0]));
}
if (isNaN(Number(centerCoordsArray[1]))) {
} else {
centerLngArray.push(Number(centerCoordsArray[1]));
}
}
var centerLatSum = centerLatArray.reduce(function(a, b) { return a + b; });
var centerLngSum = centerLngArray.reduce(function(a, b) { return a + b; });
var centerLat = centerLatSum / filteredtextCoordinatesArray.length ;
var centerLng = centerLngSum / filteredtextCoordinatesArray.length ;
console.log(centerLat);
console.log(centerLng);
var mapOpt = {
zoom:8,
center: {lat: centerLat, lng: centerLng}
};