Google Maps API v3에서 여러 마커가있는 자동 중심지도
이것은 3 개의 핀 / 마커가있는 맵을 표시하는 데 사용하는 것입니다.
<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>
내가 찾는 것은로지도의 중심을 "수동으로"찾지 않아도되는 방법입니다 center: new google.maps.LatLng(41.923, 12.513)
. 세 좌표를 중심으로지도를 자동으로 배치하는 방법이 있습니까?
LatLngBounds
두 지점에서 명시 적으로 하나를 만드는 대신 빈을 확장하여 더 쉬운 방법이 있습니다 . (자세한 내용은 이 질문 을 참조하십시오)
다음과 같이 코드에 추가되어야합니다.
//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);
});
이런 식으로 임의의 수의 포인트를 사용할 수 있으며 순서를 미리 알 필요는 없습니다.
jsFiddle 데모 : http://jsfiddle.net/x5R63/
라티 투딘 최소 및 경도 최소를 계산해야한다고 생각합니다. 다음은 포인트를 중앙에 배치하는 데 사용할 함수가있는 예입니다.
//Example values of min & max latlng values
var lat_min = 1.3049337;
var lat_max = 1.3053515;
var lng_min = 103.2103116;
var lng_max = 103.8400188;
map.setCenter(new google.maps.LatLng(
((lat_max + lat_min) / 2.0),
((lng_max + lng_min) / 2.0)
));
map.fitBounds(new google.maps.LatLngBounds(
//bottom left
new google.maps.LatLng(lat_min, lng_min),
//top right
new google.maps.LatLng(lat_max, lng_max)
));
To find the exact center of the map you'll need to translate the lat/lon coordinates into pixel coordinates and then find the pixel center and convert that back into lat/lon coordinates.
You might not notice or mind the drift depending how far north or south of the equator you are. You can see the drift by doing map.setCenter(map.getBounds().getCenter()) inside of a setInterval, the drift will slowly disappear as it approaches the equator.
You can use the following to translate between lat/lon and pixel coordinates. The pixel coordinates are based on a plane of the entire world fully zoomed in, but you can then find the center of that and switch it back into lat/lon.
var HALF_WORLD_CIRCUMFERENCE = 268435456; // in pixels at zoom level 21
var WORLD_RADIUS = HALF_WORLD_CIRCUMFERENCE / Math.PI;
function _latToY ( lat ) {
var sinLat = Math.sin( _toRadians( lat ) );
return HALF_WORLD_CIRCUMFERENCE - WORLD_RADIUS * Math.log( ( 1 + sinLat ) / ( 1 - sinLat ) ) / 2;
}
function _lonToX ( lon ) {
return HALF_WORLD_CIRCUMFERENCE + WORLD_RADIUS * _toRadians( lon );
}
function _xToLon ( x ) {
return _toDegrees( ( x - HALF_WORLD_CIRCUMFERENCE ) / WORLD_RADIUS );
}
function _yToLat ( y ) {
return _toDegrees( Math.PI / 2 - 2 * Math.atan( Math.exp( ( y - HALF_WORLD_CIRCUMFERENCE ) / WORLD_RADIUS ) ) );
}
function _toRadians ( degrees ) {
return degrees * Math.PI / 180;
}
function _toDegrees ( radians ) {
return radians * 180 / Math.PI;
}
I use the method above to set the map boundaries, then, instead of resetting the zoom level, I just calculate the average LAT and average LON and set the center point to that location. I add up all the lat values into latTotal and all the lon values into lontotal and then divide by the number of markers. I then set the map center point to those average values.
latCenter = latTotal / markercount; lonCenter = lontotal / markercount;
i had a situation where i can't change old code, so added this javascript function to calculate center point and zoom level:
//input
var tempdata = ["18.9400|72.8200-19.1717|72.9560-28.6139|77.2090"];
function getCenterPosition(tempdata){
var tempLat = tempdata[0].split("-");
var latitudearray = [];
var longitudearray = [];
var i;
for(i=0; i<tempLat.length;i++){
var coordinates = tempLat[i].split("|");
latitudearray.push(coordinates[0]);
longitudearray.push(coordinates[1]);
}
latitudearray.sort(function (a, b) { return a-b; });
longitudearray.sort(function (a, b) { return a-b; });
var latdifferenece = latitudearray[latitudearray.length-1] - latitudearray[0];
var temp = (latdifferenece / 2).toFixed(4) ;
var latitudeMid = parseFloat(latitudearray[0]) + parseFloat(temp);
var longidifferenece = longitudearray[longitudearray.length-1] - longitudearray[0];
temp = (longidifferenece / 2).toFixed(4) ;
var longitudeMid = parseFloat(longitudearray[0]) + parseFloat(temp);
var maxdifference = (latdifferenece > longidifferenece)? latdifferenece : longidifferenece;
var zoomvalue;
if(maxdifference >= 0 && maxdifference <= 0.0037) //zoom 17
zoomvalue='17';
else if(maxdifference > 0.0037 && maxdifference <= 0.0070) //zoom 16
zoomvalue='16';
else if(maxdifference > 0.0070 && maxdifference <= 0.0130) //zoom 15
zoomvalue='15';
else if(maxdifference > 0.0130 && maxdifference <= 0.0290) //zoom 14
zoomvalue='14';
else if(maxdifference > 0.0290 && maxdifference <= 0.0550) //zoom 13
zoomvalue='13';
else if(maxdifference > 0.0550 && maxdifference <= 0.1200) //zoom 12
zoomvalue='12';
else if(maxdifference > 0.1200 && maxdifference <= 0.4640) //zoom 10
zoomvalue='10';
else if(maxdifference > 0.4640 && maxdifference <= 1.8580) //zoom 8
zoomvalue='8';
else if(maxdifference > 1.8580 && maxdifference <= 3.5310) //zoom 7
zoomvalue='7';
else if(maxdifference > 3.5310 && maxdifference <= 7.3367) //zoom 6
zoomvalue='6';
else if(maxdifference > 7.3367 && maxdifference <= 14.222) //zoom 5
zoomvalue='5';
else if(maxdifference > 14.222 && maxdifference <= 28.000) //zoom 4
zoomvalue='4';
else if(maxdifference > 28.000 && maxdifference <= 58.000) //zoom 3
zoomvalue='3';
else
zoomvalue='1';
return latitudeMid+'|'+longitudeMid+'|'+zoomvalue;
}
Here's my take on this in case anyone comes across this thread:
This helps protect against non-numerical data destroying either of your final variables that determine lat
and lng
.
It works by taking in all of your coordinates, parsing them into separate lat
and lng
elements of an array, then determining the average of each. That average should be the center (and has proven true in my test cases.)
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}
};
'IT story' 카테고리의 다른 글
matplotlib의 반전 컬러 맵 (0) | 2020.05.01 |
---|---|
C #에서 참조로 속성 전달 (0) | 2020.05.01 |
C #보다 F #을 사용하는 것이 어떤 영역에서 더 적합 할 수 있습니까? (0) | 2020.05.01 |
기계적 인조 인간; (0) | 2020.05.01 |
10 진수 / 더블이 정수인지 확인하는 방법? (0) | 2020.05.01 |