通过经纬度获取中心点,以及绑定的缩放级别

beyond1993

于 2011-01-31 22:28:32 发布

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分类专栏： googleMap 文章标签： 360

本文链接：https://blog.youkuaiyun.com/beyond1993/article/details/83836621

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本文介绍了一种根据经纬度计算地图中心点及适宜缩放级别的方法。通过计算最大和最小经纬度之间的差异，确定地图显示的最佳缩放级别。

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//根据经纬度的距离获取地图的缩放级
function getRoom(diff){
	var room =    new Array(0,  1,  2, 3, 4, 5, 6,7,8,  9,   10,  11,  12,  13, 14);
	var diffArr = new Array(360,180,90,45,22,11,5,2.5,1.25,0.6,0.3,0.15,0.07,0.03,0);
	for(var i = 0; i < diffArr.length; i ++){
		if((diff - diffArr[i]) >= 0){
			return room[i];
		}
	}	
	return 14;
}


function getCenterPoint(maxJ,minJ,maxW,minW){//通过经纬度获取中心位置和缩放级别
	if(maxJ==minJ&&maxW==minW)return [maxJ,maxW,0];
	var diff = maxJ - minJ;
	if(diff < (maxW - minW))diff = maxW - minW;
	diff = parseInt(10000 * diff)/10000;	
	var centerJ = minJ*1000000+1000000*(maxJ - minJ)/2;
	var centerW = minW*1000000+1000000*(maxW - minW)/2;
         var zoom = getRoom(diff);
	return [centerJ/1000000,centerW/1000000,zoom];
}