来自Openstreetmap的缓存瓷砖

我想在html中制作离线地图应用程序。我正在使用Openlayers 3 来自Openstreetmap的缓存瓷砖

我可以保存从Openstreetmap瓷砖，这不是问题。

我的问题：

当我有这样一个正方形： LonLat1，LonLat2

1------ 

-------2

我如何计算我有瓷砖抢，多zoomlevels

缓存全部：'{z}/{x}/{y} .png'tiles

后来我在离线状态下使用它们，像这样

var map = new ol.Map({ 
    target: 'map', 
    layers: [new ol.layer.Tile({ 
      source: new ol.source.XYZ({ 
      url: '{z}/{x}/{y}.png' 
     }) 
    })], ...

来源

2014-05-14 Johan Hoeksma

答案在这里http://stackoverflow.com/questions/21513646/how-to-get-xyz-coordinates-of-tile-by-click-on-leaflet-map作者：Lukas Vrabel解决了这个问题，尽管在传单中，数学是一样的。 –

这看起来不错。 Tommorow我会尽力为测试做一个jsfiddle。 –

如果您计划直接从osm.org下载，您还应该了解Tile使用策略：http://wiki.openstreetmap.org/wiki/Tile_usage_policy – Tordanik

我解决了问题Tnx to John Barca的链接！

制造一个小提琴：

Fiddle

Fiddle Full Screen

包括ol3.js/ol3.css/jquery的

HTML：

的javascript：

var map; 
var icons = []; 
var selMarker; 
var click = 0; 
var tmpCoord; 

map = new ol.Map({ 
    layers: [ 
    new ol.layer.Tile({source: new ol.source.OSM()}), 
    new ol.layer.Vector({  
     source: new ol.source.Vector({features:icons}) 
    })], 
    renderer: "canvas", 
    target: 'map', 
    view: new ol.View2D({ 
     zoom: 11 
    }) 
}); 

map.getView().setCenter(transform(5, 52)); 

map.on("click", function(evt){ 
    var coordinate = evt.coordinate; 
    if(click == 0) { 
     tmpCoord = coordinate; 
     click++; 
    } else { 
     addVierkant(tmpCoord, coordinate); 
     click = 0; 
     getAllTiles(tmpCoord, coordinate); 
    } 
    addCircle(coordinate); 
}) 

function transform(lng, lat) { 
    return ol.proj.transform([lng, lat], 'EPSG:4326', 'EPSG:3857'); 
} 

function transform2(lng, lat) { 
    return ol.proj.transform([lng, lat], 'EPSG:3857', 'EPSG:4326'); 
} 

function addCircle(c) { 
    var source = map.getLayers().getAt(1).getSource(); 
    var iconFeature = new ol.Feature({ 
     geometry: new ol.geom.Circle(c, 300), 
    }); 
    iconFeature.setStyle(getCircleStyle()); 
    icons[icons.length] = iconFeature; 
    icons[icons.length - 1][0] = "circle"; 
    source.addFeature(iconFeature); 
    return iconFeature; 
} 

function addVierkant(c1, c2) { 
    var source = map.getLayers().getAt(1).getSource(); 
    p1 = c1; 
    p2 = [c1[0], c2[1]]; 
    p3 = c2; 
    p4 = [c2[0], c1[1]]; 
    var coords = [p1,p2,p3,p4,p1]; 
    var iconFeature = new ol.Feature({ 
     geometry: new ol.geom.LineString(coords), 
    }); 
    iconFeature.setStyle(getVierkantStyle()); 
    icons[icons.length] = iconFeature; 
    icons[icons.length - 1][0] = "vierkant"; 
    source.addFeature(iconFeature); 
    return iconFeature; 
} 

function getCircleStyle() { 
    var iconStyle = new ol.style.Style({ 
     stroke: new ol.style.Stroke({ 
      color: 'rgba(0,0,0,0.3)', 
      width: 4 
     }), 
     fill: new ol.style.Fill({ 
      color: 'rgba(255,255,255,0.9)' 
     }) 
    }); 
    return iconStyle; 
} 

function getVierkantStyle() { 
    var iconStyle = new ol.style.Style({ 
     stroke: new ol.style.Stroke({ 
      color: 'rgba(255,255,255,0.9)', 
      width: 2 
     }) 
    }); 
    return iconStyle; 
} 

function getAllTiles(coord1, coord2) { 
    out1 = getTileURL(coord1, 10); 
    out2 = getTileURL(coord2, 10); 

    $("#output").html("zoom ------ " + out1[0] + "<br>from " + out1[1] + " to " + out2[1] + "<br>from " + out1[2] + " to " + out2[2]); 

} 


function getTileURL(coord, zoom) { 
     cor = transform2(coord[0], coord[1]); 
     lon = cor[0]; 
     lat = cor[1]; 
     var out = []; 
     var xtile = parseInt(Math.floor((lon + 180)/360 * (1<<zoom))); 
     var ytile = parseInt(Math.floor((1 - Math.log(Math.tan(lat.toRad()) + 1/Math.cos(lat.toRad()))/Math.PI)/2 * (1<<zoom))); 
     log(">> " + zoom + "/" + xtile + "/" + ytile); 
     out[0] = zoom; 
     out[1] = xtile; 
     out[2] = ytile; 
     return out; 
} 


function log(text) { 
    console.log(text); 
} 

if (typeof(Number.prototype.toRad) === "undefined") { 
    Number.prototype.toRad = function() { 
    return this * Math.PI/180; 
    } 
}

来源

2014-05-20 15:30:09

来自Openstreetmap的缓存瓷砖

回答

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