瓦片地图坐标相关计算
在给定level下，把行号tileY和列号tileX
转换为2进制，然后
行列交叉存储，再
转换为4进制，即得到了相应的quadkey。譬如Level 3的第6行第4列的Tile计算：tileY = 5 = 101 ，tileX = 3 = 011；quadkey = 100111 = 213（4进制） = “213”。
那么，下面我们给出Ｃ＃的代码实现
///
<summary>
///
Converts tile XY coordinates into a QuadKey at a specified level of detail.
///
</summary>
///
<param name="tileX">
Tile X coordinate.
</param>
///
<param name="tileY">
Tile Y coordinate.
</param>
///
<param name="levelOfDetail">
Level of detail, from 1 (lowest detail)
///
to 23 (highest detail).
</param>
///
<returns>
A string containing the QuadKey.
</returns>
public
static
string
TileXYToQuadKey(
int
tileX,
int
tileY,
int
levelOfDetail)
{
StringBuilder quadKey
=
new
StringBuilder();
for
(
int
i
=
levelOfDetail; i
>
0
; i
--
)
{
char
digit
=
'
0
'
;
int
mask
=
1
<<
(i
-
1
);
if
((tileX
&
mask)
!=
0
)
{
digit
++
;
}
if
((tileY
&
mask)
!=
0
)
{
digit
++
;
digit
++
;
}
quadKey.Append(digit);
}
return
quadKey.ToString();
}
接下来是反算代码
///
<summary>
///
Converts a QuadKey into tile XY coordinates.
///
</summary>
///
<param name="quadKey">
QuadKey of the tile.
</param>
///
<param name="tileX">
Output parameter receiving the tile X coordinate.
</param>
///
<param name="tileY">
Output parameter receiving the tile Y coordinate.
</param>
///
<param name="levelOfDetail">
Output parameter receiving the level of detail.
</param>
public
static
void
QuadKeyToTileXY(
string
quadKey,
out
int
tileX,
out
int
tileY,
out
int
levelOfDetail)
{
tileX
=
tileY
=
0
;
levelOfDetail
=
quadKey.Length;
for
(
int
i
=
levelOfDetail; i
>
0
; i
--
)
{
int
mask
=
1
<<
(i
-
1
);
switch
(quadKey[levelOfDetail
-
i])
{
case
'
0
'
:
break
;
case
'
1
'
:
tileX
|=
mask;
break
;
case
'
2
'
:
tileY
|=
mask;
break
;
case
'
3
'
:
tileX
|=
mask;
tileY
|=
mask;
break
;
default
:
throw
new
ArgumentException(
"
Invalid QuadKey digit sequence.
"
);
}
}
}
上面所讲到的是针对ｂｉｎｇｍａｐ的情况，ｇｏｏｇｌｅｅａｒｔｈ采用QRTS编码，即Q、R、T、S分别代表0、1、2、3，运算方法类似，代码如下
/// <summary>
/// 由tileX、tileY和level求解quadkey
/// </summary>
/// <param name="tileX"></param>
/// <param name="tileY"></param>
/// <param name="levelOfDetail"></param>
/// <returns></returns>
private static string TileXYToQuadKey(int tileX, int tileY, int levelOfDetail)
{
var quadKey = new StringBuilder();
for (int i = levelOfDetail; i > 0; i--)
{
char digit = '0';
//掩码，最高位设为1，其他位设为0
int mask = 1 << (i - 1);
//与运算取得tileX的最高位，若为1，则加1
if ((tileX & mask) != 0)
{
digit++;
}
//与运算取得tileY的最高位，若为1，则加2
if ((tileY & mask) != 0)
{
digit++;
digit++;
}
//也即2*y+x
quadKey.Append(digit);
}
return quadKey.ToString();
}
protected static string QuadKeyNumberToAlpha(string base4)
{
return base4.Replace('0', 'q').Replace('1', 'r').Replace('2', 't').Replace('3', 's');
}
private static string QuadKeyNumberToAlphaUrl(string url, int tilePositionX, int tilePositionY, int zoom)
{
string quadKey = TileXYToQuadKey(tilePositionX, tilePositionY, zoom);
//获取最后一位，选择4个服务器中的一台
string str3 = quadKey.Substring(quadKey.Length - 1, 1);
string str4 = QuadKeyNumberToAlpha(quadKey);
url = string.Format(url, str4, str3);
return url;
}
Google卫星
地图是由256x256大小的jpg图片拼接而成，每块图片的URL
格式为“
http://kh.google.com/kh?v=3&t=trstrq”样。
参数v选择4台服务器中的一台，起到均衡负载的作用，参数t是“qrst”4个字符排列而成的字符串。为获取某经纬度的URL，就需要把经纬度转化为“qrst”字符串。 Google卫星地图在zoom=1时，全球就为一个256x256的图片，它的中心经纬度为(0,0)，URL为“
http://kh.google.com/kh?v=3&t=t”。zoom=2时裂化为4块，每块的编号为：左上”t=tq”,右上”t=tr”,右下“t=ts”,左下”t=tt”。依此类推，每放大一倍，每一小块都裂分为四，从左上到右下顺时针按qrst编号，裂分后的编码为裂分前的编号上小块的编号。
代码如下：
function GetQuadtreeAddress(long, lat)
{
var PI = 3.1415926535897;
var digits = 18; // how many digits precision
// now convert to normalized square coordinates
// use standard equations to map into mercator projection
var x = (180.0 + parseFloat(long)) / 360.0;
var y = -parseFloat(lat) * PI / 180; // convert to radians
y = 0.5 * Math.log((1+Math.sin(y)) / (1 - Math.sin(y)));
y *= 1.0/(2 * PI); // scale factor from radians to normalized
y += 0.5; // and make y range from 0 - 1
var quad = "t"; // google addresses start with t
var lookup = "qrts"; // tl tr bl br
while (digits–)
{
// make sure we only look at fractional part
x -= Math.floor(x);
y -= Math.floor(y);
quad = quad + lookup.substr((x >= 0.5 ? 1 : 0) + (y >= 0.5 ? 2 : 0), 1);
// now descend into that square
x *= 2;
y *= 2;
}
return quad;
}