Server 2012，geography.STContains，结果错误吗？

Question

我正在使用Server 2012，geography.STContains (文档)，我不明白为什么下面的代码会失败。

如果我切换到geometry，它就能工作。

有人能解释一下吗？

//丹尼尔

declare @geo geography
set @geo = geography::STPolyFromText('POLYGON ((17.519133203852 59.8297423369731, 17.5190071588812 59.8296936773323, 17.5189979955459 59.8298203729009, 17.5191345140461 59.8298223425544, 17.519133203852 59.8297423369731))', 4326)

-- Is not within
declare @p1 geography
set @p1 = geography::STPointFromText('POINT(17.5184709839477 59.829925754067)', 4326)

-- Is within
declare @p2 geography
set @p2 = geography::STPointFromText('POINT(17.519060 59.829774)', 4326) 

select
    @geo.STContains(@p1), -- should be 0 is 1
    @geo.STContains(@p2) -- should be 1 is 0

更新了：--如果我将它们倒置--它工作得很好，但是我不明白这一点：

declare @geo geography
set @geo = geography::STPolyFromText('POLYGON ((17.519133203852 59.8297423369731, 17.5190071588812 59.8296936773323, 17.5189979955459 59.8298203729009, 17.5191345140461 59.8298223425544, 17.519133203852 59.8297423369731))', 4326)

select
    @geo.STAsText() Polygon,
    @geo.STPointN(1).STAsText() Point1,
    @geo.STPointN(1).Lat Point1Latitud,
    @geo.STPointN(1).Long Point1Longitude

其结果是：

Polygon 
POLYGON ((17.519133203852 59.8297423369731, 17.5190071588812 59.8296936773323, 17.5189979955459 59.8298203729009, 17.5191345140461 59.8298223425544, 17.519133203852 59.8297423369731))

Point1  
POINT (17.519133203852 59.8297423369731)

Point1Latitud   
59,8297423369731

Point1Longitude
17,519133203852

Answer 1

现在我发现了这个问题。用户画多边形，从右下角开始，顺时针方向移动.如果我从最大纬度重新排序，然后用long排序反方向，lat是有效的。为它找个帮手，但只有当你“知道这是错的”时才能做到：

if(sqlGeography.EnvelopeAngle() > 90)
    sqlGeography ? sqlGeography.ReorientObject();

只需组合一个小的解决方案，它将修复我的值：https://github.com/danielwertheim/GeographyFactory

还有一篇关于它的博客：http://danielwertheim.se/sqlgeography-in-sql-server-2012-polygon-must-start-on-correct-position/

关于真正的“问题”，左手法则：

http://danielwertheim.se/sqlgeography-in-sql-server-2012-polygon-must-start-on-correct-position-no/