如何获取另一个 XML 命名空间值?
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【中文标题】如何获取另一个 XML 命名空间值?【英文标题】:How can I get another XML namespace value? 【发布时间】:2021-11-03 00:39:22 【问题描述】:我是 XML 新手,尝试阅读 the GML file 并发现您可以使用 XDocuments
加载 GML 文件。
如果我单独使用命名空间,它就可以工作。 工作代码:
XDocument doc = XDocument.Load(_gmlFilePath);
XNamespace agiv = "http://www.agiv.be/agiv";
XNamespace gml = "http://www.opengis.net/gml";
var agivQuery = doc.Descendants(agiv + "CrabAdr").Select(e =>new
Id = (int) e.Element(agiv + "ID"),
StraatId = (int) e.Element(agiv + "STRAATNMID"),
StraatNm = (string) e.Element(agiv + "STRAATNM"),
HuisNr = (string) e.Element(agiv + "HUISNR"),
HuisNrLabel = (string) e.Element(agiv + "HNRLABEL"),
Niscode = (int) e.Element(agiv + "NISCODE"),
Gemeente = (string) e.Element(agiv + "GEMEENTE"),
Postcode = (int) e.Element(agiv + "POSTCODE")
);
var coordQuery = doc.Descendants(gml + "coord").Select(e =>new
X = (decimal) e.Element(gml + "X"),
Y = (decimal) e.Element(gml + "Y")
);
foreach(var c in agivQuery)
Console.WriteLine(c);
Console.WriteLine("\n --- \n");
foreach (var c in coordQuery)
Console.WriteLine(c);
我想要一个变量中的所有内容 (agivQuery
)
并尝试过。代码无效:
var agivQuery = doc.Descendants(agiv + "CrabAdr").Select(e =>new
Id = (int) e.Element(agiv + "ID"),
StraatId = (int) e.Element(agiv + "STRAATNMID"),
StraatNm = (string) e.Element(agiv + "STRAATNM"),
HuisNr = (string) e.Element(agiv + "HUISNR"),
HuisNrLabel = (string) e.Element(agiv + "HNRLABEL"),
Niscode = (int) e.Element(agiv + "NISCODE"),
Gemeente = (string) e.Element(agiv + "GEMEENTE"),
Postcode = (int) e.Element(agiv + "POSTCODE"),
X = (decimal) e.Element(gml + "X"),
Y = (decimal) e.Element(gml + "Y")
);
foreach(var c in agivQuery)
Console.WriteLine(c);
Console.WriteLine("\n --- \n");
//foreach (var c in coordQuery)
//
// Console.WriteLine(c);
//
这是 GML 文件的 30 行:
<?xml version="1.0" encoding="utf-8"?>
<agiv:FeatureCollection xmlns:agiv="http://www.agiv.be/agiv" xmlns:gml="http://www.opengis.net/gml" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="CrabAdr.xsd">
<gml:boundedBy>
<gml:Box srsName="EPSG:31370">
<gml:coordinates />
</gml:Box>
</gml:boundedBy>
<gml:featureMember>
<agiv:CrabAdr>
<agiv:ID>2000000022</agiv:ID>
<agiv:STRAATNMID>37468</agiv:STRAATNMID>
<agiv:STRAATNM>Vuggelberg</agiv:STRAATNM>
<agiv:HUISNR>6</agiv:HUISNR>
<agiv:APPTNR />
<agiv:BUSNR />
<agiv:HNRLABEL>6</agiv:HNRLABEL>
<agiv:NISCODE>24134</agiv:NISCODE>
<agiv:GEMEENTE>Scherpenheuvel-Zichem</agiv:GEMEENTE>
<agiv:POSTCODE>3270</agiv:POSTCODE>
<agiv:HERKOMST>afgeleidVanGebouw</agiv:HERKOMST>
<gml:pointProperty>
<gml:Point srsName="EPSG:31370" xmlns:gml="http://www.opengis.net/gml">
<gml:coord>
<gml:X>190413.49</gml:X>
<gml:Y>183795.53</gml:Y>
</gml:coord>
</gml:Point>
</gml:pointProperty>
</agiv:CrabAdr>
</gml:featureMember>
【问题讨论】:
【参考方案1】:请尝试以下方法。
c#
void Main()
const string _gmlFilePath = @"e:\temp\GML.xml";
XDocument doc = XDocument.Load(_gmlFilePath);
XNamespace agiv = doc.Root.GetNamespaceOfPrefix("agiv");
XNamespace gml = doc.Root.GetNamespaceOfPrefix("gml");
var agivQuery = doc.Descendants(agiv + "CrabAdr").Select(e => new
Id = (int)e.Element(agiv + "ID"),
StraatId = (int)e.Element(agiv + "STRAATNMID"),
StraatNm = (string)e.Element(agiv + "STRAATNM"),
HuisNr = (string)e.Element(agiv + "HUISNR"),
HuisNrLabel = (string)e.Element(agiv + "HNRLABEL"),
Niscode = (int)e.Element(agiv + "NISCODE"),
Gemeente = (string)e.Element(agiv + "GEMEENTE"),
Postcode = (int)e.Element(agiv + "POSTCODE"),
X = (decimal)e.Element(gml + "pointProperty")
.Descendants(gml + "X").FirstOrDefault(),
Y = (decimal)e.Element(gml + "pointProperty")
.Descendants(gml + "Y").FirstOrDefault()
);
foreach (var c in agivQuery)
Console.WriteLine(c);
【讨论】:
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