XML 文件如下所示
<osm version="0.6" generator="CGImap 0.0.2">
<bounds minlat="54.0889580" minlon="12.2487570" maxlat="54.0913900" maxlon="12.2524800"/>
<node id="298884269" lat="54.0901746" lon="12.2482632" user="SvenHRO" uid="46882" visible="true" version="1" changeset="676636" timestamp="2008-09-21T21:37:45Z"/>
<node id="261728686" lat="54.0906309" lon="12.2441924" user="PikoWinter" uid="36744" visible="true" version="1" changeset="323878" timestamp="2008-05-03T13:39:23Z"/>
<way id="11103644" visible="true" version="22" changeset="57021161" timestamp="2018-03-09T06:06:21Z" user="thetornado76" uid="1624486">
<nd ref="5058358201"/>
<nd ref="5058358200"/>
<nd ref="459672758"/>
<nd ref="459672757"/>
<nd ref="459672756"/>
<tag k="bicycle" v="no"/>
<tag k="destination" v="Airport"/>
</way>
<way id="11103973" visible="true" version="1" changeset="410956" timestamp="2007-11-05T15:10:28Z" user="DaveHansenTiger" uid="7168">
<nd ref="98983980"/>
<nd ref="98983981"/>
<tag k="tiger:source" v="tiger_import_dch_v0.6_20070809"/>
<tag k="tiger:tlid" v="116306702"/>
<tag k="tiger:upload_uuid" v="bulk_upload.pl-bcfe2a6c-15d8-4c58-8cb2-f0d6e1120159"/>
</way>
</osm>
这是我到目前为止
//Removes the "tag" element from xml
osmFile.Root.Elements().Elements().Where(x => x.Name == "tag").Remove();
//Each iteration of query has the an entire <way> elements and all its children
var query = from c in osmFile.Root.Elements("way")
select c;
我的问题是,如何才能使遍历查询的每次迭代仅具有特定方式的 nd 元素的所有属性值?
例如,如果我这样做
foreach (var x in query)
MessageBox.Show(x.ToString());
第一次迭代将显示
5058358201
5058358200
459672758
459672757
459672756
第二个会显示
98983980
98983981
另外,有没有办法在不首先删除所有标签元素的情况下实现这一点?
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