![](/img/trans.png)
[英]How to remove an element from an xml using Xdocument when we have multiple elements with same name but different attributes
[英]How to parse XML using c# and XDocument into an object when multiple elements have the same name?
我正在使用SharePoint 2013的REST API来解析XML,但是在解析XML的特定块时遇到了麻烦。 这是一些XML的截断的示例,我可能会回过头来解析没有问题:
<feed xml:base="http://server/DesktopApplications/_api/" xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml">
<id>2dff4586-6b7d-4186-ae63-4d048f74a112</id>
<title />
<updated>2014-03-19T15:32:15Z</updated>
<entry m:etag=""19"">
<id>http://server/DesktopApplications/_api/Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')</id>
<category term="SP.List" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
<link rel="edit" href="Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')" />
<title />
<updated>2014-03-19T15:32:15Z</updated>
<author>
<name />
</author>
<content type="application/xml">
<m:properties>
<d:BaseTemplate m:type="Edm.Int32">101</d:BaseTemplate>
<d:BaseType m:type="Edm.Int32">1</d:BaseType>
<d:Id m:type="Edm.Guid">0ac46109-38d9-4070-96b7-f90085b56f1e</d:Id>
<d:ListItemEntityTypeFullName>SP.Data.Shared_x0020_DocumentsItem</d:ListItemEntityTypeFullName>
<d:Title>Documents</d:Title>
</m:properties>
</content>
</entry>
...
</feed>
我可以使用XDocument
将其解析为以下内容:
private static readonly XNamespace d = "http://schemas.microsoft.com/ado/2007/08/dataservices";
private static readonly XNamespace m = "http://schemas.microsoft.com/ado/2007/08/dataservices/metadata";
...
List<KLList> lists = doc.Descendants(m + "properties").Select(
list => new KLList()
{
Id = list.Element(d + "Id").Value,
Title = list.Element(d + "Title").Value,
ListItemEntityTypeFullName = list.Element(d + "ListItemEntityTypeFullName").Value,
BaseType = (BaseType)Convert.ToInt32(list.Element(d + "BaseType").Value),
ListTemplateType = (ListTemplateType)Convert.ToInt32(list.Element(d + "BaseTemplate").Value)
}).ToList();
当我扩展查询以获取列表的根文件夹的服务器相对URL时,我得到了这样的XML,但我遇到了麻烦:
<feed xml:base="http://server/DesktopApplications/_api/" xmlns="http://www.w3.org/2005/Atom" xmlns:d="http://schemas.microsoft.com/ado/2007/08/dataservices" xmlns:m="http://schemas.microsoft.com/ado/2007/08/dataservices/metadata" xmlns:georss="http://www.georss.org/georss" xmlns:gml="http://www.opengis.net/gml">
<id>a01c22dc-a87f-4253-91d1-0a6ef8efa6d0</id>
<title />
<updated>2014-03-19T15:27:11Z</updated>
<entry m:etag=""19"">
<id>http://server/DesktopApplications/_api/Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')</id>
<category term="SP.List" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
<link rel="edit" href="Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')" />
<link rel="http://schemas.microsoft.com/ado/2007/08/dataservices/related/RootFolder" type="application/atom+xml;type=entry" title="RootFolder" href="Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')/RootFolder">
<m:inline>
<entry>
<id>http://server/DesktopApplications/_api/Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')/RootFolder</id>
<category term="SP.Folder" scheme="http://schemas.microsoft.com/ado/2007/08/dataservices/scheme" />
<link rel="edit" href="Web/Lists(guid'0ac46109-38d9-4070-96b7-f90085b56f1e')/RootFolder" />
<title />
<updated>2014-03-19T15:27:11Z</updated>
<author>
<name />
</author>
<content type="application/xml">
<m:properties>
<d:ServerRelativeUrl>/DesktopApplications/Shared Documents</d:ServerRelativeUrl>
</m:properties>
</content>
</entry>
</m:inline>
</link>
<title />
<updated>2014-03-19T15:27:11Z</updated>
<author>
<name />
</author>
<content type="application/xml">
<m:properties>
<d:BaseTemplate m:type="Edm.Int32">101</d:BaseTemplate>
<d:BaseType m:type="Edm.Int32">1</d:BaseType>
<d:Id m:type="Edm.Guid">0ac46109-38d9-4070-96b7-f90085b56f1e</d:Id>
<d:ListItemEntityTypeFullName>SP.Data.Shared_x0020_DocumentsItem</d:ListItemEntityTypeFullName>
<d:Title>Documents</d:Title>
</m:properties>
</content>
</entry>
...
</feed>
我尝试使用基本相同的代码,因为我希望ServerRelativeUrl
位于同一对象上:
List<KLList> lists = doc.Descendants(m + "properties").Select(
list => new KLList()
{
Id = list.Element(d + "Id").Value,
Title = list.Element(d + "Title").Value,
ListItemEntityTypeFullName = list.Element(d + "ListItemEntityTypeFullName").Value,
BaseType = (BaseType)Convert.ToInt32(list.Element(d + "BaseType").Value),
ListTemplateType = (ListTemplateType)Convert.ToInt32(list.Element(d + "BaseTemplate").Value),
RelativeUrl = list.Element(d + "ServerRelativeUrl").Value
}).ToList();
但是,这不再起作用。 doc.Descendants(m + "properties")
返回的第一件事是
<m:properties>
<d:ServerRelativeUrl>/DesktopApplications/Shared Documents</d:ServerRelativeUrl>
</m:properties>
因此尝试同时设置其他属性会引发对象引用异常,因为其他元素不在此处。
如何解析该XML,以便将所有想要的值放入一个对象中? 我宁愿不必单独调用以获取每个列表的根文件夹url。
更新:
我在下面发布了答案,但我觉得那里有更好的方法。 如果答案比我的要好,请随时发布答案,我会将您的答案标记为答案。
您想要做的是在获取元素值之前检查该元素是否存在,因为目前您只是假设它们将始终存在。
在此之前,有人问过这个问题(我没有足够的声誉来发表此评论): 解析XML时检查是否存在一个元素
我想出了一种方法来满足我的需要,但我觉得必须有更好的方法...
private readonly XNamespace a = "http://www.w3.org/2005/Atom";
private readonly XNamespace d = "http://schemas.microsoft.com/ado/2007/08/dataservices";
private readonly XNamespace m = "http://schemas.microsoft.com/ado/2007/08/dataservices/metadata";
List<KLList> lists = doc.Descendants(a + "entry").Where(element => element.Attribute(m + "etag") != null).Select(
list => new KLList()
{
Id = list.Descendants(d + "Id").FirstOrDefault().Value,
Title = list.Descendants(d + "Title").FirstOrDefault().Value,
ListItemEntityTypeFullName = list.Descendants(d + "ListItemEntityTypeFullName").FirstOrDefault().Value,
BaseType = (BaseType)Convert.ToInt32(list.Descendants(d + "BaseType").FirstOrDefault().Value),
ListTemplateType = (ListTemplateType)Convert.ToInt32(list.Descendants(d + "BaseTemplate").FirstOrDefault().Value),
RelativeUrl = list.Descendants(d + "ServerRelativeUrl").FirstOrDefault().Value
}).ToList();
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.