使用 XmlDocument 读取 XML 属性

发布于 2024-07-22 21:31:54 字数 489 浏览 3 评论 0原文

如何使用 C# 的 XmlDocument 读取 XML 属性？

我有一个 XML 文件，看起来有点像这样：

<?xml version="1.0" encoding="utf-8" ?>
<MyConfiguration xmlns="http://tempuri.org/myOwnSchema.xsd" SuperNumber="1" SuperString="whipcream">
    <Other stuff />
</MyConfiguration>

How will I read the XML attribute SuperNumber and SuperString?

目前我正在使用 XmlDocument，并且使用 XmlDocument 的 GetElementsByTagName() 获取中间的值，效果非常好。我只是不知道如何获得属性？

原文

How can I read an XML attribute using C#'s XmlDocument?

I have an XML file which looks somewhat like this:

<?xml version="1.0" encoding="utf-8" ?>
<MyConfiguration xmlns="http://tempuri.org/myOwnSchema.xsd" SuperNumber="1" SuperString="whipcream">
    <Other stuff />
</MyConfiguration>

How would I read the XML attributes SuperNumber and SuperString?

Currently I'm using XmlDocument, and I get the values in between using XmlDocument's GetElementsByTagName() and that works really well. I just can't figure out how to get the attributes?

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回忆凄美了谁 2024-07-29 21:32:01

如果您的 XML 包含命名空间，则可以执行以下操作以获得属性的值：

var xmlDoc = new XmlDocument();

// content is your XML as string
xmlDoc.LoadXml(content);

XmlNamespaceManager nsmgr = new XmlNamespaceManager(new NameTable());

// make sure the namespace identifier, URN in this case, matches what you have in your XML 
nsmgr.AddNamespace("ns", "urn:oasis:names:tc:SAML:2.0:protocol");

// get the value of Destination attribute from within the Response node with a prefix who's identifier is "urn:oasis:names:tc:SAML:2.0:protocol" using XPath
var str = xmlDoc.SelectSingleNode("/ns:Response/@Destination", nsmgr);
if (str != null)
{
    Console.WriteLine(str.Value);
}

有关 XML 命名空间的更多信息此处和此处。

If your XML contains namespaces, then you can do the following in order to obtain the value of an attribute:

var xmlDoc = new XmlDocument();

// content is your XML as string
xmlDoc.LoadXml(content);

XmlNamespaceManager nsmgr = new XmlNamespaceManager(new NameTable());

// make sure the namespace identifier, URN in this case, matches what you have in your XML 
nsmgr.AddNamespace("ns", "urn:oasis:names:tc:SAML:2.0:protocol");

// get the value of Destination attribute from within the Response node with a prefix who's identifier is "urn:oasis:names:tc:SAML:2.0:protocol" using XPath
var str = xmlDoc.SelectSingleNode("/ns:Response/@Destination", nsmgr);
if (str != null)
{
    Console.WriteLine(str.Value);
}

More on XML namespaces here and here.

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还在原地等你 2024-07-29 21:32:00

假设您的示例文档位于字符串变量 doc 中

> XDocument.Parse(doc).Root.Attribute("SuperNumber")
1

Assuming your example document is in the string variable doc

> XDocument.Parse(doc).Root.Attribute("SuperNumber")
1

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没有伤那来痛 2024-07-29 21:31:59

XmlDocument.Attributes 也许？（其中有一个方法 GetNamedItem 可能会做你想要的事情，尽管我总是只是迭代属性集合）

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嘿哥们儿 2024-07-29 21:31:58

我有一个 Xml 文件 books.xml

<ParameterDBConfig>
    <ID Definition="1" />
</ParameterDBConfig>

程序：

XmlDocument doc = new XmlDocument();
doc.Load("D:/siva/books.xml");
XmlNodeList elemList = doc.GetElementsByTagName("ID");     
for (int i = 0; i < elemList.Count; i++)     
{
    string attrVal = elemList[i].Attributes["Definition"].Value;
}

现在，attrVal 的值为 ID。

I have an Xml File books.xml

<ParameterDBConfig>
    <ID Definition="1" />
</ParameterDBConfig>

Program:

XmlDocument doc = new XmlDocument();
doc.Load("D:/siva/books.xml");
XmlNodeList elemList = doc.GetElementsByTagName("ID");     
for (int i = 0; i < elemList.Count; i++)     
{
    string attrVal = elemList[i].Attributes["Definition"].Value;
}

Now, attrVal has the value of ID.

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梦罢 2024-07-29 21:31:58

您可以迁移到 XDocument 而不是 XmlDocument，然后使用 Linq（如果您喜欢该语法）。就像是：

var q = (from myConfig in xDoc.Elements("MyConfiguration")
         select myConfig.Attribute("SuperString").Value)
         .First();

You can migrate to XDocument instead of XmlDocument and then use Linq if you prefer that syntax. Something like:

var q = (from myConfig in xDoc.Elements("MyConfiguration")
         select myConfig.Attribute("SuperString").Value)
         .First();

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你在我安 2024-07-29 21:31:56

您应该查看 XPath。一旦开始使用它，您会发现它比迭代列表更高效且更容易编码。它还可以让您直接获得您想要的东西。

那么代码将类似于“

string attrVal = doc.SelectSingleNode("/MyConfiguration/@SuperNumber").Value;

请注意，XPath 3.0 已于 2014 年 4 月 8 日成为 W3C 推荐标准”。

You should look into XPath. Once you start using it, you'll find its a lot more efficient and easier to code than iterating through lists. It also lets you directly get the things you want.

Then the code would be something similar to

string attrVal = doc.SelectSingleNode("/MyConfiguration/@SuperNumber").Value;

Note that XPath 3.0 became a W3C Recommendation on April 8, 2014.

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挖个坑埋了你 2024-07-29 21:31:56

XmlNodeList elemList = doc.GetElementsByTagName(...);
for (int i = 0; i < elemList.Count; i++)
{
    string attrVal = elemList[i].Attributes["SuperString"].Value;
}

XmlNodeList elemList = doc.GetElementsByTagName(...);
for (int i = 0; i < elemList.Count; i++)
{
    string attrVal = elemList[i].Attributes["SuperString"].Value;
}

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~没有更多了~