XML C# xml-deserialization xml-serialization

将 XML 反序列化为对象数组或单个对象

发布于 2024-09-06 02:15:08 字数 1014 浏览 4 评论 0原文

我正在尝试编写一个通用方法，可用于将 xml 反序列化为对象数组。

给定的 XML 如下所示：

<people>
    <person>
        <someElement>content</someElement>
    </person>
    <person>
        <someElement>more content</someElement>
    </person>
</people>

在下面的代码中显示为 xmlDoc。和 person 类作为 T

XmlNodeReader reader = new XmlNodeReader(xmlDoc.DocumentElement);
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T[]), new XmlRootAttribute(xmlDoc.DocumentElement.Name));
results = xmlSerializer.Deserialize(reader) as T[];

这按预期工作，并返回带有 2 个条目的 person[] 。

然而，在我正在使用的 API 中，如果只返回 1 个结果，它只会返回：

<person>
    <someElement>content</someElement>
</person>

并且我的反序列化会失败。 person[] 保持为空。

关于实现这一点的最佳方法有什么想法吗？

编辑

我正在考虑在两者之间运行 XSLT，并传递 T 的名称，如果它与根节点匹配，然后添加一个包装节点？

原文

I'm trying to write a generic method that can be used to deserialize xml to an array of objects.

Given XML that looks like so:

<people>
    <person>
        <someElement>content</someElement>
    </person>
    <person>
        <someElement>more content</someElement>
    </person>
</people>

Shown in the below code as xmlDoc. And a person class as T

XmlNodeReader reader = new XmlNodeReader(xmlDoc.DocumentElement);
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T[]), new XmlRootAttribute(xmlDoc.DocumentElement.Name));
results = xmlSerializer.Deserialize(reader) as T[];

This works as expected, and returns person[] with 2 entries.

However, in the API I am working with, if there is only 1 result returned, it just returns:

<person>
    <someElement>content</someElement>
</person>

And my deserialization breaks down. person[] remains empty.

Any thoughts on the best way to implement this?

Edit

I'm contemplating running an XSLT in between, and passing the name of T in, if it matches the root node, then add a wrapping node?

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筱果果 2024-09-13 02:15:08

我最终使用 XSLT 来确保我所查找的节点不是根节点。

基本上我有一个 XSLT 文件，其中包含：（

<xsl:template match="/">
    <rootNode>
        <xsl:apply-templates select="node()|@*"/>
    </rootNode>
</xsl:template>

不确定这个 XSLT 是否理想，希望有一些评论）。

这封装了来自 api 的入站 XML。我之前提到的 Person 类应用了一个 [XmlType("person")] 属性，我可以执行以下操作：

//using reflection to look what the XmlType has been declared on this type
var typeAttributes = Attribute.GetCustomAttribute(typeof(T), typeof(XmlTypeAttribute));

//determine an xpath query to find this type of elements parent
string xPathtoTypeName = string.Format("//{0}/parent::node()", ((XmlTypeAttribute)typeAttributes).TypeName);

//use the above xpath query to find the parent node.
var parentNode = transformedDocument.SelectSingleNode(xPathtoTypeName);

//deserialize as before
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T[]), new XmlRootAttribute(parentNode.Name));
XmlNodeReader nodeReader = new XmlNodeReader(parentNode);
results = xmlSerializer.Deserialize(nodeReader) as T[];

I ended up using XSLT to ensure the node(s) I was after weren't the root.

Basically I've a XSLT file containing:

<xsl:template match="/">
    <rootNode>
        <xsl:apply-templates select="node()|@*"/>
    </rootNode>
</xsl:template>

(Not sure if this XSLT is ideal, would love some comments).

This wraps a around my inbound XML from the api. My previously mentioned Person class has a [XmlType("person")] attribute applied to it, armed with that I can do:

//using reflection to look what the XmlType has been declared on this type
var typeAttributes = Attribute.GetCustomAttribute(typeof(T), typeof(XmlTypeAttribute));

//determine an xpath query to find this type of elements parent
string xPathtoTypeName = string.Format("//{0}/parent::node()", ((XmlTypeAttribute)typeAttributes).TypeName);

//use the above xpath query to find the parent node.
var parentNode = transformedDocument.SelectSingleNode(xPathtoTypeName);

//deserialize as before
XmlSerializer xmlSerializer = new XmlSerializer(typeof(T[]), new XmlRootAttribute(parentNode.Name));
XmlNodeReader nodeReader = new XmlNodeReader(parentNode);
results = xmlSerializer.Deserialize(nodeReader) as T[];

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