Deserialization XmlArray not populating

Question

我正在尝试使用C＃从API中验证XML响应。我无法获得ItemList元素的值。除了ItemList之外，我还能获得其他元素。示例XML在下面，在其下方分类，最后是序列化函数。

XML：

<Root xmlns="http://whatever">
    <Child1>
        <ChildElement1>16f09999</ChildElement1>
    </Child1>
    <ItemList>
        <Element1>
        ...
        </Element1>
        <Item>
            <ItemElem1>12345567</ItemElem1>
            <ItemElem2>12356465784</ItemElem2>
           ...
            <ItemAttributes>
            ...
            </ItemAttributes>
        </Item>
        <Item>
        ...
        </Item>
    </ItemList>
</Root>
[Serializable()]
[System.Xml.Serialization.XmlRoot("Root")]
public class ResponseClass
{
    [System.Xml.Serialization.XmlElement("Child1")]
    public OperationRequest OperationRequest { get; set; }
    [System.Xml.Serialization.XmlElement("ItemList")]
    public ItemList ItemList { get; set; }
}

类：

[Serializable()]
[System.Xml.Serialization.XmlRoot(ElementName = "ItemList")]
public class ItemList
{
    [XmlArray("ItemList")]
    [XmlArrayItem("Item", typeof(Item))]
    public Item item { get; set; }
    [System.Xml.Serialization.XmlElement("Element1")]
    public int Element1 { get; set; }
}

序列化器：

XmlSerializer serializer = new XmlSerializer(typeof(ResponseClass));
StreamReader reader = new StreamReader(path);
response = (ResponseClass)serializer.Deserialize(reader);
reader.Close();

Answer 1

请尝试以下内容。由于ItemList/Item不是XmlArray，因此您需要将其定义为简单的XmlElement。

[Serializable()]
[System.Xml.Serialization.XmlRoot(ElementName = "Root", Namespace = "http://whatever")]
public class ResponseClass
{
    [System.Xml.Serialization.XmlElement("Child1")]
    public OperationRequest OperationRequest { get; set; }
    [System.Xml.Serialization.XmlElement("ItemList")]
    public ItemList ItemList { get; set; }
}
[Serializable()]
[System.Xml.Serialization.XmlRoot(ElementName = "ItemList", Namespace = "http://whatever")]
public class ItemList
{
    [System.Xml.Serialization.XmlElement("Item")]
    public Item[] Items { get; set; }
    [System.Xml.Serialization.XmlElement("Element1")]
    public int Element1 { get; set; }
}