C＃将XML反序列化为对象：XML文档中存在错误（3、2） [英] C# Deserialize XML to object: There is an error in XML document (3, 2)

问题描述

我正在尝试将XML反序列化为C＃对象

I'm trying to deserialize XML into a C# object

我遇到了错误：XML文档中出现错误（3、2）。

I am getting the error:There is an error in XML document (3, 2).

似乎无法修复它！这是代码：

Cannot seem to fix it! Here is the code:

XSD是：

<?xml version="1.0" encoding="utf-8" ?>
<!--Created with Liquid XML Studio 2012 Developer Edition (Trial) 10.0.2.3955     (http://www.liquid-technologies.com)-->
<xs:schema xmlns:tns="http://www.adamroe.com/xsd/cameras.xsd" elementFormDefault="qualified" targetNamespace="http://www.adamroe.com/xsd/cameras.xsd"     xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="CameraBase">
    <xs:complexType>
        <xs:sequence minOccurs="1" maxOccurs="1">
            <xs:element name="Cameras">
                <xs:complexType>
                    <xs:sequence minOccurs="0" maxOccurs="unbounded">
                        <xs:element name="Camera" type="tns:CameraType" />
                    </xs:sequence>
                </xs:complexType>
            </xs:element>
        </xs:sequence>
    </xs:complexType>
</xs:element>
<xs:complexType name="CameraType">
    <xs:sequence>
        <xs:element name="Make" type="xs:string" />
        <xs:element name="Model" type="xs:string" />
        <xs:element name="Variable1" type="xs:double" />
        <xs:element name="Variable2" type="xs:double" />
    </xs:sequence>
</xs:complexType>
</xs:schema>

XML是：

<?xml version="1.0" encoding="utf-8"?>
<!-- Created with Liquid XML Studio 2012 Developer Edition (Trial) 10.0.2.3955     (http://www.liquid-technologies.com) -->
<aroe:CameraBase xmlns:aroe="http://www.adamroe.com/xsd/cameras.xsd"         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"         xsi:schemaLocation="http://www.adamroe.com/xsd/cameras.xsd     C:\Users\Adam\Desktop\Cameras.xsd">
<aroe:Cameras>
    <aroe:Camera>
    <aroe:Make>SONY</aroe:Make>
    <aroe:Model>DSC-W130</aroe:Model>
    <aroe:Variable1>0.6352</aroe:Variable1>
    <aroe:Variable2>22.375</aroe:Variable2>
</aroe:Camera>
<aroe:Camera>
    <aroe:Make>Panasonic</aroe:Make>
    <aroe:Model>DMC-FX30</aroe:Model>
    <aroe:Variable1>0.8869</aroe:Variable1>
    <aroe:Variable2>24.73</aroe:Variable2>
</aroe:Camera>
<aroe:Camera>
    <aroe:Make>Olympus</aroe:Make>
    <aroe:Model>X450</aroe:Model>
    <aroe:Variable1>0.6003</aroe:Variable1>
    <aroe:Variable2>20.654</aroe:Variable2>
</aroe:Camera>
<aroe:Camera>
    <aroe:Make>Fujifilm</aroe:Make>
    <aroe:Model>FinePix S9600</aroe:Model>
    <aroe:Variable1>1.0024</aroe:Variable1>
    <aroe:Variable2>35.704</aroe:Variable2>
</aroe:Camera>
<aroe:Camera>
    <aroe:Make>Canon</aroe:Make>
    <aroe:Model>EOS 400D</aroe:Model>
    <aroe:Variable1>1.5143</aroe:Variable1>
    <aroe:Variable2>69.409</aroe:Variable2>
</aroe:Camera>
</aroe:Cameras>
</aroe:CameraBase>

类为：

public class Camera
{
    public string Make;
    public string Model;
    public double Variable1;
    public double Variable2;
}

反序列化代码：

public class PopulateXML
{
    public void DeserializeObject(string filenameXML)
    {
        Console.WriteLine("Reading with XmlReader");

        // Create an instance of the XmlSerializer specifying type and namespace.
        XmlSerializer serializer = new
        XmlSerializer(typeof(List<Camera>));

        // A FileStream is needed to read the XML document.
        FileStream fs = new FileStream(filenameXML, FileMode.Open);
        XmlReader reader = new XmlTextReader(fs);

        // Declare an object variable of the type to be deserialized.
        List<Camera> i;

        // Use the Deserialize method to restore the object's state.
        i = (List<Camera>)serializer.Deserialize(reader);
    }
}

Main：

PopulateXML x = new PopulateXML();
        // Read a purchase order.
        x.DeserializeObject("Cameras.xml");

抛出异常：i =（List）serializer.Deserialize（reader）;

The exception is thrown on: i = (List)serializer.Deserialize(reader);

推荐答案

XmlSerializer serializer = new XmlSerializer(typeof(Camera));

Stream fs = File.OpenRead(filenameXMLpath);

// Use the Deserialize method to restore the object's state.
Camera cam = serializer.Deserialize(fs) as Camera;

这绝对适合您。在您的情况下，它不起作用，因为您将其类型强制转换为不正确的数据类型，即List。每当开发人员类型将反序列化的对象转换为不正确的数据类型时，通常都会弹出此错误。

This will definitely work for you. In your case it was not working because you were type casting it to the incorrect data type i.e., List. This error generally pops-up whenever developer type cast deserialized object to incorrect data type.

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