Question

我有一个返回一些xml的Web服务。没有相应的xsd。

我在Visual Studio中使用选择性粘贴->粘贴XML作为类功能来生成要与XmlSerializer一起使用的类：

代码执行该行时

--> XmlSerializer xmlSerialize = new XmlSerializer(typeof(table)); <--
    table t = (table)xmlSerialize.Deserialize(new StringReader(soapResult));

它引发异常：

System.InvalidOperationException: 'Unable to generate a temporary class (result=1).
error CS0030: Cannot convert type 'string[]' to 'string'
error CS0029: Cannot implicitly convert type 'string' to 'string[]'

生成的类如下：

// NOTE: Generated code may require at least .NET Framework 4.5 or .NET Core/Standard 2.0.
    /// <remarks/>
    [System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
    [System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
    public partial class table
    {

        private string[] labelField;

        private string[] classnameField;

        private string[] datatypeField;

        private string[][] rowField;


        /// <remarks/>
        [System.Xml.Serialization.XmlArrayItemAttribute("d", IsNullable = false)]
        public string[] label
        {
            get
            {
                return this.labelField;
            }
            set
            {
                this.labelField = value;
            }
        }

        /// <remarks/>
        [System.Xml.Serialization.XmlArrayItemAttribute("d", IsNullable = false)]
        public string[] classname
        {
            get
            {
                return this.classnameField;
            }
            set
            {
                this.classnameField = value;
            }
        }

        /// <remarks/>
        [System.Xml.Serialization.XmlArrayItemAttribute("d", IsNullable = false)]
        public string[] datatype
        {
            get
            {
                return this.datatypeField;
            }
            set
            {
                this.datatypeField = value;
            }
        }

        /// <remarks/>
        [System.Xml.Serialization.XmlArrayItemAttribute("d", typeof(string), IsNullable = true)]
        public string[][] row
        {
            get
            {
                return this.rowField;
            }
            set
            {
                this.rowField = value;
            }
        }
    }

问题出在行字段，其他行都可以。如果我改变

[System.Xml.Serialization.XmlArrayItemAttribute("d", typeof(string), IsNullable = true)]

到

[System.Xml.Serialization.XmlArrayItemAttribute("d", typeof(string[]), IsNullable = true)]

错误消失了，但是我的对象只有一个空字符串数组。

要反序列化（为了简化而压缩）的XML数据如下

<?xml version="1.0" encoding="UTF-8"?>
<table>
   <label>
      <d>JobDefinition.SearchName</d>
      <d>Job.JobDefinition</d>
   </label>
   <classname>
      <d>JobDefinition.SearchName</d>
      <d>Job.JobDefinition</d>
   </classname>
   <datatype>
      <d>String</d>
      <d>obj.JobDefinition</d>
   </datatype>
   <row>
      <d>AB_DEFG_QA_RUN</d>
      <d>A_JOB_Run</d>
   </row>
   <row>
      <d>AB_DEFG_QA_RUN</d>
      <d>B_JOB_Run</d>
   </row>
</table>

Answer 1

您的问题看上去与从this answer到How to serialize List<List<object>>?的最终更新中描述的问题相同，即xsd.exe（因此粘贴XML例如类）在为包含重复元素的XML推断正确的XML数据模型时遇到麻烦，这些重复元素包含嵌套的重复元素，此处为<row>和<d>：

<table>
   <row>
      <d>Value of repeating element inside a repeating element.</d>
      <d>Value of repeating element inside a repeating element.</d>
   </row>
   <row>
      <d>Value of repeating element inside a repeating element.</d>
      <d>Value of repeating element inside a repeating element.</d>
   </row>
</table>

该答案的建议是使用其他代码生成工具或手动构造类。我将您的XML上传到https://xmltocsharp.azurewebsites.net/，在合并重复的类之后，获得了以下数据模型：

[XmlRoot(ElementName = "row")]
public class Row
{
    [XmlElement(ElementName = "d")]
    public List<string> D { get; set; }
}

[XmlRoot(ElementName = "table")]
public class Table
{
    [XmlElement(ElementName = "label")]
    public Row Label { get; set; }
    [XmlElement(ElementName = "classname")]
    public Row Classname { get; set; }
    [XmlElement(ElementName = "datatype")]
    public Row Datatype { get; set; }

    [XmlElement(ElementName = "row")]
    public List<Row> Row { get; set; }
}

使用此方法，我可以成功地反序列化和重新序列化XML，而不会丢失数据。注意：

代码生成工具为Label，Classname，Datatype和Row生成了截然不同但相同的类。我决定将它们全部合并为单个Row类型，并将其用于Table的所有相关成员。

提琴here。

XmlSerializer和将XML粘贴为类

1 个答案: