請耐心等待，因為我是新手。我需要將 JSON 字串轉換為 XML 進行一些修改
{"computerid":123456,"computername":"mycomputer","computermodel":"mymodel"}
到
<?xml version="1.0"?>
<SqlMultiFilter xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Parameter>
<Filter>
<ParamName>computerid</ParamName>
<ParamValues>
<ParamValue>123456</ParamValue>
</ParamValues>
</Filter>
<Filter>
<ParamName>computername</ParamName>
<ParamValues>
<ParamValue>mycomputer</ParamValue>
</ParamValues>
</Filter>
<Filter>
<ParamName>computermodel</ParamName>
<ParamValues>
<ParamValue>mymodel</ParamValue>
</ParamValues>
</Filter>
</Parameter>
</SqlMultiFilter>
我以這種方式創建了一個公共課程
[XmlRoot("SQLMultiFilter", Namespace = "http://www.cpandl.com",
IsNullable = false)]
public class SQLMultiFilter
{
[XmlArrayAttribute("Parameter")]
public string ParamName;
[XmlArrayAttribute("ParamValues")]
public string ParamValue;
}
當我發布我的 json 時，我在這方面遇到了錯誤
XmlSerializer serializer = new XmlSerializer(typeof(SQLMultiFilter));
錯誤
Inner Exception 1:
InvalidOperationException: There was an error reflecting field 'ParamName'.
Inner Exception 2:
InvalidOperationException: For non-array types, you may use the following attributes: XmlAttribute, XmlText, XmlElement, or XmlAnyElement.
我哪里出錯了？
提前致謝
uj5u.com熱心網友回復：
[XmlRoot("SQLMultiFilter", Namespace = "http://www.cpandl.com", IsNullable = false)]
public class SqlMultiFilter
{
public List<Filter> Parameter { get; set; }
}
public class Filter
{
public string ParamName;
[XmlArrayItem("ParamValue")]
public List<string> ParamValues { get; set; }
}
這兩個類將允許您重現所示模式的 XML。
var serializer = new XmlSerializer(typeof(SqlMultiFilter));
var sqlMultiFilter = new SqlMultiFilter
{
Parameter = new List<Filter>
{
new Filter { ParamName = "computerid", ParamValues = new List<string> { "123456" } },
new Filter { ParamName = "computername", ParamValues = new List<string> { "mycomputer" } },
new Filter { ParamName = "computermodel", ParamValues = new List<string> { "mymodel" } }
}
};
var settings = new XmlWriterSettings { Indent = true };
using (var xmlWriter = XmlWriter.Create(Console.Out, settings))
serializer.Serialize(xmlWriter, sqlMultiFilter);
會給出想要的結果。
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