104

我有一个名为WebserviceType我从 XSD 文件中的工具 xsd.exe 获得的类。

现在我想将一个WebServiceType对象的实例反序列化为一个字符串。我怎样才能做到这一点？

该MethodCheckType对象有一个WebServiceType数组作为参数。

我的第一次尝试就像我序列化它：使用 aXmlSerializer和 a StringWriter（序列化时我使用了 a StringReader）。

这是我序列化WebServiceType对象的方法：

XmlSerializer serializer = new XmlSerializer(typeof(MethodCheckType));
        MethodCheckType output = null;
        StringReader reader = null;

        // catch global exception, logg it and throw it
        try
        {
            reader = new StringReader(path);
            output = (MethodCheckType)serializer.Deserialize(reader);
        }
        catch (Exception)
        {
            throw;
        }
        finally
        {
            reader.Dispose();
        }

        return output.WebService;

编辑：

也许我可以换一种说法：我得到了这个MethodCheckType对象的一个​​实例，另一方面，我得到了序列化这个对象的 XML 文档。现在我想将此实例转换为字符串形式的 XML 文档。在此之后，我必须证明两个字符串（XML 文档）是否相同。这是我必须做的，因为我对第一种方法进行了单元测试，在该方法中，我将 XML 文档读入 aStringReader并将其序列化为MethodCheckType对象。

4

5 回答 5

214

这是两种方式的转换方法。this = 你的类的实例

public string ToXML()
    {
        using(var stringwriter = new System.IO.StringWriter())
        { 
            var serializer = new XmlSerializer(this.GetType());
            serializer.Serialize(stringwriter, this);
            return stringwriter.ToString();
        }
    }

 public static YourClass LoadFromXMLString(string xmlText)
    {
        using(var stringReader = new System.IO.StringReader(xmlText))
        {
            var serializer = new XmlSerializer(typeof(YourClass ));
            return serializer.Deserialize(stringReader) as YourClass ;
        }
    }
于 2012-07-12T08:56:36.740 回答
81

我意识到这是一篇非常古老的帖子，但在查看了 LB 的回复后，我想到了如何改进已接受的答案并使其适用于我自己的应用程序。这是我想出的：

public static string Serialize<T>(T dataToSerialize)
{
    try
    {
        var stringwriter = new System.IO.StringWriter();
        var serializer = new XmlSerializer(typeof(T));
        serializer.Serialize(stringwriter, dataToSerialize);
        return stringwriter.ToString();
    }
    catch
    {
        throw;
    }
}

public static T Deserialize<T>(string xmlText)
{
    try
    {
        var stringReader = new System.IO.StringReader(xmlText);
        var serializer = new XmlSerializer(typeof(T));
        return (T)serializer.Deserialize(stringReader);
    }
    catch
    {
        throw;
    }
}

这些方法现在可以放置在静态帮助程序类中，这意味着无需对需要序列化的每个类重复代码。

于 2014-02-10T18:31:15.030 回答
21 
    public static string Serialize(object dataToSerialize)
    {
        if(dataToSerialize==null) return null;

        using (StringWriter stringwriter = new System.IO.StringWriter())
        {
            var serializer = new XmlSerializer(dataToSerialize.GetType());
            serializer.Serialize(stringwriter, dataToSerialize);
            return stringwriter.ToString();
        }
    }

    public static T Deserialize<T>(string xmlText)
    {
        if(String.IsNullOrWhiteSpace(xmlText)) return default(T);

        using (StringReader stringReader = new System.IO.StringReader(xmlText))
        {
            var serializer = new XmlSerializer(typeof(T));
            return (T)serializer.Deserialize(stringReader);
        }
    }
于 2014-04-14T11:49:34.527 回答
1 
 public static class XMLHelper
    {
        /// <summary>
        /// Usage: var xmlString = XMLHelper.Serialize<MyObject>(value);
        /// </summary>
        /// <typeparam name="T">Kiểu dữ liệu</typeparam>
        /// <param name="value">giá trị&lt;/param>
        /// <param name="omitXmlDeclaration">bỏ qua declare</param>
        /// <param name="removeEncodingDeclaration">xóa encode declare</param>
        /// <returns>xml string</returns>
        public static string Serialize<T>(T value, bool omitXmlDeclaration = false, bool omitEncodingDeclaration = true)
        {
            if (value == null)
            {
                return string.Empty;
            }
            try
            {
                var xmlWriterSettings = new XmlWriterSettings
                {
                    Indent = true,
                    OmitXmlDeclaration = omitXmlDeclaration, //true: remove <?xml version="1.0" encoding="utf-8"?>
                    Encoding = Encoding.UTF8,
                    NewLineChars = "", // remove \r\n
                };

                var xmlserializer = new XmlSerializer(typeof(T));

                using (var memoryStream = new MemoryStream())
                {
                    using (var xmlWriter = XmlWriter.Create(memoryStream, xmlWriterSettings))
                    {
                        xmlserializer.Serialize(xmlWriter, value);
                        //return stringWriter.ToString();
                    }

                    memoryStream.Position = 0;
                    using (var sr = new StreamReader(memoryStream))
                    {
                        var pureResult = sr.ReadToEnd();
                        var resultAfterOmitEncoding = ReplaceFirst(pureResult, " encoding=\"utf-8\"", "");
                        if (omitEncodingDeclaration)
                            return resultAfterOmitEncoding;
                        return pureResult;
                    }
                }
            }
            catch (Exception ex)
            {
                throw new Exception("XMLSerialize error: ", ex);
            }
        }

        private static string ReplaceFirst(string text, string search, string replace)
        {
            int pos = text.IndexOf(search);

            if (pos < 0)
            {
                return text;
            }

            return text.Substring(0, pos) + replace + text.Substring(pos + search.Length);
        }
    }
于 2020-09-29T02:46:45.283 回答
0

这是我的解决方案，对于任何列表对象，您都可以使用此代码转换为 xml 布局。KeyFather 是您的主要标签，而 KeySon 是您的 Forech 的起点。

public string BuildXml<T>(ICollection<T> anyObject, string keyFather, string keySon)
    {
        var settings = new XmlWriterSettings
        {
            Indent = true
        };
        PropertyDescriptorCollection props = TypeDescriptor.GetProperties(typeof(T));
        StringBuilder builder = new StringBuilder();
        using (XmlWriter writer = XmlWriter.Create(builder, settings))
        {
            writer.WriteStartDocument();
            writer.WriteStartElement(keyFather);
            foreach (var objeto in anyObject)
            {
                writer.WriteStartElement(keySon);
                foreach (PropertyDescriptor item in props)
                {
                    writer.WriteStartElement(item.DisplayName);
                    writer.WriteString(props[item.DisplayName].GetValue(objeto).ToString());
                    writer.WriteEndElement();
                }
                writer.WriteEndElement();
            }
            writer.WriteFullEndElement();
            writer.WriteEndDocument();
            writer.Flush();
            return builder.ToString();
        }
    }
于 2019-02-01T13:41:56.533 回答