将C#类序列化为XML文本的更简单方法

Question

在尝试回答另一个问题时,我将C#对象序列化为XML字符串.这真是令人惊讶的困难; 这是我能提出的最短的代码片段:

var yourList = new List<int>() { 1, 2, 3 };
var ms = new MemoryStream();
var xtw = new XmlTextWriter(ms, Encoding.UTF8);
var xs = new XmlSerializer(yourList.GetType());
xs.Serialize(xtw, yourList);
var encoding = new UTF8Encoding();
string xmlEncodedList = encoding.GetString(ms.GetBuffer());

结果还可以:

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfInt
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <int>1</int>
    <int>2</int>
    <int>3</int>
</ArrayOfInt>

但是片段比我想象的要复杂得多.我无法相信你必须知道编码和MemoryStream这个简单的任务.

是否有更短的方法将对象序列化为XML字符串？

Answer 1

稍短一些:-)

var yourList = new List<int>() { 1, 2, 3 };
using (var writer = new StringWriter())
{
    new XmlSerializer(yourList.GetType()).Serialize(writer, yourList);
    var xmlEncodedList = writer.GetStringBuilder().ToString();
}

虽然这种先前的方法存在缺陷值得指出.它会生成一个utf-16标头,因为我们使用StringWriter,因此它不完全等同于您的代码.要获取utf-8头,我们应该使用MemoryStream和XmlWriter,这是一行额外的代码:

var yourList = new List<int>() { 1, 2, 3 };
using (var stream = new MemoryStream())
{
    using (var writer = XmlWriter.Create(stream))
    {
        new XmlSerializer(yourList.GetType()).Serialize(writer, yourList);
        var xmlEncodedList = Encoding.UTF8.GetString(stream.ToArray());
    }
}