今天重构代码时，想把如下xml文件嵌入程序集中，在运行时读取：

<?xml version="1.0" encoding="utf-8"?>
<Convertors xmlns="http://tempuri.org/~vs24E.xsd">
    <Convertor>
        <Name>1</Name>
        <Category>1</Category>
        <Description>1</Description>
    </Convertor>
    <Convertor>
        <Name>2</Name>
        <Category>2</Category>
        <Description>2</Description>
    </Convertor>
    <Convertor>
        <Name>3</Name>
        <Category>3</Category>
        <Description>3</Description>
    </Convertor>
</Convertors>
到处找了一番，都是关于读取.txt和.resx类型的嵌入资源的，后来灵光一现，试出以下方法：

private static ConvertorData GetConvertorData()
        {
            Assembly assembly = typeof(ConvertorProvider).Assembly ;
            System.IO.Stream stream = assembly.GetManifestResourceStream("TextConvertor.Convertor.xml") ;

            ConvertorData data = new ConvertorData() ;
            data.ReadXml(stream) ;
            return data ;
        }
大概是先得到Assembly对象，然后得到流对象，以后就好办了，要不读到XmlDocument，要不读到根据xml文件生成的数据集中。