序列化多个对象

Question

我的序列化代码是这样的..

public class slab
{
    public int lowerlimit {get; set;}
    public int upperlimit { get; set; }
    public int percentage { get; set; }

}

public class Details
{
    static void Main(string[] args)
    {
        slab s= new slab();
        s.lowerlimit = 0;
        s.upperlimit = 200000;
        s.percentage = 0;
        XmlSerializer serializer = new XmlSerializer(s.GetType());
        StreamWriter writer = new StreamWriter(@"filepath");
        serializer.Serialize(writer.BaseStream, s);
    }
}

它工作正常,我得到输出文件:

<?xml version="1.0"?>
<slab xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <lowerlimit>0</lowerlimit>
    <upperlimit>200000</upperlimit>
    <percentage>0</percentage>
</slab>

但是,如何序列化多个对象？我想得到一个输出文件

<slabs>
    <slab>
        <lowerlimit>0</lowerlimit>
        <upperlimit>200000</upperlimit>
        <percentage>0</percentage>
    </slab>
    <slab>
        <lowerlimit>200000</lowerlimit>
        <upperlimit>500000</upperlimit>
        <percentage>10</percentage>
    </slab>
    <slab>
        <lowerlimit>500000</lowerlimit>
        <upperlimit>1000000</upperlimit>
        <percentage>20</percentage>
    </slab>
    <slab>
        <lowerlimit>1000000</lowerlimit>
        <upperlimit>0</upperlimit>
        <percentage>30</percentage>
    </slab>
</slabs>

Answer 1

实际上,所需的输出格式不是有效的XML,因为XML文件总是需要单个根元素.你可以将你的slabs放入一个列表(List<Slab> slabs = new List<Slab>();)并序列化,但你可能得到这样的输出:

<slabs>
    <slab>
    <lowerlimit>0</lowerlimit>
    <upperlimit>200000</upperlimit>
    <percentage>0</percentage>
    </slab>

    <slab>
    <lowerlimit>200000</lowerlimit>
    <upperlimit>500000</upperlimit>
    <percentage>10</percentage>
    </slab>

    <slab>
    <lowerlimit>500000</lowerlimit>
    <upperlimit>1000000</upperlimit>
    <percentage>20</percentage>
    </slab>

    <slab>
    <lowerlimit>1000000</lowerlimit>
    <upperlimit>0</upperlimit>
    <percentage>30</percentage>
    </slab>
</slabs>

编辑
序列化的另一种方法是这样,告诉序列化器更多关于根元素:

List<Slab> slabs = new List<Slab>();
slabs.Add(...);

XmlSerializer serializer = new XmlSerializer(slabs.GetType(), new XmlRootAttribute("slabs"));
StreamWriter writer = new StreamWriter(@"filepath");
serializer.Serialize(writer.BaseStream, slabs);