2014-01-29 43 views
0

我正在用xmlserialize類創建一個xml文件,我得到的輸出幾乎是我所需要的。 這是我得到自定義XmlSerialize輸出

<?xml version="1.0" encoding="UTF-8"?> 
<pdf file-name="C:\Users\d.sassoli\Desktop\media\mergiato.pdf" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> 
<document> 
    <document start-page="1"> 
    <SheetChange> 
    <SheetChange sheet-type="A4" page-number="1"/> 
    <SheetChange sheet-type="A3" page-number="2"/> 
    </SheetChange> 
</document> 
</pdf> 

這就是我想要

<?xml version="1.0" encoding="UTF-8"?> 
<root xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"> 
<pdf file-name="C:\Users\d.sassoli\Desktop\media\mergiato.pdf" > 
    <document start-page="1"> 
    <SheetChange sheet-type="A4" page-number="1"/> 
    <SheetChange sheet-type="A3" page-number="2"/> 
</pdf> 

這些都是我的課,我確定我的XML

[XmlRootAttribute("pdf", IsNullable = false)] 
public class Pdf 
{ 
    private Pdf() { } 
    public static Pdf Builder() 
    { 
     Pdf pdf = new Pdf(); 
     return pdf; 
    } 
    [XmlAttribute("file-name")] 
    public List<string> FileName { get; set; } 
    public List<document> document; 

} 
public class document 
{ 

    private document() { } 
    public static document Builder() 
    { 
     document doc = new document(); 
     return doc; 
    } 
    [XmlAttribute("last-page")] public string LastPage; 
    [XmlAttribute("start-page")] public string StartPage; 
    public List<SheetChange> SheetChange; 

} 
public class SheetChange 
{ 
    private SheetChange() { } 
    public static SheetChange Builder() 
    { 
     SheetChange sheetChange = new SheetChange(); 
     return sheetChange; 
    } 
    [XmlAttribute("page-number")] 
    public string PageNumber { get; set; } 
    [XmlAttribute("sheet-type")] 
    public string SheetType { get; set; } 
} 

我的結構嘗試擺弄xmlattribute但沒有達到我想要的。

回答