Question

有没有人知道如何在深度优先遍历中跟踪xsd中的路径例如：如果我有这个架构，

 <xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"          targetNamespace="urn:books"            xmlns:bks="urn:books">
  <xsd:element name="books" type="bks:BooksForm"/>
  <xsd:complexType name="BooksForm">
    <xsd:sequence>
      <xsd:element name="book"                   type="bks:BookForm"                   minOccurs="0"                   maxOccurs="unbounded"/>
     </xsd:sequence>
  </xsd:complexType>
  <xsd:complexType name="BookForm">
    <xsd:sequence>
      <xsd:element name="author"   type="xsd:string"/>
      <xsd:element name="title"    type="xsd:string"/>
      <xsd:element name="genre"    type="xsd:string"/>
      <xsd:element name="price"    type="xsd:float" />
      <xsd:element name="pub_date" type="xsd:date" />
      <xsd:element name="review"   type="xsd:string"/>
    </xsd:sequence>
    <xsd:attribute name="id"   type="xsd:string"/>
  </xsd:complexType>
</xsd:schema>

如何在c＃

中获得此输出

path1 : books.book.author
path2 : books.book.title

..等等

用于任何架构结构

有没有人对这个或任何好的起点有所了解提前谢谢

感谢您的回复我试着编写你的建议，但我仍然陷入困境

   while (r.Read())
    {               
            switch (r.Name)
            {
                case "xsd:element":
                    myStringBuilder.Append(r.GetAttribute("name"));
                    break;
                case "xsd:complexType":
                    checkComplex(r);
                    wholepath += r.GetAttribute("name");

                    //this will only concatenate complex elements only
                    Console.WriteLine("checkComplexcaller{1}", wholepath);
                    break;
            }

        }

用于递归部分..我做了

   public static void checkComplex(/*what I should send here*/)
    {

        if (r.GetAttribute("name") == "xsd:complexType")
        {
            //What I sould to do for this recursive                    

        }
        else if (r.GetAttribute("name") == "xsd:element")
        {
            myStringBuilder.Append(r.GetAttribute("name"));
        }
    }

如何正确设置路径

Answer 1

如果我做对了:(伪代码）

(make a xml document of the xsd)
foreach node of type xsd:element: 
   track path to the node as input
   apend it with values of 'name' and 'type' attributes
   if its a complex type: apply this recursively on all elements

C＃中XSD的路径/节点顺序

1 个答案: