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Xml 显示没有编号的节点_Xml_Xslt - Fatal编程技术网

Xml 显示没有编号的节点

Xml 显示没有编号的节点,xml,xslt,Xml,Xslt,我有一个创建XML节点的代码,如下所示 foreach(var books in booksCollection) { XmlText bookTitle = xmlDoc.CreateTextNode(books.bookTitle); bookTitle.AppendChild(bookTitle); XmlText companyTitle = xmlDoc.Creat

我有一个创建XML节点的代码,如下所示

foreach(var books in booksCollection)
            {
                XmlText bookTitle = xmlDoc.CreateTextNode(books.bookTitle);
                bookTitle.AppendChild(bookTitle);

                XmlText companyTitle = xmlDoc.CreateTextNode(books.companyTitle);
                CompanyTitle.AppendChild(companyName);

                XmlText author = xmlDoc.CreateTextNode(books.author);
                Author.AppendChild(author);
            }
我使用的是一个列表,因为有很多数据。 现在,在XSLT中使用“position()”可以根据列表计数打印数据。 也就是说,如果有3个项目,它将打印出3个项目,如下所示:

  <xsl:for-each select="bookInfo" >  // my code
                <xsl:number value="position()"/>
                <xsl:value-of select="bookTitle" /><br/>
                <xsl:value-of select="companyTitle" /> <br/>
                <xsl:value-of select="author" /><br/>
            </xsl:for-each>

除非我遗漏了什么,否则就把这行删掉

 <xsl:number value="position()"/>

xsl:number只输出数字,但不控制选择的元素

 <xsl:number value="position()"/>