*************请问关于xquery的问题

[gamecoder001]

<employee>
<Sname>张三</Sname>
<V1>1</V1>
<V2>21</V2>
<V3>43</V3>
<V4>32</V4>
<V5>8</V5>
<V6>1</V6>
</employee>
<employee>
<Sname>李四</Sname>
<V1>54</V1>
<V2>61</V2>
<V3>23</V3>
<V4>32</V4>
<V5>2</V5>
<V6>2</V6>
</employee>
<employee>
<Sname>张五</Sname>
<V1>23</V1>
<V2>21</V2>
<V3>23</V3>
<V4>22</V4>
<V5>5</V5>
<V6>2</V6>
</employee>
<employee>
<Sname>张六</Sname>
<V1>23</V1>
<V2>52</V2>
<V3>16</V3>
<V4>33</V4>
<V5>1</V5>
<V6>3</V6>
</employee>

我一个这样的xml文档

我想求出对于每一个sname 后面的v1-v6的求和
请问应该怎么做啊
得到的结果大概是这样的
<emp Sname="" Amount="" />
...

[cyy1981]

用户xpath取出值来加就可以了吧

[孙晓军82]

假设当前处理的是employee节点
xpath "sum(./node()[starts-with('V')]/text())";

[gamecoder001]

顶

[gamecoder001]

能写出详细的代码吗

[孙晓军82]

sorry!上面的xpath写得有错误.

如果c#操作的话,
XPathDocument doc = new XPathDocument("employees.xml");
XPathNavigator nav = doc.CreateNavigator();
XPathExpression expression = nav.Compile("sum(./node()[starts-with(name(), 'V')]/text()) ");
XPathNodeIterator iter = nav.Select("/employees/employee");
if (iter.MoveNext())
{
object obj = iter.Current.Evaluate(expression);
int sum = Convert.ToInt32(obj); //result is 106.即是张三下面所有v开头节点数值得和。
}

如果用xsl的话就是。
<xsl:value-of select="sum(./node()[starts-with(name(), 'V')]/text())"/>
就可以了，注意下相对路径。

[mengxj85]

Mark