急,200分帮忙读一个XML，对于做过的不难

[tongzhenhua]

string src=@"
<?xml version="1.0" encoding="GBK"?>
<ExInfo>
<ItemInfo DeptCode="05" WinCode="B-C63201" ItemCode="050" Action="1">
<BaseInfo>
<Field FieldName="projinfo_proj_name">培训中心</Field>
<Field FieldName="projinfo_proj_year">2004</Field>
</BaseInfo>
<UnitInfo UnitNo="2" UnitKind="ConstructUnit">
<Field FieldName="unit_proj_id">200441521</Field>
<Field FieldName="unit_corporator">庞星</Field>
</UnitInfo>
<UnitInfo UnitNo="4" UnitKind="DesiUnit">
<Field FieldName="unit_proj_id">200441521</Field>
<Field FieldName="unit_unit_id">1436</Field>
</UnitInfo>
</ItemInfo>
</ExInfo>
"
谁能帮忙写个c#
把这个XML输出成
BaseInfo
培训中心
2004
UnitInfo
200441521
庞星
UnitInfo
200441521
1436

要求给出完整代码。直接运行就能通过的。

我的是
XmlDocument doc = new XmlDocument();
doc.LoadXml(src);
然后后面怎么写也取不出东西:(

[JGood]

用ASP.NET实现：
楼主试一下：

        string src = @"<?xml   version=""1.0""   encoding=""GBK""?> 
<ExInfo> 
    <ItemInfo   DeptCode=""05""   WinCode=""B-C63201""   ItemCode=""050""   Action=""1""> 
        <BaseInfo> 
            <Field   FieldName=""projinfo_proj_name""> 培训中心 </Field> 
            <Field   FieldName=""projinfo_proj_year""> 2004 </Field> 
        </BaseInfo> 
        <UnitInfo   UnitNo=""2""   UnitKind=""ConstructUnit""> 
            <Field   FieldName=""unit_proj_id""> 200441521 </Field> 
            <Field   FieldName=""unit_corporator""> 庞星 </Field> 
        </UnitInfo> 
        <UnitInfo   UnitNo=""4""   UnitKind=""DesiUnit""> 
            <Field   FieldName=""unit_proj_id""> 200441521 </Field> 
            <Field   FieldName=""unit_unit_id""> 1436 </Field> 
        </UnitInfo> 
    </ItemInfo> 
</ExInfo> 
";
        //结果保存字符串
        System.Text.StringBuilder result = new System.Text.StringBuilder();

        //加载XML
        System.Xml.XmlDocument doc = new System.Xml.XmlDocument();
        doc.LoadXml(src);

        //使用DOM解析
        System.Xml.XmlNode nodeExInfo = doc.DocumentElement.FirstChild;
        foreach (XmlElement ele in nodeExInfo.ChildNodes)
        {
            result.Append(ele.Name.ToString()).Append("<br />");
            foreach (XmlElement nodeField in ele.ChildNodes)
            {
                result.Append("    ").Append(nodeField.FirstChild.InnerText).Append("<br />");
            }

        }

        //打印结果
        Response.Write(result.ToString());

[vicqqq]

            XmlDocument doc = new XmlDocument();
            doc.LoadXml(str);
            XmlElement element = doc.DocumentElement;
            XmlNode node = element.FirstChild;
            XmlNodeList nodelist = node.ChildNodes;
            foreach (XmlNode n in nodelist)
            {
                Response.Write(n.Name + "<br />");
                foreach (XmlNode n1 in n.ChildNodes)
                    Response.Write("  "+n1.InnerText + "<br />");
            }

[JGood]

使用这个方法：GetAttribute("FieldName")