读取XML某子节点的所有值

Question

<?xml version="1.0" encoding="UTF-8"?>
-<cat> -<status> <code>1</code> <message>Get domain list success.</message> <created_at>2011-08-07 01:46:03</created_at> </status> -<info> <domain_total>3</domain_total> </info> -<domains> -<item> <id>240481</id> <name>xx1.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>13</records> <updated_on>2011-06-21 13:17:25</updated_on> </item> -<item> <id>397274</id> <name>xx2.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>18</records> <updated_on>2011-06-21 13:43:17</updated_on> </item> -<item> <id>240340</id> <name>xx3.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>189</records> <updated_on>2011-08-06 20:21:24</updated_on> </item> </domains> </cat>
我需要domains下的三条item的子节点,包括下面的属性值
这个用dataset好像无法实现?这有三级子节点了.

Answer 1

-<cat> -<status> 这不对吧。。

Answer 2

void Main()
{
string xml=@"<?xml version=""1.0"" encoding=""UTF-8""?>
<cat> <status> <code>1</code> <message>Get domain list success.</message> <created_at>2011-08-07 01:46:03</created_at> </status> -<info> <domain_total>3</domain_total> </info> -<domains> -<item> <id>240481</id> <name>xx1.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>13</records> <updated_on>2011-06-21 13:17:25</updated_on> </item> -<item> <id>397274</id> <name>xx2.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>18</records> <updated_on>2011-06-21 13:43:17</updated_on> </item> -<item> <id>240340</id> <name>xx3.com</name> <grade>D_Free</grade> <status>enable</status> -<ext_status>
<![CDATA[]]>
</ext_status> <records>189</records> <updated_on>2011-08-06 20:21:24</updated_on> </item> </domains> </cat>";
   DataSet ds=new DataSet();
   XElement data=XElement.Parse(xml);
   var query=from x in data.Descendants("domains")
              from item in x.Elements("item")
              select new
  {
     id=item.Element("id").Value,
 name=item.Element("name").Value,
 //下边的以此类推，自己添加，你懂的
  };
Console.WriteLine(query);
}

Answer 3

 DataSet ds=new DataSet(); 这句可以删除掉

Answer 4

        string xml = @"<?xml version='1.0' encoding='utf-8'?> <cat>
    <status>
        <code>1</code>
        <message>Get domain list success.</message>
        <created_at>2011-08-07 01:46:03</created_at>
    </status> <info>
        <domain_total>3</domain_total>
    </info> <domains>
        <item>
            <id>240481</id> <name>xx1.com</name> <grade>D_Free</grade> <status>enable</status> <ext_status> <![CDATA[]]> </ext_status> <records>13</records> <updated_on>2011-06-21 13:17:25</updated_on>
        </item> <item>
            <id>397274</id> <name>xx2.com</name> <grade>D_Free</grade> <status>enable</status> <ext_status> <![CDATA[]]> </ext_status> <records>18</records> <updated_on>2011-06-21 13:43:17</updated_on>
        </item> <item>
            <id>240340</id> <name>xx3.com</name> <grade>D_Free</grade> <status>enable</status> <ext_status> <![CDATA[]]> </ext_status> <records>189</records> <updated_on>2011-08-06 20:21:24</updated_on>
        </item>
    </domains>
</cat>";
        XmlDocument dom = new XmlDocument();
        dom.LoadXml(xml);        XmlNodeList list = dom.SelectNodes(@"/cat/domains/item/child::*/child::text()");
        
        for (int i = 0; i < list.Count; i++)
        {
            Response.Write(list[i].OuterXml);
        }

Answer 5

OuterXml
这个关键字好像找不到,用你上面的语句,
winform

Answer 6

更正下这个XElement data = XElement.Load(path);
                    var query = from x in data.Descendants("domains")
                                from item in x.Elements("item")
                                select new
                                {
                                    id=item.Element("id").Value,
                                    name=item.Element("name").Value                                };
                    foreach (var item in query)
                    {
                        MessageBox.Show(item.id);
                        MessageBox.Show(item.name);
                    }这样才正确
XElement.Parse(strig ..)这里能只能是xml格式的字符串吧?你声明了string xml,是对的.接路径只能
XElement.Load(string path)不过 Console.WriteLine(query);
这个有点不太明白,能直接这样显示出来么?

Answer 7

Console.WriteLine(query);
这不是在VS内写的代码
是在一个小工具内写的测试代码 在这个工具内可以直接输出结果