所以我有以下xml文件，我的Windows窗体来自网站。
<WebserviceResult xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="">
<data xmlns="">
<ArrayOfMatches>
<Match>
<TimeSlot>17:00:00</TimeSlot>
<Game_Id>18</Game_Id>
<ArrayOfTeams>
<Team>
<Team_id>14</Team_id>
<Team_Name>New Zealand</Team_Name>
<TeamPlayers>
<Player>
<PlayerId>10</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
<Player>...</Player>
<Player>...</Player>
<Player>...</Player>
</TeamPlayers>
</Team>
<Team>
<Team_id>16</Team_id>
<Team_Name>west indies</Team_Name>
<TeamPlayers>...</TeamPlayers>
</Team>
</ArrayOfTeams>
</Match>
</ArrayOfMatches>
</data>
这只是整个事情的片段，这里的想法是两个团队将面对彼此，分配一个Game_Id
目前我的想法是将Game_Id Team_Id和Team_Name加载到一行中的数据集中，每个Game_Id（有xml文件中的多个游戏）。团队名称最终将以Team1 V Team2的形式出现在组合框中;然后，当选择一个时，Player个元素将被加载到两个单独的列表框中。
我目前能够做的是用TimeSlot＆amp;填充不同的数据集。 Game_Id，使用TeamPlayers的xml文件中的所有DS.ReadXML(xmlRead)，但不是我想要的方式，并将Game_Id放在一个组合框中：
XmlElement root = xdoc.DocumentElement;
XmlNodeList nodeList = xdoc.GetElementsByTagName("Match");
foreach (XmlNode xn in nodeList)
{
cb_matches.Items.Add(xn.InnerText);
}
这些都有效但不是我想做的事。
对此事的任何帮助都将不胜感激，我觉得我已尽可能多地提供信息。
EDITED
<Date> </Date>每个网络服务只出现一次，当天有多个<Game_Id>。它用于确保使用正确的XML文件。
EDITED
<data xmlns="">
<ArrayOfMatches>
<Match>...</Match>
<Match>...</Match>
</ArrayOfMatches>
</data>
EDITED
所以我仍然比这更难挣扎。
+-----+-------+---------+------------+---------+-----------+
|Time |Game_Id|Team_A_Id|Team_A_Name |Team_B_Id|Team_B_Name|
|17:00|18 |14 |South Africa|16 |West Indies|
|19:00|19 |18 |New Zealand |12 |England |
这是我想要的结构DataTable
现在我一直在玩xml-linq等，但我似乎不能像那样说出来。
XDocument xdoc = XDocument.Load(@"c:/matchdata.xml");
IEnumerable<XElement> games = xdoc.Descendants("ArrayOfMatches");
var matches = games
.Select(x => new Games()
{
GameID = x.Element("Game_Id").Value,
TeamAID = x.Element("Team_Id").Value,
TeamAName = x.Element("Team_Name").Value,
TeamBID = x.Element("Team_Id").Value,
TeamBName = x.Element("Team_Name").Value
});
dataGridView1.DataSource = xdoc;
这会编译，但不显示gridview中的任何数据，也不会显示任何错误。
答案 0 :(得分：1)
您可以将整个XML反序列化为对象并轻松使用它 （我从你的xml片段中删除了第一行，而匹配是一个条目，因为如果有多个匹配，则不清楚Date会去哪里，
<data>
<ArrayOfMatches>
<Date>2017-04-04</Date>
<Match>
<TimeSlot>17:00:00</TimeSlot>
<Game_Id>18</Game_Id>
<ArrayOfTeams>
<Team>
<Team_id>14</Team_id>
<Team_Name>New Zealand</Team_Name>
<TeamPlayers>
<Player>
<PlayerId>10</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
<Player>
<PlayerId>11</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
<Player>
<PlayerId>12</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
</TeamPlayers>
</Team>
<Team>
<Team_id>15</Team_id>
<Team_Name>New Zealand</Team_Name>
<TeamPlayers>
<Player>
<PlayerId>10</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
<Player>
<PlayerId>11</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
<Player>
<PlayerId>12</PlayerId>
<PlayerName>Trent</PlayerName>
<PlayerSurname>Boult</PlayerSurname>
</Player>
</TeamPlayers>
</Team>
</ArrayOfTeams>
</Match>
</ArrayOfMatches>
</data>
但你应该明白这一点）
现在您可以使用result了。
通过抓取诸如game_id，球队阵列，球员等属性。
public class data
{
public DateTime Date { get; set; }
public Match[] ArrayOfMatches { get; set; }
}
public class ArrayOfMatches
{
public Match[] Match { get; set; }
}
public class Match
{
public String TimeSlot { get; set; }
public int Game_Id { get; set; }
public Team[] ArrayOfTeams { get; set; }
}
public class Team {
public int Team_id { get; set; }
public string Team_Name { get; set; }
public Player[] TeamPlayers { get; set; }
}
public class Player
{
public int PlayerId { get; set; }
public string PlayerName { get; set; }
public string PlayerSurname { get; set; }
}
class Program
{
static void Main(string[] args)
{
using (FileStream fileStream = new FileStream(@"C:\temp\input.xml", FileMode.Open))
{
XmlSerializer serializer = new XmlSerializer(typeof(data));
data result = (data)serializer.Deserialize(fileStream);
}
}
}
答案 1 :(得分：0)
data wanted;
format collect $1000.;
retain collect;
set begin;
if id = bill then collect = catx("/",collect,product);
if id ~= bill then collect = catx("-",collect,product);
run;
这样一来，var dt = new DataTable();
dt.Columns.Add(new DataColumn("TimeSlot", typeof(string)));
dt.Columns.Add(new DataColumn("GameID", typeof(string)));
dt.Columns.Add(new DataColumn("TeamAID", typeof(string)));
dt.Columns.Add(new DataColumn("TeamAName", typeof(string)));
dt.Columns.Add(new DataColumn("TeamBID", typeof(string)));
dt.Columns.Add(new DataColumn("TeamBName", typeof(string)));
XDocument xdoc = XDocument.Load(@"c:/matchdata.xml");
var games = from i in xdoc.Descendants("Match")
select new //creates a few new anonymous types to store values to dump into the datatable
{
TimeSlot = (string)i.Element("TimeSlot"),
GameID = (string)i.Element("Game_Id"),
TeamAID = (string)i.Element("ArrayOfTeams").Element("Team").Element("Team_id"),
TeamAName = (string)i.Element("ArrayOfTeams").Element("Team").Element("Team_Name"),
TeamBID = (string)i.Element("ArrayOfTeams").Elements("Team").Skip(1).First().Element("Team_id"), //skips a node to collect info for the next team.
TeamBName = (string)i.Element("ArrayOfTeams").Elements("Team").Skip(1).First().Element("Team_Name") //skips a node to collect info for the next team.
};
foreach (var item in games)
{
dt.Rows.Add(item.TimeSlot, item.GameID, item.TeamAID, item.TeamAName, item.TeamBID, item.TeamBName); //adds a new row of data into the datatable for every <Match> </Match>
}
可以成为dt的来源，您可以看到结果。这可能不是最好的方法，但至少我已经学到了一些东西，对更好/更有效的评论将被研究。