将 XML 文件中的数据排序到数据集中

Question

所以我有以下 xml 文件，我的 windows 表单是从网站获取的。

<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>

这只是整个事情的一小部分，这里的想法是两支队伍要面对面，分配一个 Game_Id

我目前的想法是将 Game_Id Team_Id 和 Team_Name 全部加载到一行数据集中，对于每个 Game_Id ( xml 文件中有不止一款游戏）。团队名称最终将以 Team1 V Team2 的形式出现在组合框中； then when one is selected the Player elements will be loaded into 2 separate list boxes.

我现在能做的是用 TimeSlot 和 Game_Id 填充不同的数据集，xml 文件中的所有 TeamPlayers 使用 DS.ReadXML(xmlRead)，但不是我想要的方式，并将 Game_Id 放入组合框中：

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，但不是我想要做的。

如有任何帮助，我们将不胜感激，我确实觉得我已经尽可能多地在此处提供了信息。

已编辑

<Date> </Date> 每个网络服务只出现一次，当天有多个 <Game_Id>。它用于确保使用了正确的 XML 文件。

已编辑

<data xmlns="">
    <ArrayOfMatches>
        <Match>...</Match>
        <Match>...</Match>
    </ArrayOfMatches>
</data>

已编辑

所以我仍然在为此苦苦挣扎。

+-----+-------+---------+------------+---------+-----------+
|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    |

这就是我想要的 DataTable

结构

现在我一直在玩 xml-linq 等，但我似乎不能那样布置它。

        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;

这可以编译，但不会在 gridview 中显示任何数据，也不会显示任何错误。

Answer 1

您可以将整个 XML 反序列化为一个对象并轻松使用它（我从你的 xml 代码片段中删除了第一行，匹配项是一个条目，因为如果有多个匹配项，则不清楚日期将去哪里，

<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>

但你应该明白了）

现在您可以使用 result。通过获取 game_id、球队数组、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);
            }

        }
    }

Answer 2

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>
        }

如此有效，dt可以成为 gridview 的来源，您可以看到结果。这可能不是最好的方法，但至少我学到了一些东西，将查看对 better/more 高效的评论。

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