Dictionary<string, List<ClassObj>> LINQ To XML 查询

Question

我尝试将我的 XML 文件放入带有通用列表的字典中。如何使用正确的键将正确的查询列表合并到我的字典中？代替.ToList() .ToDictionary 不行吗？

<?xml version="1.0"?>
<customers>
  <cust ID="1" DeviceID="1" Name="Bob" Latitude="10" Longitude="58" Device="HW1.0"> </cust>
  <cust ID="2" DeviceID="2" Name="Jack" Latitude="28" Longitude="20" Device="HW2.2"> </cust>
</customers>

//XML attribute Name is Dict Key

public class Customers
{
    public int Longitude { get; set; }
    public int Latitude { get; set; }
    public int DeviceID { get; set; }
    public string Device { get; set; }
}

class Program
{

    private static Dictionary<string, List<Customers>> ReadXmlToDict(string filename)
    {
        // Should be Key = Xml Attribute Name Value, List of class 
        Dictionary<string, List<Customers>> dict = new Dictionary<string, List<Customers>>();

        XDocument xdoc = XDocument.Load(filename);
        var querylist = (from row in xdoc.Descendants("cust")
                         select new Customers()
                         {
                             //Name = (string)row.Attribute("Name"),  // Wrong here should be the Dic key
                             DeviceID = (int)row.Attribute("DeviceID"), // list value
                             Longitude = (int)row.Attribute("Longitude"),  // list value
                             Latitude = (int)row.Attribute("Latitude"), // list value
                             Device = (string)row.Attribute("Device")  // list value
                         }).ToList();

        return null; // null for test To.List and not Dict
    }

Answer 1

这就是我将如何实现它，我认为它可以实现您正在寻找的东西。你有一个名为Customers 的类，然后想用一个键存储这些客户的列表……我不遵循这个逻辑。

我创建了一个名为Customer 的类，其中包含单个客户的信息。由于您要返回 Dictionary<string, Customer>，其中字符串是 xml 中的唯一属性 Name，因此您的字典的值不会是 List<Customer>。也许如果您有多个同名客户，您会使用它，但为什么不将密钥设置为（我假设）真正唯一的标识符，DeviceID？

namespace TestConsole
{
    class Customer
    {
        public int DeviceID;
        public int Longitude;
        public int Latitude;
        public string Device;
    }
    class Program
    {

        private static Dictionary<string, Customer> ReadXmlToDictionary(string filename)
        {
            var dict = new Dictionary<string, Customer>();

            var doc = XDocument.Load(@"C:\test.xml");

            dict = doc.Descendants("cust")
                .ToDictionary(
                    row => (string)row.Attribute("Name"),
                    row => new Customer {
                        Device = (string)row.Attribute("Device"),
                        DeviceID = (int)row.Attribute("DeviceID"),
                        Latitude = (int)row.Attribute("Latitude"),
                        Longitude = (int)row.Attribute("Longitude")
                });

            return dict;
        }

        static void Main(string[] args)
        {
            ReadXmlToDictionary(null);
        }
    }
}

编辑：认为与性能相关的答案很有趣，因此决定对这个单级 xml 进行尝试（使用 ID 作为唯一标识符）。结果如下：

1019 Descendants took 0.0030257 seconds.
1019 Elements took 0.0028348 seconds.
10000 Descendants took 0.0098942 seconds.
10000 Elements took 0.0101478 seconds.
100000 Descendants took 0.0873025 seconds.
100000 Elements took 0.1223577 seconds.

编辑： 创建 xsd 并从中生成一个类后，您可以这样使用它：

var parsed = XDocument.Parse(doc.ToString());

var serializer = new XmlSerializer(typeof(Xsds.customers));

var typedPayload = serializer.Deserialize(doc.Root.CreateReader());

var xmlAsClass = (TestConsole.Xsds.customers)typedPayload;

dict = xmlAsClass.cust
    .ToDictionary(
        row => (int)row.ID,
        row => new Customer {
            Device = row.Device,
            DeviceID = row.DeviceID,
            Latitude = row.Latitude,
            Longitude = row.Longitude,
            Name = row.Name
        });

Answer 2

您可以使用ToDictionary() 扩展方法轻松完成。但在性能方面，使用Elements() 方法比使用Descendants() 好得多；如需进一步阅读，请阅读此博客文章： WHY (OR WHEN) YOU SHOULD/SHOULDN’T USE DESCENDANTS() METHOD

您的查询将如下所示：

var customersDictionary = 
    xDoc.Root
        .Elements("cust")
        .ToDictionary(xe => 
                        xe.Attribute("Name").Value, xe => 
                        new Customers
                        {
                            DeviceID = (int)xe.Attribute("DeviceID"), 
                            Longitude = (int)xe.Attribute("Longitude"),  
                            Latitude = (int)xe.Attribute("Latitude"),
                            Device = (string)xe.Attribute("Device")
                        });