Question

<?xml version="1.0" standalone="yes"?>
<CompanyInfo>
     <Employee name="Jon" deptId="123">
      <Region name="West">
        <Area code="96" />
      </Region>
      <Region name="East">
        <Area code="88" />
      </Region>
     </Employee>
</CompanyInfo>  

public class Employee
{
    public string EmployeeName { get; set; }
    public string DeptId { get; set; }
    public List<string> RegionList {get; set;}
}

public class Region
{
    public string RegionName { get; set; }
    public string AreaCode { get; set; }
}

我正在尝试阅读这些XML数据，到目前为止我已经尝试过这个：

XDocument xml = XDocument.Load(@"C:\data.xml");
var xElement = xml.Element("CompanyInfo");
if (xElement != null)
    foreach (var child in xElement.Elements())
    {
        Console.WriteLine(child.Name);  
        foreach (var item in child.Attributes())
        {
            Console.WriteLine(item.Name + ": " + item.Value);
        }

        foreach (var childElement in child.Elements())
        {
            Console.WriteLine("--->" + childElement.Name);
            foreach (var ds in childElement.Attributes())
            {
                Console.WriteLine(ds.Name + ": " + ds.Value);
            }
            foreach (var element in childElement.Elements())
            {
                Console.WriteLine("------->" + element.Name);
                foreach (var ds in element.Attributes())
                {
                    Console.WriteLine(ds.Name + ": " + ds.Value);
                }
            }
        }                
    }

这使我能够获取每个节点，其属性名称和值，因此我可以将这些数据保存到数据库中的相关字段中，但这似乎是一个漫长的方式和不灵活，例如，如果XML结构改变所有那些foreach语句需要重新访问，也很难以这种方式过滤数据，我需要编写一些if语句来过滤数据（例如，只从West获取员工等等）

我正在寻找一种更灵活的方式，使用linq，类似这样：

List<Employees> employees =
              (from employee in xml.Descendants("CompanyInfo")
               select new employee
               {
                   EmployeeName = employee.Element("employee").Value,
                   EmployeeDeptId = ?? get data,
                   RegionName = ?? get data,
                   AreaCode = ?? get data,,
               }).ToList<Employee>();

但我不确定如何从子节点获取值并应用过滤（仅限某些员工）。这可能吗？任何帮助表示赞赏。

由于

Answer 1

var employees = (from e in xml.Root.Elements("Employee")
                 let r = e.Element("Region")
                 where (string)r.Attribute("name") == "West"
                 select new Employee
                 {
                     EmployeeName = (string)e.Attribute("employee"),
                     EmployeeDeptId = (string)e.Attribute("deptId"),
                     RegionName = (string)r.Attribute("name"),
                     AreaCode = (string)r.Element("Area").Attribute("code"),
                 }).ToList();

但是当XML文件结构发生变化时，它仍然需要进行查询修订。

修改的

每位员工查询多个地区：

var employees = (from e in xml.Root.Elements("Employee") select new Employee { EmployeeName = (string)e.Attribute("employee"), DeptId = (string)e.Attribute("deptId"), RegionList = e.Elements("Region") .Select(r => new Region { RegionName = (string)r.Attribute("name"), AreaCode = (string)r.Element("Area").Attribute("code") }).ToList() }).ToList();

然后，您可以过滤来自特定区域的员工的列表：

var westEmployees = employees.Where(x => x.RegionList.Any(r => r.RegionName == "West")).ToList();

Answer 2

您可以跟踪结构：

from employee in xml
      .Element("CompanyInfo")       // must be root
      .Elements("Employee")         // only directly children of CompanyInfo

或不太严格

from employee in xml.Descendants("Employee")    // all employees at any level

然后获取您想要的信息：

       select new Employee
       {
           EmployeeName = employee.Attribute("name").Value,
           EmployeeDeptId = employee.Attribute("deptId").Value,
           RegionName = employee.Element("Region").Attribute("name").Value,
           AreaCode = employee.Element("Region").Element("Area").Attribute("code").Value,
       }

如果有关于多个区域的其他信息，请假设List<Region> Regions属性：

       select new Employee
       {
           EmployeeName = employee.Attribute("name").Value,
           EmployeeDeptId = employee.Attribute("deptId").Value,
           //RegionName = employee.Element("Region").Attribute("name").Value,
           //AreaCode = employee.Element("Region").Element("Area").Attribute("code").Value,
           Regions = (from r in employee.Elements("Region") select new Region 
                      {
                         Name = r.Attribute("name").Value,
                         Code = r.Element("Area").Attribute("code").Value,
                      }).ToList();
       }

Answer 3

您可以在一个查询中进行选择，然后在第二个中进行过滤，或者将它们组合到一个查询中：

两个问题：

        // do te transformation
        var employees =
          from employee in xml.Descendants("CompanyInfo").Elements("Employee")
          select new
          {
              EmployeeName = employee.Attribute("name").Value,
              EmployeeDeptId = employee.Attribute("deptId").Value,
              Regions = from region in employee.Elements("Region")
                        select new
                            {
                                Name = region.Attribute("name").Value,
                                AreaCode = region.Element("Area").Attribute("code").Value,
                            }
          };

        // now do the filtering
        var filteredEmployees = from employee in employees
                                from region in employee.Regions
                                where region.AreaCode == "96"
                                select employee;

组合一个查询（相同输出）：

          var employees2 =
          from selectedEmployee2 in
              from employee in xml.Descendants("CompanyInfo").Elements("Employee")
              select new
              {
                  EmployeeName = employee.Attribute("name").Value,
                  EmployeeDeptId = employee.Attribute("deptId").Value,
                  Regions = from region in employee.Elements("Region")
                            select new
                                {
                                    Name = region.Attribute("name").Value,
                                    AreaCode = region.Element("Area").Attribute("code").Value,
                                }
              }
          from region in selectedEmployee2.Regions
          where region.AreaCode == "96"
          select selectedEmployee2;

但是你应该考虑添加一件小事。为了提高稳健性，您需要检查元素和属性的存在，然后选择将如下所示：

 var employees =
          from employee in xml.Descendants("CompanyInfo").Elements("Employee")
          select new
          {
              EmployeeName = (employee.Attribute("name") != null) ? employee.Attribute("name").Value : string.Empty,
              EmployeeDeptId = (employee.Attribute("deptId") != null) ? employee.Attribute("deptId").Value : string.Empty,
              Regions = (employee.Elements("Region") != null)?
                        from region in employee.Elements("Region")
                        select new
                            {
                                Name = (region.Attribute("name")!= null) ? region.Attribute("name").Value : string.Empty,
                                AreaCode = (region.Element("Area") != null && region.Element("Area").Attribute("code") != null) ? region.Element("Area").Attribute("code").Value : string.Empty,
                            }
                        : null
          };

使用LINQ解析XML以获取子元素

3 个答案: