用于联接三个表的 Lambda 表达式

本文关键字：三个 Lambda 表达式 用于 | 更新日期: 2023-09-27 18:37:27

>员工表 

EmpID
Emp_First_Name
Manager_ID
Department_ID
RoleID

部门表

Department_ID
Department_N

阿梅

角色表

RoleID
RoleName

连接三个表

var join = from u in db.TBL_Employees
                       join v in db.TBL_Departments
                       on u.Department_ID equals v.Department_ID
                       join x in db.TBL_Employees
                       on u.Manager_ID equals x.Emp_ID
                       join z in db.TBL_Roles
                       on u.RoleID equals z.RoleID
                       select new
                       {
                           Name = u.Emp_First_Name,
                           Department = v.Department_Name,
                           Manager = x.Emp_First_Name,
                           Role = z.RoleName
                       };

此查询工作正常。但是我想在 Lambda 表达式中编写相同的查询。如何使用 Lambda 表达式显示相同的输出？

用于联接三个表的 Lambda 表达式

你可以这样做作为连续的Joins，每个都用新的连接继续前面的投影，但正如注释中提到的，这很快就会变得混乱且难以遵循： （特别是如果您习惯于使用 Sql 连接语法）

var result = db.TBL_Employees
   .Join(db.TBL_Departments, u => u.Department_ID, v => v.Department_ID, 
         (u, v) => new {Employee = u, Department = v})
   .Join(db.TBL_Employees, ed => ed.Employee.Manager_ID, x => x.Emp_ID, 
         (ed, x) => new {EmployeeDepartment = ed, Manager = x})
   .Join(db.TBL_Roles, edm => edm.EmployeeDepartment.Employee.RoleID, z => z.RoleID, 
         (edm, z) => new {EmployeeDepartmentManager = edm, Role = z})
.Select(edmr => new
{
  Name = edmr.EmployeeDepartmentManager.EmployeeDepartment.Employee.Emp_First_Name,
  Department = edmr.EmployeeDepartmentManager.EmployeeDepartment.Department.Department_Name,
  Manager = edmr.EmployeeDepartmentManager.Manager.Emp_First_Name,
  Role = edmr.Role.RoleName
});

（我保留了您的原始别名以实现可追溯性，并使用您的模式为中间匿名投影添加了新别名，例如 edmr是EmployeeDepartmentManagerRole）

但是，我建议您确保强制执行表中似乎存在的外键关系，然后在 Linq2Sql DBML 模型中将这些关系作为可导航关系进行拉取。启用延迟加载或设置适当的预先加载LoadsWith DataContext 选项后，您将能够将查询和投影简化为：

var result = db.TBL_Employees
   .Select(e => new
{      
    Name = e.Emp_First_Name,
    Department = e.Department.Department_Name,
    Manager = e.Manager.Emp_First_Name,
    Role = e.Role.RoleName
});
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