Question

我有三个相关的表格：

Employee(EmployeeId, EmployeeName)

Skill(SkillId, SkillName)

EmployeeSkill(EmployeSkillId, EmployeeId, SkillId)

EmployeSkillId是一种身份。

数据库中的行如下：

员工表：

EmployeeId | EmployeeNumber | EmployeeName
---------- | -------------- | ------------
        1  |         10015  |  John Doe

技能表：

SkillId | SkillName
------- | ---------
     1  |  .NET
     2  |  SQL
     3  |  OOD
     4  |  Leadership

EmployeeSkill表：

EmployeeSkillId | EmployeeId | SkillId
--------------- | ---------- | -------
             1  |         1  |      1
             2  |         1  |      2
             3  |         1  |      3
             4  |         1  |      4

如果我的员工在EmployeSkill注册了三项技能，我希望能够得到如下结果：

John Doe, "Skill-1, Skill2, Skill-3"

即，将该员工的技能名称连接成一个字符串。

我尝试了以下操作，但它无效。

var query = from emp in Employee.All()
            from es in emp.EmployeeSkills
            join sk in Skill.All() on es.SkillId equals sk.SkillId
            group sk by new {emp.EmployeeName} into g
            select new TestEntity
            {
                Name = g.Key.EmployeeName,
                Skills = g.Aggregate(new StringBuilder(),
                                     (sb, grp_row) => sb.Append(grp_row.SkillName))
                          .ToString()
            };

技能名称的汇总列表将返回空白。我怎么能这样做？

Answer 1

听起来你可以将联接作为选择的一部分：

var query = from emp in Employee.All()
            select new TestEntity {
                Name = emp.EmployeeName,
                Skills = string.Join(", ", 
                       (from es in emp.EmployeeSkills
                        join sk in Skill.All() on es.SkillId equals sk.SkillId
                        select sk.SkillName)) };

现在，这将分别为每个人单独进行连接，这不是非常有效。另一种选择是首先建立从技能ID到技能名称的映射：

var skillMap = Skill.All().ToDictionary(sk => sk.SkillId,
                                        sk => sk.SkillName);

然后主要查询很简单：

var query = from emp in Employee.All()
            select new TestEntity {
                Name = emp.EmployeeName,
                Skills = string.Join(", ",
                       emp.EmployeeSkills.Select(sk => skillMap[sk.SkillId]))};

最终有很多方法可以给这只猫上皮 - 例如，如果你想坚持原来的方法，这仍然是可行的。我会这样做：

var query = from emp in Employee.All()
            from es in emp.EmployeeSkills
            join sk in Skill.All() on es.SkillId equals sk.SkillId
            group sk.SkillName by emp into g
            select new TestEntity
            {
                Name = g.Key.EmployeeName,
                Skills = string.Join(", ", g)
            };

此时它与原始查询非常相似，当然只使用string.Join而不是Aggregate。如果所有这三种方法都带有空技能列表，那么我怀疑你的数据有问题。对我来说，为什么你的第一个查询会“成功”但是有一个空的技能列表并不明显。

编辑：好的，这是一个简短的（-ish）但完整的工作示例：

using System;
using System.Collections.Generic;
using System.Linq;

public class Employee
{
    public int EmployeeId { get; set; }
    public string EmployeeName { get; set; }

    public static List<Employee> All { get; set; }

    public IEnumerable<EmployeeSkill> EmployeeSkills
    {
        get 
        { 
            return EmployeeSkill.All
                           .Where(x => x.EmployeeId == EmployeeId);
        }
    }
}

public class Skill
{
    public string SkillName { get; set; }
    public int SkillId { get; set; }

    public static List<Skill> All { get; set; }
}

public class EmployeeSkill
{
    public int SkillId { get; set; }
    public int EmployeeId { get; set; }

    public static List<EmployeeSkill> All { get; set; }
}

class Test
{
    static void Main()
    {
        Skill.All = new List<Skill>
        {
            new Skill { SkillName = "C#", SkillId = 1},
            new Skill { SkillName = "Java", SkillId = 2},
            new Skill { SkillName = "C++", SkillId = 3},
        };

        Employee.All = new List<Employee>
        {
            new Employee { EmployeeName = "Fred", EmployeeId = 1 },
            new Employee { EmployeeName = "Ginger", EmployeeId = 2 },
        };

        EmployeeSkill.All = new List<EmployeeSkill>
        {
            new EmployeeSkill { SkillId = 1, EmployeeId = 1 },
            new EmployeeSkill { SkillId = 2, EmployeeId = 1 },
            new EmployeeSkill { SkillId = 2, EmployeeId = 2 },
            new EmployeeSkill { SkillId = 3, EmployeeId = 2 },
        };

        var query = from emp in Employee.All
            from es in emp.EmployeeSkills
            join sk in Skill.All on es.SkillId equals sk.SkillId
            group sk.SkillName by emp.EmployeeName into g
            select new
            {
                Name = g.Key,
                Skills = string.Join(", ", g)
            };

        foreach (var result in query)
        {
            Console.WriteLine(result);
        }
    }
}

结果：

{ Name = Fred, Skills = C#, Java }
{ Name = Ginger, Skills = Java, C++ }

如何聚合连接？

员工表：

技能表：

EmployeeSkill表：

1 个答案: