41

我正在尝试使用 dapper 的多重映射功能来返回专辑列表以及相关的艺术家和流派。

public class Artist
{
public virtual int ArtistId { get; set; }
public virtual string Name { get; set; }
}    


public class Genre
{
public virtual int GenreId { get; set; }
public virtual string Name { get; set; }
public virtual string Description { get; set; }
}


public class Album
{
public virtual int AlbumId { get; set; }
public virtual int GenreId { get; set; }
public virtual int ArtistId { get; set; }
public virtual string Title { get; set; }
public virtual decimal Price { get; set; }
public virtual string AlbumArtUrl { get; set; }
public virtual Genre Genre { get; set; }
public virtual Artist Artist { get; set; }
}


var query = @"SELECT AL.Title, AL.Price, AL.AlbumArtUrl, GE.Name, GE.[Description], AR.Name FROM Album AL INNER JOIN Genre GE ON AL.GenreId = GE.GenreId INNER JOIN Artist AR ON AL.ArtistId = AL.ArtistId";

var res = connection.Query<Album, Genre, Artist, Album>(query, (album, genre, artist) => { album.Genre = genre; album.Artist = artist; return album; }, commandType: CommandType.Text, splitOn: "ArtistId, GenreId");

我已经检查了有关此的解决方案，但没有一个有效。谁能让我知道我哪里出错了？

谢谢@Alex :) 但我仍然感到震惊。这就是我所做的：

CREATE TABLE Artist
(
ArtistId INT PRIMARY KEY IDENTITY(1,1)
,Name VARCHAR(50)
)

CREATE TABLE Genre
(
    GenreId INT PRIMARY KEY IDENTITY(1,1)
    ,Name VARCHAR(20)
    ,[Description] VARCHAR(1000)
)

CREATE TABLE Album
(
    AlbumId INT PRIMARY KEY IDENTITY(1,1)
    ,GenreId INT FOREIGN KEY REFERENCES Genre(GenreId)
    ,ArtistId INT FOREIGN KEY REFERENCES Artist(ArtistId)
    ,Title VARCHAR(100)
    ,Price FLOAT
    ,AlbumArtUrl VARCHAR(300) 
)

INSERT INTO Artist(Name) VALUES ('Jayant')
INSERT INTO Genre(Name,[Description]) VALUES ('Rock','Originally created during school days. The year was.....I guess 1998')
DECLARE @gen_id INT
        ,@art_id INT
SET @gen_id = (SELECT MAX(GenreId) FROM Genre)
SET @art_id = (SELECT MAX(ArtistId) FROM Artist)
INSERT INTO Album(GenreId,ArtistId,Title,Price,AlbumArtUrl) VALUES (@gen_id,@art_id,'I go mad for you',200,'http://asha4u.com/IGoMad')

根据您的建议，我将查询更改为：

var query = @"SELECT AL.AlbumId, AL.Title, AL.Price, AL.AlbumArtUrl, GE.GenreId, GE.Name, GE.Description, AR.ArtistId, AR.Name FROM Album AL INNER JOIN Artist AR ON AR.ArtistId = AL.ArtistId INNER JOIN Genre GE ON GE.GenreId = AL.GenreId";

var res = connection.Query<Album, Genre, Artist, Album>(query, (album, genre, artist) => { album.Genre = genre; album.Artist = artist; return album; }, commandType: CommandType.Text, splitOn: "GenreId, ArtistId");

现在我对 GenreId 和 ArtistId 使用 splitOn。我仍然得到同样的错误。请帮忙。

4

3 回答 3

55

您需要在选择查询中包含要拆分的列。您的只是选择所有其他属性-因此Dapper找不到匹配的列来拆分对象。

您的查询可能应该是这样的：

var query = @"SELECT AlbumId, Title, Price, AlbumArtUrl, GenreId, Name, Description , ArtistId, Name ......" etc

Sam 为多映射和 splitOn 选项写了一个很好的答案： https ://stackoverflow.com/a/7478958/1028323

编辑：如果您的查询如上所述，则必须拆分 GenreId 和 ArtistId。

 AlbumId, Title, Price, AlbumArtUrl | GenreId, Name, Description | ArtistId, Name

这些管道用于您尝试映射的新 POCO 的开始。所以SplitOn参数将是 GenreId 和 ArtistId。

Edit2：问题是你的 POCO Album。您将ArtistIdand指定GenreId为属性，但它们基本上属于它们各自的POCO's.

public class Album
    {
        public virtual int AlbumId { get; set; }
        public virtual string Title { get; set; }
        public virtual decimal Price { get; set; }
        public virtual string AlbumArtUrl { get; set; }
        public virtual Genre Genre { get; set; }
        public virtual Artist Artist { get; set; }
    }

和

var sql = @"SELECT AL.AlbumId
                 , AL.Title
                 , AL.Price
                 , AL.AlbumArtUrl
                 , GE.GenreId
                 , GE.Name
                 , GE.Description
                 , AR.ArtistId
                 , AR.Name 
            FROM Album AL 
      INNER JOIN Artist AR ON AR.ArtistId = AL.ArtistId 
      INNER JOIN Genre GE ON GE.GenreId = AL.GenreId";

using (var conn = connFactory.OpenConnection())
{
    var res = conn.Query<Album, Genre, Artist, Album>(sql, (album, genre, artist) =>
    {
        album.Genre = genre;
        album.Artist = artist;
        return album;
    }, splitOn: "GenreId,ArtistId");
}

应该做的伎俩。无论如何，您不需要GenreId，ArtistId因为您在Albums.

于 2013-06-18T12:54:49.160 回答
3

我遇到了同样的问题。这是技巧和示例。

public abstract class BaseEntity
{
    [Key]
    public int Id { get; set; }
}

public class Category : BaseEntity
{
    public string Name { get; set; }
}


public class Status : BaseEntity
{
    public string Name { get; set; }
}

public class User : BaseEntity
{
    public string Name { get; set; }
    public string Surname { get; set; }
    public bool Active { get; set; }
}


public class TodoItem : BaseEntity
{
    public string Title { get; set; }
    public string Message { get; set; }
    public Status Status { get; set; }
    public Category Category { get; set; }
    public User User { get; set; }
    public DateTime CreatedOn { get; set; }
}

使用

            string sql = @"select 
                          t.Id,
                          t.Title,
                          t.Message,
                          t.CreatedOn,
                          s.Id as Id,
                          s.Name,

                          c.Id as Id,
                          c.Name,

                          u.Id as Id,
                          u.Name,
                          u.Surname,
                          u.Active

                         from ToDoItem t
                        inner join [Status] s on (t.StatusId = s.Id)
                        inner join [Category] c on (t.CategoryId = c.Id)
                        inner join [User] u on (t.AssignUserId = u.Id)";
            var result = connection.Query<TodoItem, Status, Category, User, TodoItem>
                (sql, (todoItem, status, category, user) =>
            {
                todoItem.Status = status;
                todoItem.Category = category;
                todoItem.User = user;
                return todoItem;
            },splitOn: "Id,Id,Id,Id");

这是技巧splitOn：“Id，Id，Id，Id”

于 2019-10-14T12:31:45.987 回答
0

我也有这个问题，发现问题是我没有为选择中的每个表选择 .Id 列

于 2021-09-16T18:08:37.133 回答