使用多映射API时,如果您具有Id“,”splitOn以外的键,请确保设置splitOn参数

dapper

我正在尝试使用dapper的多映射功能来返回相册和相关艺术家和流派的列表。

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");

我已经检查过有关这方面的解决方案,但是没有工作。谁能告诉我哪里出错了?

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");

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

热门答案

您需要在选择查询中包含要拆分的列。您只需选择所有其他属性 - 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,您将必须拆分GenreId和ArtistId。

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

管道用于启动您要映射的新POCO。因此, SplitOn参数将是GenreId和ArtistId。

编辑2:问题是你的POCO Album 。您将ArtistIdGenreId指定为属性,但它们基本上属于各自的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; }
    }

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GenreIdArtistId ,因为您在Albums引用了这些对象。



许可下: CC-BY-SA with attribution
不隶属于 Stack Overflow
这个KB合法吗? 是的,了解原因
许可下: CC-BY-SA with attribution
不隶属于 Stack Overflow
这个KB合法吗? 是的,了解原因