2017-08-07 111 views
1

我有一个目标:小巧玲珑多映射

public class Species : IEntity<int> 
{ 
    public int Id { get; set; } 
    public string Name { get; set; } 
    public SpeciesCategory Category { get; set; } 
    public WetlandIndicator WetlandIndicator { get; set; } 
} 

public class SpeciesCategory : IEntity<int> 
{ 
    public int Id { get; set; } 
    public string Name { get; set; } 
} 

public class WetlandIndicator : IEntity<string> 
{ 
    public string Id { get; set; } 
    public string Designation { get; set; } 
    public bool Status { get; set; } 
} 

然而,当我用小巧玲珑来调用下面的查询:

SELECT 
    [Species].*, 
    [SpeciesType].*, 
    [WetlandIndicator].Code AS Id, 
    [WetlandIndicator].Designation 
FROM 
    ((([Watershed].[Vegetation].[Species] INNER JOIN [Vegetation].[SpeciesCategory] 
     ON [Watershed].[Vegetation].[Species].[SpeciesCategoryId] = [Vegetation].[SpeciesCategory].[Id]) INNER JOIN [Watershed].[Vegetation].[SpeciesType] 
     ON [Watershed].[Vegetation].[Species].[SpeciesTypeId] = [Vegetation].[SpeciesType].[Id]) INNER JOIN [Watershed].[Vegetation].[WetlandIndicator] 
     ON [Vegetation].[Species].[WetlandIndicatorCode] = [Vegetation].[WetlandIndicator].[Code]) 

我收到使用多映射时确保,请确保您使用splitOn属性。我是谁,但我仍然收到错误。所以我假设我有一些类型的使用错误或合成文本错误。 ,不断的错误是因为代码如下:

public async Task<IEnumerable<SpeciesDomain>> GetAllSpecies(string query) => 
     await dbConnection.QueryAsync<Species, SpeciesCategory, WetlandIndicator, SpeciesDomain>(query, (species, speciesCategory, wetlandIndicator) => 
     { 
      species.SpeciesCategory = speciesCategory; 
      species.WetlandIndicator = wetlandIndicator; 
      return species; 
     }, splitOn: "Id, Code"); 

重要提示:默认情况下小巧玲珑利用标识,这就是为什么我改名守则ID,但即使有代码或重命名我仍然收到了多映射错误。

回答

2

看起来你只需要从splitOn中删除'代码':“Id,Code”。您的查询将其重命名为“Id”。

Dapper还使用“Id”作为默认值,因此不需要指定。

Dapper能够通过假设 您的Id列被命名为Id或id来拆分返回的行。如果您的主键不同于 或者您想要在除Id以外的其他位置拆分该行,请使用 可选splitOn参数。

下面是一个快速测试来验证:

using (var conn = new SqlConnection(@"Data Source=.\sqlexpress;Integrated Security=true; Initial Catalog=foo")) 
{ 
    var result = conn.Query<Species, SpeciesCategory, WetlandIndicator, Species>(
     "select Id = 11, Name = 'Foo', Id = 22, Name = 'Bar', Id = 33, Designation = 'House Cat' ", 
     (species, speciesCategory, wetlandIndicator) => 
    { 
     species.Category = speciesCategory; 
     species.WetlandIndicator = wetlandIndicator; 
     return species; 
    }).First(); 

    Assert.That(result.Id, Is.EqualTo(11)); 

    Assert.That(result.Category.Id, Is.EqualTo(22)); 
    Assert.That(result.Category.Name, Is.EqualTo("Bar")); 

    Assert.That(result.WetlandIndicator.Id, Is.EqualTo(33)); 
    Assert.That(result.WetlandIndicator.Designation, Is.EqualTo("House Cat")); 
} 

更新演示不同领域和类型

public class Species 
{ 
    public int Id { get; set; } 
    public string Name { get; set; } 
    public SpeciesCategory Category { get; set; } 
    public WetlandIndicator WetlandIndicator { get; set; } 
} 

public class SpeciesCategory 
{ 
    public int Id { get; set; } 
    public string Name { get; set; } 
} 

public class WetlandIndicator 
{ 
    public string Code { get; set; } 
    public string Designation { get; set; } 
    public bool Status { get; set; } 
} 

using (var conn = new SqlConnection(@"Data Source=.\sqlexpress;Integrated Security=true; Initial Catalog=foo")) 
{ 
    var result = conn.Query<Species, SpeciesCategory, WetlandIndicator, Species>(
     "select Id = 11, Name = 'Foo', Id = 22, Name = 'Bar', Code = 'X', Designation = 'House Cat' ", 
     (species, speciesCategory, wetlandIndicator) => 
    { 
     species.Category = speciesCategory; 
     species.WetlandIndicator = wetlandIndicator; 
     return species; 
    }, splitOn: "Id, Code").First(); 

    Assert.That(result.Id, Is.EqualTo(11)); 

    Assert.That(result.Category.Id, Is.EqualTo(22)); 
    Assert.That(result.Category.Name, Is.EqualTo("Bar")); 

    Assert.That(result.WetlandIndicator.Code, Is.EqualTo("X")); 
    Assert.That(result.WetlandIndicator.Designation, Is.EqualTo("House Cat")); 
} 
0

分裂所以,我发现它失败的主要原因。 Dapper不喜欢splitOn参数在intstring之间交替。通过迫使他们统一起来,它就起作用了。我注意到的另一个项目是,如果你有一个名为Code的列,例如镜像一个主键但SQL没有设置为关系标识符,它也会出错。

经过纠正后,没有问题。

+0

只是为了澄清他人:如果您使用不同数据类型的字段进行拆分,Dapper不会失败。我会更新我的演示答案。 –

+0

我需要澄清,如果你有字段:'代码'和一个名为'代码'问题的Id列。 – Greg