LINQ比較エラー

Question

LINQを使用してレコードを更新しようとすると、比較エラーメッセージが表示されます。

var tools = from tl in MVCWebsite.MvcApplication.DataContext.tblTools 
               where tl.pk_fk_Environment == model.Environment 
               && tl.pk_fk_Classification == model.Classification 
               && tl.pk_ToolNumber == model.ToolNumber 
               select tl; 
     Models.Database.tblTool updatingTool = tools.First(); 
     //try 
     //{ 

     if (createPerson(model.ToolEngineer, true) != null) 
      updatingTool.fk_Engineer = model.ToolEngineer; 
     else 
      updatingTool.fk_Engineer = null; 

     updatingTool.fk_BuiltBy = model.ToolBuiltBy; 

     if (createPerson(model.ToolBuiltBy, true) != null) 
      updatingTool.fk_BuiltBy = model.ToolBuiltBy; 
     else 
      updatingTool.fk_BuiltBy = null; 
     if (createPerson(model.ToolDesignedBy, true) != null) 
      updatingTool.fk_Designer = model.ToolDesignedBy; 
     else 
      updatingTool.fk_Designer = null; 
     updatingTool.DateOfBuild = model.DateOfBuild; 
     updatingTool.Machine_Name_Primary = model.ToolPrimaryMachineName; 
     updatingTool.Machine_Description_Primary = model.ToolPrimaryMachineDescription; 
     updatingTool.Machine_Name_Secondary = model.ToolSecondaryMachineName; 
     updatingTool.Machine_Description_Secondary = model.ToolSecondaryMachineDescription; 
     updatingTool.MERNumber = model.MERNumber; 
     updatingTool.AssetNumber = model.AssetNumber; 
     updatingTool.Additional_Cavities = model.AdditionalFields.Cavities; 
     updatingTool.Additional_Gate = model.AdditionalFields.TypeOfGate; 
     updatingTool.Additional_Shrinkage = model.AdditionalFields.Shrinkage; 
     updatingTool.Additional_DieClearance = model.AdditionalFields.DieClearance; 
     updatingTool.Additional_Field1 = model.AdditionalFields.Additional1; 
     updatingTool.Additional_Field2 = model.AdditionalFields.Additional2; 
     updatingTool.Additional_Field3 = model.AdditionalFields.Additional3; 
     updatingTool.Additional_Field4 = model.AdditionalFields.Additional4; 
     updatingTool.Additional_OtherInformation = model.AdditionalFields.OtherInformation; 

（以下作成ユーザー関数がある）

private Models.Database.tblPerson createPerson(string user) 
    { 
     if (user == null || user == "") 
      return null; 
     var people = from p in MVCWebsite.MvcApplication.DataContext.tblPersons 
        where p.pk_PersonID == user 
        select p; 
     if (people.Count() == 1) 
     { 
      return people.First(); 
     } 
     else 
     { 
      Models.UserInformation ui; 
      ui = MVCWebsite.MvcApplication.DisplayUser(user); 
      if (!ui.Exists) 
       return null; 
      tblPerson person = new tblPerson() 
      { 
       pk_PersonID = user, 
       FirstName = ui.GivenName, 
       LastName = ui.Surname, 
       Email = ui.EmailAddress 
      }; 
      return person; 
     } 
    } 

私は私のコードは、現時点では、クリーンであるかどうか、私はこのエラーを取得していますし、それを困らせるために開始される程度悩まされていないです私。

CREATE TABLE [dbo].[tblTool](
    [pk_fk_Environment] [varchar](2) COLLATE Latin1_General_CI_AS NOT NULL, 
    [pk_fk_Classification] [varchar](3) COLLATE Latin1_General_CI_AS NOT NULL, 
    [pk_fk_Style] [varchar](4) COLLATE Latin1_General_CI_AS NULL, 
    [pk_ToolNumber] [int] NOT NULL, 
    [DateOfBuild] [datetime] NULL, 
    [fk_Engineer] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_Designer] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_BuiltBy] [varchar](20) COLLATE Latin1_General_CI_AS NULL, 
    [MERNumber] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [AssetNumber] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_Material] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_LocationCompany] [varchar](20) COLLATE Latin1_General_CI_AS NULL, 
    [fk_LocationCountry] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_LocationRegion] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [fk_Status] [varchar](10) COLLATE Latin1_General_CI_AS NULL, 
    [Machine_Name_Primary] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Machine_Description_Primary] [text] COLLATE Latin1_General_CI_AS NULL, 
    [Machine_Name_Secondary] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Machine_Description_Secondary] [text] COLLATE Latin1_General_CI_AS NULL, 
    [OldToolID] [varchar](100) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Cavities] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Gate] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Shrinkage] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_DieClearance] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Field1] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Field2] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Field3] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_Field4] [varchar](50) COLLATE Latin1_General_CI_AS NULL, 
    [Additional_OtherInformation] [text] COLLATE Latin1_General_CI_AS NULL, 
CONSTRAINT [PK_tblTool] PRIMARY KEY CLUSTERED 
(
    [pk_fk_Environment] ASC, 
    [pk_fk_Classification] ASC, 
    [pk_ToolNumber] ASC 
)WITH (PAD_INDEX = OFF, IGNORE_DUP_KEY = OFF) ON [PRIMARY] 
) ON [PRIMARY] TEXTIMAGE_ON [PRIMARY] 

それに既にある同じ情報でフィールドを更新するとき、私はエラーを取得：

をここで

は、メインテーブルスクリプトを作成します（あなたは、フィールドの設定を見ることができるようですこれが起こっされる可能性がありますなぜ

SQL Server does not handle comparison of NText, Text, Xml, or Image data types. 

は誰もが知っていますか？

おかげで、 オリバー

Answer 1

は VARCHAR(MAX)データ型には非推奨 TEXTデータ型からあなた Additional_OtherInformation、 Machine_Description_Primary、および Machine_Description_Secondary列を変更

。列のマッピングを更新すると、すべての設定が完了しているはずです。