通过实体框架使用现有的存储过程为Oracle数据库生成身份 [英] Generate identity for an Oracle database through Entity Framework using an exisiting stored procedure

问题描述

如何通过Entity Framework自动生成Oracle数据库的身份？

How to automatically generate identity for an Oracle database through Entity Framework?

我有一个可以调用并生成不在上下文中的列的函数我通过Entity Framework显式调用存储过程吗？我正在使用存储库模式。

I have a function that I could call and generate the column which is not in the context how do I explicitly call the stored procedure through Entity Framework? I am using a repository pattern.

随机数生成器插入记录（我通过UDF获取主键，并将其传递给要插入的实体） / p>

Random Number generator to insert a record (where I get the primary key through a UDF and pass this to the entity to insert).

推荐答案

1）在Oracle中创建序列

 CREATE SEQUENCE dummy_test_seq
  MINVALUE 1
  MAXVALUE 999999999999999999999999999
  START WITH 1
  INCREMENT BY 1;

2）创建属性

   sealed public class CommonUtilities
    {
      #region Sequences
       public static int DummyTestSeq
        {
         get
          {              
            using (Entities ctx = new Entities()) 
             { 
               return Convert.ToInt32(ctx.Database.SqlQuery<decimal>("SELECT dummy_test_seq.NEXTVAL FROM DUAL").ToList().Single()); 
              }  
            }
         }
    #endregion
}

3）获取序列

   public int InsertTable1()
    {
      using (Entities ctx = new Entities())
        {
            ctx.tabel1.Add(new tabel1()
            {
                SEQ = CommonUtilities.DummyTestSeq,
                Date= DateTime.Now
            });
            return ctx.SaveChanges();
        }
     }

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