实体框架代码首先为映射到相同类的不同表返回相同数据

Question

I have a class Timer which I want to use with different tables that have the same structure so I am passing in the table name.

public class TimerContext : DbContext
{
    public DbSet<Timer> Timers { get; set; }

    private readonly string _tableName;

    public TimerContext(string tableName) : base("name=fooDb")
    {
        _tableName = tableName;
    }

    protected override void OnModelCreating(DbModelBuilder modelBuilder)
    {
        modelBuilder.Entity<Timer>().ToTable(_tableName);

        base.OnModelCreating(modelBuilder);
    }
}

However when I pass in two different table names they return the same data. prevTimers contains the exact same data as currTimers . How do I get the unique data from each table? Why do I get the same data for two different tables?

var currTimers = new TimerContext(currentTimerTableName).Timers.ToList();
var prevTimers = new TimerContext(previousTimerTableName).Timers.ToList();

Answer 1

EF will call the OnModelCreating method to create an in-memory copy of the model as soon as it is needed. Once it has done this, this copy will be used. In your case, the code to generate prevTimers uses this in-memory model which is mapped to the current timers table . If you place a breakpoint on the OnModelCreating method, you should see that it is only called one time.

All that said, it is possible to dig deep and interrogate the in-memory model (you have to use the old school context type, ObjectContext vs. DbContext). Using some code from Rowan Miller here , you can find which table is mapped to each entity set. Using this code, each item in the tables variable has a read / write Table property that contains the database table name. I haven't tried setting this but it certainly seems plausible. Of course, you would need to alter the model somewhere else outside of the OnModelCreating method, say in the constructor, so that the code will fire each time an instance of the context is created.

* UPDATE *

Because I am always interested in learning new things, I couldn't leave this alone and threw together a test application. Unfortunately, it looks like you cannot set the property (despite it being a read / write property) as it throws an InvalidOperationException stating that the item is read-only. Perhaps there is another way but I have not found it...yet.

* UPDATE *

The solution is actually much simpler that what I had first mentioned. A couple of the constructors of the DbContext class accepts an instance of the DbCompiledModel class as one of its parameters. Using the DbModelBuilder class, you can build the same code that you would normally put in the OnModelCreating method. You can call the Build method of this class to create an instance of the DbModel class. You can call the Compile method of this class to create an instance of the DbCompiledModel class. The only real trick is that the Build method requires some additional information (I used an instance of the DbProviderInfo class but I think you could also use an actual connection but that would probably incur a hit to the database). I have tested this and this one does indeed work as desired.

Something like...

DbModelBuilder builder = null;

builder = new DbModelBuilder();
builder.Entity<TestEntity>().ToTable(tableName);

DbModel model1 = null;

model1 = builder.Build(new DbProviderInfo("System.Data.SqlClient", "2012"));

builder.Entity<TestEntity>().ToTable(anotherTableName);

DbModel model2 = null;

model2 = builder.Build(new DbProviderInfo("System.Data.SqlClient", "2012"));

DbCompiledModel compiledModel1 = null;
DbCompiledModel compiledModel2 = null;

compiledMdoel1 = model1.Compile();
compiledMdoel2 = model2.Compile();

TestContext context1 = null;
TestContext context2 = null;

context1 = new TestContext(compiledModel1);
context2 = new TestContext(compiledModel2);

Of course, the constructor of the TestContext class would have to pass the compiled model instance to the base constructor.

Answer 2

cannot add comment, but I just want to add one thing that happened to me

my code was like this:

DbModelBuilder builder = new DbModelBuilder();
this.OnModelCreating(builder);
var model = builder.Build(this.Database.Connection);

I was thought that when i pass the current DbConnection object to this method, it will somehow "inherit" of all connection settings, but seems I was wrong. After debugging a while, i just realize it generate some weird connection string for me, which always results in cannot find database issue. So my solution is, when instantiate the "TestContext" (as in Jason Richmeier's answer ), pass nameOrConnectionString as first parameter and the compiled model as second, and it solved my issue.

I wonder, since EF keep a in-memory copy of certain model, is manually create another one will just create a new copy in memory? And if my code need to do this a lot of times, it is going to keep creating new models in memory and finally ends with a memory overflow?