In table A I've got a composite of 3 columns as a primary key. I want to have only one of these three columns as a foreign key in table B, just to make sure that the value that I insert into table B's column exists in table A.
Currently from what I've read it looks like in Entity Framework I have to add all three columns of composite PK, which is not really what I need. The latest answer that I've found was of 2015, maybe since then something changed?
I know that I can add a manual check on each insert/update call, but I don't want to do that, maybe there is more elegant way.
Related
The entity A has an OneToMany relation with the entity B. I would like to use a composite primary key for the entity B, where one of the columns is the same as the the primary key of the entity A and the second column should be generated. How can the second column be started from 1 whenever the the other involving column within the composite primary key changes?
I use Oracle data bank and could not find any solution.
Thanks in advance!
Hi all,
Can any understand what's going on here?
The case is:
There are 2 tables, called "matricula" and "pagament" with a 1:1 relationship cardinality.
Table matricula primary key composed by 3 fields "edicio","curs" and "estudiant".
Table pagament primary key, the same as above. Furthermore, it references matricula.
As shown, trying to insert a row in pagament table is rejected because it does not exists a row in table matricula. However, asking for this row returns one result.
What am I missing?
Thanks you all
Carles
The problem is that the order of the fields in both tables is not the same, and, moreover, the restriction of the foreign key in table pagament, said that
foreign key (estudiant,curs,edicio) references matricula
without specifying the matricula fields.
It's been solved by setting this restriction as
foreign key (estudiant,curs,edicio) references matricula(estudiant,curs,edicio)
I have a junction table with and idenity primary key columns to realize a many to many relationship. Visual Studio automatically detects it as a many to many relationship and the junction table is not an entity.
How can i realize it that also this table is generated as an entity? I need this for breeze.js .
You just need to add additional columns (or properties) to that table (or model).
You said that your table has acolumn named ID and it's the primary key withe IsIdentity set to true. It must works, I'm using this approach...
There must be a problem or missing with your table definition. However, if all are OK, just add a nullable column in your table and update your model from database. The problem will go away.
I am using SQL Server 2012 and I want to create a "changes" table - it will be populated with data from other table when the second table columns values are changed.
I am adding to the "changes" table "datatime2", and "rowversion" columns in order to track when the changes are made.
Is it ok to use "rowversion" as primary key?
I have read here that it will be changed, if the current row is updated and that's why it is not a good candidate for "primary key" making foreign keys invalid.
Anyway, if it won't be used as a foreign key and the rows of "changes" table will never be updated (only new rows will be inserted) is it ok to use the "rowversion" as PK or I should use additional column?
Some good info here:
Careful reading of the MSDN page also shows that duplicate rowversion values are possible if SELECT INTO statements are used improperly. Something to watch out for there.
I would stick with an Identity field in the original data, carried over into the change tracking table that has its own Identity field.
If I have 2 tables 1 with a composite primary key where one of the keys is also a foreign key in another table:
Table 1:
A (PK, FK - maps to X in Table 2)
B (PK)
C
Table 2:
X (PK)
Y
Because A is both the PK in table 1 and FK in table 2, when I use EF to generate the entity model, I have both a Scalar AND a Navigation property for A in table 1. I cannot seem to remove A as a scalar (I think because it is a primary key).
The problem I am having is that if I create a table1Entity and set A's scalar property to a new value, A's navigation property will not be changed automatically (and vice versa).
Ideally I just want A to expose the navigation property - which is the way it behaves if A was not also part of the composite primary key anyway. Is there any way to achieve this?
Am I correct in assuming that Table1 derives from Table2? If so, I would do it like so:
(I'd also change the PK for both tables to the same name, since they probably have the same meaning - for the instance of this, I'll use the example ID)
First, create the model with the default relationships (I usually just import the two tables from the database)
In the designer, right click the base type, add inheritance, select the derived type.
Delete the one to zero or one association
Then, since the base type already has column ID, delete it from the derived type.
Go to table mapping for the derived type, and map the ID property to the ID of the table.
Well, not really. Create the view with schemabinding and create a clustered index on the view (SQL Server 2008 or later, earlier versions I'm not sure can do that). The clustered index will be recognised as a primary key, thus tricking EF(VS) into believing the view is a real table.
Have you expicity set the Ids of the composite key and referenced these in your configuration?
i.e
public class Table1
{
public Table2 A{get;set}
public int AId {get;set;}
public int BId {get;set;}
}
I assume you'll need something like:
HasKey(pc => new { pc.AId, pc.BId});
HasRequired(x => x.A).WithMany().HasForeignKey(x => x.AId);
Instead of mapping to table 1 directly, add a view to your database that's got all of table 1's fields, plus an extra copy of A (A2).
Then, map the scalar key to A2 and the nav key to A.
(You'll run into a problem where if you use a view, Visual Studio can't find a primary key; fix this by manually editing the XML of the edmx file and adding a <Key><PropertyRef ... /></Key> to the <EntityType> for table A)
I know - it's hacky and horrible... but hey - it works!