SQL 2008 - Foreign Key Constraints in the INFORMATION_SCHEMA View

Question

I am writing a C # unit test that checks the row properties for an ORM class for a target database, always SQL 2008 and the class to which the data is attached.

Verifying that the specified foreign key is valid in the database is simple:

    static private bool ConstraintExsits(string table, string column, ConstraintType constraintType)
    {
        string constraintTypeWhereClause;
        switch (constraintType)
        {
            case ConstraintType.PrimaryKey:
                constraintTypeWhereClause = "PRIMARY KEY";
                break;
            case ConstraintType.ForeignKey:
                constraintTypeWhereClause = "FOREIGN KEY";
                break;
            default:
                throw new ArgumentOutOfRangeException("constraintType");
        }

        var cmd = new SqlCommand(
                           @"SELECT a.CONSTRAINT_NAME
                            FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS a 
                            JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE b on a.CONSTRAINT_NAME = b.CONSTRAINT_NAME
                            WHERE a.TABLE_NAME = @table AND b.COLUMN_NAME = @column AND a.CONSTRAINT_TYPE = '" + constraintTypeWhereClause + "'",
                           Connection);
        cmd.Parameters.AddWithValue("@table", table.Trim('[').Trim(']'));
        cmd.Parameters.AddWithValue("@column", column.Trim('[').Trim(']'));
        return !string.IsNullOrEmpty((string)cmd.ExecuteScalar());
    }

Now take the following external relations:

My question is: how can I request relationships from the "main base / unique key table" and the "primary / unique key columns"? I do not see these links in INFORMATION_SCHEMA views.

Thanks J

Answer 1

This is the SQL I was after!

SELECT 
FK_Table  = FK.TABLE_NAME, 
FK_Column = CU.COLUMN_NAME, 
PK_Table  = PK.TABLE_NAME, 
PK_Column = PT.COLUMN_NAME, 
Constraint_Name = C.CONSTRAINT_NAME 
FROM 
INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS C 
INNER JOIN 
INFORMATION_SCHEMA.TABLE_CONSTRAINTS FK 
    ON C.CONSTRAINT_NAME = FK.CONSTRAINT_NAME 
INNER JOIN 
INFORMATION_SCHEMA.TABLE_CONSTRAINTS PK 
    ON C.UNIQUE_CONSTRAINT_NAME = PK.CONSTRAINT_NAME 
INNER JOIN 
INFORMATION_SCHEMA.KEY_COLUMN_USAGE CU 
    ON C.CONSTRAINT_NAME = CU.CONSTRAINT_NAME 
INNER JOIN 
( 
    SELECT 
        i1.TABLE_NAME, i2.COLUMN_NAME 
    FROM 
        INFORMATION_SCHEMA.TABLE_CONSTRAINTS i1 
        INNER JOIN 
        INFORMATION_SCHEMA.KEY_COLUMN_USAGE i2 
        ON i1.CONSTRAINT_NAME = i2.CONSTRAINT_NAME 
        WHERE i1.CONSTRAINT_TYPE = 'PRIMARY KEY' 
) PT 
ON PT.TABLE_NAME = PK.TABLE_NAME 

Answer 2

Jaimie , (, ) , UNIQUE_CONSTRAINT_NAME. :

SELECT 
    FK = fk.name, 
    FKTable = QUOTENAME(OBJECT_SCHEMA_NAME(fkcol.[object_id])) 
        + '.' + QUOTENAME(OBJECT_NAME(fkcol.[object_id])),
    FKCol = fkcol.name,
    ' references => ',
    PKTable = QUOTENAME(OBJECT_SCHEMA_NAME(pkcol.[object_id])) 
        + '.' + QUOTENAME(OBJECT_NAME(pkcol.[object_id])),
    PKCol = pkcol.name
FROM sys.foreign_keys AS fk
INNER JOIN sys.foreign_key_columns AS fkc
ON fk.[object_id] = fkc.constraint_object_id
INNER JOIN sys.columns AS fkcol
ON fkc.parent_object_id = fkcol.[object_id]
AND fkc.parent_column_id = fkcol.column_id
INNER JOIN sys.columns AS pkcol
ON fkc.referenced_object_id = pkcol.[object_id]
AND fkc.referenced_column_id = pkcol.column_id
ORDER BY fkc.constraint_column_id;

: http://sqlblog.com/blogs/aaron_bertrand/archive/2011/11/03/the-case-against-information-schema-views.aspx