SQL 2008 - INFORMATION_SCHEMA 视图中的外键约束答案

【问题标题】：SQL 2008 - Foreign key constraints in the INFORMATION_SCHEMA viewSQL 2008 - INFORMATION_SCHEMA 视图中的外键约束
【发布时间】：2009-11-09 10:48:52
【问题描述】：

我正在编写一个 c# 单元测试，它针对目标数据库（始终是 SQL 2008）以及数据映射到的类验证 ORM 类的字符串属性。

检查指定的外键在数据库中是否有效很容易：

    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());
    }

现在采用以下外键关系：

我的问题：如何从“主/唯一键基表”和“主/唯一键列”端查询关系？我在 INFORMATION_SCHEMA 视图中看不到这些引用。

谢谢 J

【问题讨论】：

标签： c# sql orm

【解决方案1】：

这是我所追求的 SQL！

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

【讨论】：

【解决方案2】：

如果引用的表（外键正在查看的表）具有唯一键，则 Jaimie 的答案无法正确返回所有外键，因为它使用 UNIQUE_CONSTRAINT_NAME 列。我建议：

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;

来源：

The case against INFORMATION_SCHEMA views

【讨论】：