编辑:修复了外键查询中的小错误,该错误不影响答案,但返回错误的唯一表
约束适配到任意一个
基础边表的唯一约束
列数也一样
这是一对一/无。
如果所有FK列都符合其中一个
唯一约束,如果所有列都包含,则包含更多列
对我来说,这是一个执行问题。
我制定了一个能够产生正确结果的解决方案,但由于我不是SQL专家,我担心这是相当困难的。也许我会把它作为另一个问题提出来,以供审查。
无论如何-此脚本将标识所有1:?数据库中的关系。
/*
Will identify immediate 1:? fk relationships
*/
-- TODO: puzzle: work out the set-based equivalent
SET NOCOUNT ON
BEGIN -- Get a table full of PK and UQ columns
DECLARE @unique_keys TABLE
(
-- contains PK and UQ indexes
[schema_name] NVARCHAR(128),
table_name NVARCHAR(128),
index_name NVARCHAR(128),
column_id INT,
column_name NVARCHAR(128),
is_primary_key BIT,
is_unique_constraint BIT,
is_unique BIT
)
INSERT INTO @unique_keys
(
[schema_name],
table_name,
index_name,
column_id,
column_name,
is_primary_key,
is_unique_constraint,
is_unique
)
-- selects PK and UQ indexes
SELECT S.name AS [schema_name],
T.name AS table_name,
IX.name AS index_name,
IC.column_id,
C.name AS column_name,
IX.is_primary_key,
IX.is_unique_constraint,
IX.is_unique
FROM sys.tables AS T
INNER JOIN sys.schemas AS S ON T.schema_id = S.schema_id
INNER JOIN sys.indexes AS IX ON T.object_id = IX.object_id
INNER JOIN sys.index_columns AS IC ON IX.object_id = IC.object_id
AND IX.index_id = IC.index_id
INNER JOIN sys.columns AS C ON IC.column_id = C.column_id
AND IC.object_id = C.object_id
WHERE ( IX.is_unique = 1 )
AND ( T.name <> 'sysdiagrams' )
AND IX.is_unique = 1
ORDER BY schema_name,
table_name,
index_name,
C.column_id
END
BEGIN -- Get a table full of FK columns
DECLARE @foreign_key_columns TABLE
(
constraint_name NVARCHAR(128),
base_schema_name NVARCHAR(128),
base_table_name NVARCHAR(128),
base_column_id INT,
base_column_name NVARCHAR(128),
unique_schema_name NVARCHAR(128),
unique_table_name NVARCHAR(128),
unique_column_id INT,
unique_column_name NVARCHAR(128)
)
INSERT INTO @foreign_key_columns
(
constraint_name,
base_schema_name,
base_table_name,
base_column_id,
base_column_name,
unique_schema_name,
unique_table_name,
unique_column_id,
unique_column_name
)
SELECT FK.name AS constraint_name,
S.name AS base_schema_name,
T.name AS base_table_name,
C.column_id AS base_column_id,
C.name AS base_column_name,
US.name AS unique_schema_name,
UT.name AS unique_table_name,
UC.column_id AS unique_column_id,
UC.name AS unique_column_name
FROM sys.tables AS T
INNER JOIN sys.schemas AS S ON T.schema_id = S.schema_id
INNER JOIN sys.foreign_keys AS FK ON T.object_id = FK.parent_object_id
INNER JOIN sys.foreign_key_columns AS FKC ON FK.object_id = FKC.constraint_object_id
INNER JOIN sys.columns AS C ON FKC.parent_object_id = C.object_id
AND FKC.parent_column_id = C.column_id
INNER JOIN sys.columns AS UC ON FKC.referenced_object_id = UC.object_id
AND FKC.referenced_column_id = UC.column_id
INNER JOIN sys.tables AS UT ON FKC.referenced_object_id = UT.object_id
INNER JOIN sys.schemas AS US ON UT.schema_id = US.schema_id
WHERE ( T.name <> 'sysdiagrams' )
ORDER BY base_schema_name,
base_table_name
END
DECLARE @constraint_name NVARCHAR(128),
@base_schema_name NVARCHAR(128),
@base_table_name NVARCHAR(128),
@unique_schema_name NVARCHAR(128),
@unique_table_name NVARCHAR(128)
-- The foreign key side of the constraint is always singular, we need to check from the perspective
-- of the unique side of the constraint.
-- for each FK constraint in DB
DECLARE tmpC CURSOR READ_ONLY
FOR SELECT DISTINCT
constraint_name,
base_schema_name,
base_table_name,
unique_schema_name,
unique_table_name
FROM @foreign_key_columns
OPEN tmpC
FETCH NEXT FROM tmpC INTO @constraint_name, @base_schema_name, @base_table_name, @unique_schema_name, @unique_table_name
WHILE @@FETCH_STATUS = 0
BEGIN
-- get the columns in the base side of the FK constraint
DECLARE @fkc TABLE
(
column_name NVARCHAR(128)
)
DELETE FROM @fkc
INSERT INTO @fkc ( column_name )
SELECT base_column_name
FROM @foreign_key_columns
WHERE constraint_name = @constraint_name
-- check for one to one/none
-- If the base side columns of the constraint fit into any one of the base side tables unique constraints
-- AND the column count is the same then we have a one-to-one/none and should be realized as a singular
-- object reference
-- I realize that if the base side unique constraint has more columns than the unique side unique constraint
-- AND all of those columns DO represent a 1:? that would actually qualify but it seems like an edge case and
-- beyond the scope of this question.
DECLARE @uk_schema_name NVARCHAR(128),
@uk_table_name NVARCHAR(128),
@uk_index_name NVARCHAR(128),
@is_may_have_a BIT
SET @is_may_have_a = 0
-- have to open another cursor over the unique keys of the base table - i want
-- a distinct list of unique constraints for the base table
DECLARE cKey CURSOR READ_ONLY
FOR SELECT DISTINCT
[schema_name],
table_name,
index_name
FROM @unique_keys
WHERE [schema_name] = @base_schema_name
AND table_name = @base_table_name
OPEN cKey
FETCH NEXT FROM cKey INTO @uk_schema_name, @uk_table_name, @uk_index_name
WHILE @@FETCH_STATUS = 0
BEGIN
-- get the unique constraint columns
DECLARE @pkc TABLE
(
column_name NVARCHAR(128)
)
DELETE FROM @pkc
INSERT INTO @pkc ( column_name )
SELECT column_name
FROM @unique_keys
WHERE [schema_name] = @uk_schema_name
AND table_name = @uk_table_name
AND index_name = @uk_index_name
-- if count is same and columns are same
DECLARE @count1 INT, @count2 INT
SELECT @count1 = COUNT(*) FROM @fkc
SELECT @count2 = COUNT(*) FROM @pkc
IF @count1 = @count2
BEGIN
-- select all from both on name and exclude mismatches
SELECT @count1 = COUNT(*)
FROM @fkc F
FULL OUTER JOIN @pkc P ON f.column_name = p.column_name
WHERE NOT p.column_name IS NULL AND NOT f.column_name IS NULL
IF @count1 = @count2
BEGIN
-- the base side of the fk constraint corresponds exactly to
-- at least on unique constraint making it effectively 1:?
SET @is_may_have_a = 1
BREAK
END
END
FETCH NEXT FROM cKey INTO @uk_schema_name, @uk_table_name, @uk_index_name
END
CLOSE cKey
DEALLOCATE cKey
IF @is_may_have_a = 1
PRINT 'for ' + @unique_schema_name + '.' + @unique_table_name + ' constraint ' + + @constraint_name + ' is 1:? '
FETCH NEXT FROM tmpC INTO @constraint_name, @base_schema_name, @base_table_name, @unique_schema_name, @unique_table_name
END
CLOSE tmpC
DEALLOCATE tmpC
有关更详细的测试db的结果,请参见
TSQL: Identify a 1:? relationship
