SQL Server 2008 - 獲取表約束

Question

你能幫我構建一個查詢來檢索所有表中的約束、每個表中的約束計數，並為沒有任何約束的表顯示NULL 。

這是我到目前為止：

Select  SysObjects.[Name] As [Constraint Name] ,
        Tab.[Name] as [Table Name],
        Col.[Name] As [Column Name]
From SysObjects Inner Join 
(Select [Name],[ID] From SysObjects) As Tab
On Tab.[ID] = Sysobjects.[Parent_Obj] 
Inner Join sysconstraints On sysconstraints.Constid = Sysobjects.[ID] 
Inner Join SysColumns Col On Col.[ColID] = sysconstraints.[ColID] And Col.[ID] = Tab.[ID]
order by [Tab].[Name] 

Answer 1

您應該使用當前的sys目錄視圖（如果您使用的是 SQL Server 2005或更新版本 - sysobjects視圖已被棄用，應避免使用） -在此處查看有關目錄視圖的大量 MSDN SQL Server 聯機叢書文檔。

有很多你可能感興趣的視圖：

• sys.default_constraints用於列的默認約束
• sys.check_constraints用於檢查列的約束
• sys.key_constraints用於鍵約束（例如主鍵）
• sys.foreign_keys用於外鍵關系

還有更多- 看看吧！

您可以查詢並加入這些視圖以獲取所需的信息 - 例如，這將列出表、列和在其上定義的所有默認約束：

SELECT 
    TableName = t.Name,
    ColumnName = c.Name,
    dc.Name,
    dc.definition
FROM sys.tables t
INNER JOIN sys.default_constraints dc ON t.object_id = dc.parent_object_id
INNER JOIN sys.columns c ON dc.parent_object_id = c.object_id AND c.column_id = dc.parent_column_id
ORDER BY t.Name

Answer 2

我使用以下查詢檢索SQL Server 2012中的約束信息，並且完美運行。 我希望它對你有用。

SELECT 
    tab.name AS [Table]
    ,tab.id AS [Table Id]
    ,constr.name AS [Constraint Name]
    ,constr.xtype AS [Constraint Type]
    ,CASE constr.xtype WHEN 'PK' THEN 'Primary Key' WHEN 'UQ' THEN 'Unique' ELSE '' END AS [Constraint Name]
    ,i.index_id AS [Index ID]
    ,ic.column_id AS [Column ID]
    ,clmns.name AS [Column Name]
    ,clmns.max_length AS [Column Max Length]
    ,clmns.precision AS [Column Precision]
    ,CASE WHEN clmns.is_nullable = 0 THEN 'NO' ELSE 'YES' END AS [Column Nullable]
    ,CASE WHEN clmns.is_identity = 0 THEN 'NO' ELSE 'YES' END AS [Column IS IDENTITY]
FROM SysObjects AS tab
INNER JOIN SysObjects AS constr ON(constr.parent_obj = tab.id AND constr.type = 'K')
INNER JOIN sys.indexes AS i ON( (i.index_id > 0 and i.is_hypothetical = 0) AND (i.object_id=tab.id) AND i.name = constr.name )
INNER JOIN sys.index_columns AS ic ON (ic.column_id > 0 and (ic.key_ordinal > 0 or ic.partition_ordinal = 0 or ic.is_included_column != 0)) 
                                    AND (ic.index_id=CAST(i.index_id AS int) 
                                    AND ic.object_id=i.object_id)
INNER JOIN sys.columns AS clmns ON clmns.object_id = ic.object_id and clmns.column_id = ic.column_id
WHERE tab.xtype = 'U'
ORDER BY tab.name

Answer 3

你可以得到這個查詢

唯一約束，

帶值的默認約束，

帶有引用表和列的外鍵

和主鍵約束。

Select C.*, (Select definition From sys.default_constraints Where object_id = C.object_id) As dk_definition,
(Select definition From sys.check_constraints Where object_id = C.object_id) As ck_definition,
(Select name From sys.objects Where object_id = D.referenced_object_id) As fk_table,
(Select name From sys.columns Where column_id = D.parent_column_id And object_id = D.parent_object_id) As fk_col
From sys.objects As C
Left Join (Select * From sys.foreign_key_columns) As D On D.constraint_object_id = C.object_id 
Where C.parent_object_id = (Select object_id From sys.objects Where type = 'U'
And name = 'Table Name Here');

Answer 4

SELECT
    [oj].[name] [TableName],
    [ac].[name] [ColumnName],
    [dc].[name] [DefaultConstraintName],
    [dc].[definition]
FROM
    sys.default_constraints [dc],
    sys.all_objects [oj],
    sys.all_columns [ac]
WHERE
    (
        ([oj].[type] IN ('u')) AND
        ([oj].[object_id] = [dc].[parent_object_id]) AND
        ([oj].[object_id] = [ac].[object_id]) AND
        ([dc].[parent_column_id] = [ac].[column_id])
    )

Answer 5

我試圖編輯由marc_s提供的答案，但由於某種原因沒有被接受。 它格式化 sql 以便於閱讀，包括架構並命名默認名稱，以便可以輕松地將其粘貼到其他代碼中。

  SELECT SchemaName = s.Name,
         TableName = t.Name,
         ColumnName = c.Name,
         DefaultName = dc.Name,
         DefaultDefinition = dc.Definition
    FROM sys.schemas                s
    JOIN sys.tables                 t   on  t.schema_id          = s.schema_id
    JOIN sys.default_constraints    dc  on  dc.parent_object_id  = t.object_id 
    JOIN sys.columns                c   on  c.object_id          = dc.parent_object_id
                                        and c.column_id          = dc.parent_column_id
ORDER BY s.Name, t.Name, c.name

Answer 6

這是一個獲取外鍵的腳本：

    SELECT TOP(150)
       t.[name] AS [Table],
       cols.[name] AS [Column],
       t2.[name] AS [Referenced Table],
       c2.[name] AS [Referenced Column],
       constr.[name] AS [Constraint]
  FROM sys.tables t
 INNER JOIN sys.foreign_keys constr ON constr.parent_object_id = t.object_id
 INNER JOIN sys.tables t2 ON t2.object_id = constr.referenced_object_id
 INNER JOIN sys.foreign_key_columns fkc ON fkc.constraint_object_id = constr.object_id
 INNER JOIN sys.columns cols ON cols.object_id = fkc.parent_object_id AND cols.column_id = fkc.parent_column_id
 INNER JOIN sys.columns c2 ON c2.object_id = fkc.referenced_object_id AND c2.column_id = fkc.referenced_column_id
 --WHERE t.[name] IN ('?', '?', ...)
 ORDER BY t.[Name], cols.[name]