73

您能否帮我构建一个查询，该查询检索所有表中的约束、每个表中的约束计数，并显示NULL没有任何约束的表。

这是我到目前为止所拥有的：

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]
4

6 回答 6

121

您应该使用当前的sys目录视图（如果您使用的是 SQL Server 2005或更新版本 - 这些sysobjects视图已被弃用并且应该避免使用） -在此处查看有关目录视图的大量 MSDN SQL Server 联机丛书文档。

您可能会感兴趣的视图有很多：

  • sys.default_constraints对于列的默认约束
  • sys.check_constraints检查列上的约束
  • sys.key_constraints用于键约束（例如主键）
  • sys.foreign_keys对于外键关系

还有更多- 看看吧！

您可以查询并加入这些视图以获取所需的信息 - 例如，这将列出表、列和在它们上定义的所有默认约束：

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
于 2013-01-09T06:17:52.830 回答
13

您可以使用此查询

独特的约束，

带值的默认约束，

具有引用表和列的外键

和主键约束。

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');
于 2017-08-25T19:49:50.567 回答
13

我使用以下查询来检索 SQL Server 2012 中的约束信息，并且运行良好。我希望它对你有用。

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
于 2017-04-18T17:48:02.020 回答
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])
    )
于 2014-07-02T10:56:24.487 回答
3

我试图编辑marc_s提供的答案，但由于某种原因它没有被接受。它格式化 sql 以便于阅读，包括架构并命名默认名称，以便可以轻松地将其粘贴到其他代码中。

  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
于 2016-03-23T05:24:45.947 回答
2

这是获取外键的脚本：

    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]
于 2020-02-26T17:07:05.693 回答