如何列出SQL Server表的主键？

简单的问题，你如何用T-SQL列出表的主键？ 我知道如何在桌子上获得索引，但不记得如何获得PK。

Answer 1

SELECT Col.Column_Name from INFORMATION_SCHEMA.TABLE_CONSTRAINTS Tab, INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE Col WHERE Col.Constraint_Name = Tab.Constraint_Name AND Col.Table_Name = Tab.Table_Name AND Constraint_Type = 'PRIMARY KEY' AND Col.Table_Name = '<your table name>' 

Answer 2

现在通常推荐使用SQL Server中的INFORMATION_SCHEMA的sys.*视图，所以除非你打算迁移数据库，否则我会使用这些。 以下是你如何使用sys.*视图来做到这一点：

 SELECT c.name AS column_name, i.name AS index_name, c.is_identity FROM sys.indexes i inner join sys.index_columns ic ON i.object_id = ic.object_id AND i.index_id = ic.index_id inner join sys.columns c ON ic.object_id = c.object_id AND c.column_id = ic.column_id WHERE i.is_primary_key = 1 and i.object_ID = OBJECT_ID('<schema>.<tablename>'); 

Answer 3

这是一个只使用sys -tables的解决scheme。

它列出了数据库中的所有主键。 它为每个主键返回模式，表名，列名和正确的列sorting顺序 。

如果您想获取特定表的主键，则需要在SchemaName和TableName上进行筛选。

恕我直言，这个解决scheme是非常通用的，不使用任何string文字，所以它可以在任何机器上运行。

 select s.name as SchemaName, t.name as TableName, tc.name as ColumnName, ic.key_ordinal as KeyOrderNr from sys.schemas s inner join sys.tables t on s.schema_id=t.schema_id inner join sys.indexes i on t.object_id=i.object_id inner join sys.index_columns ic on i.object_id=ic.object_id and i.index_id=ic.index_id inner join sys.columns tc on ic.object_id=tc.object_id and ic.column_id=tc.column_id where i.is_primary_key=1 order by t.name, ic.key_ordinal ; 

Answer 4

这个问题简单得多： 使用sql查询获取表主键 ：

 SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE OBJECTPROPERTY(OBJECT_ID(CONSTRAINT_SCHEMA+'.'+CONSTRAINT_NAME), 'IsPrimaryKey') = 1 AND TABLE_NAME = ' <your table name> ' 

它使用KEY_COLUMN_USAGE来确定给定表的约束
然后使用OBJECTPROPERTY( id , 'IsPrimaryKey')来确定每个是否是主键

Answer 5

我喜欢INFORMATION_SCHEMA技术，但我用过的另一个是：exec sp_pkeys'table'

Answer 6

正在使用MS SQL Server，您可以执行以下操作：

 --List all tables primary keys select * from information_schema.table_constraints where constraint_type = 'Primary Key' 

如果你想要一个特定的表，你也可以在table_name列上过滤。

Answer 7

– 这是另一个修改版本，也是关联查询的一个例子

 SELECT TC.TABLE_NAME as [Table_name], TC.CONSTRAINT_NAME as [Primary_Key] FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS TC INNER JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE CCU ON TC.CONSTRAINT_NAME = CCU.CONSTRAINT_NAME WHERE TC.CONSTRAINT_TYPE = 'PRIMARY KEY' AND TC.TABLE_NAME IN (SELECT [NAME] AS [TABLE_NAME] FROM SYS.OBJECTS WHERE TYPE = 'U') 

Answer 8

这应该列出所有的约束（主键和外键），并在查询放置表名称结束

 /* CAST IS DONE , SO THAT OUTPUT INTEXT FILE REMAINS WITH SCREEN LIMIT*/ WITH ALL_KEYS_IN_TABLE (CONSTRAINT_NAME,CONSTRAINT_TYPE,PARENT_TABLE_NAME,PARENT_COL_NAME,PARENT_COL_NAME_DATA_TYPE,REFERENCE_TABLE_NAME,REFERENCE_COL_NAME) AS ( SELECT CONSTRAINT_NAME= CAST (PKnUKEY.name AS VARCHAR(30)) , CONSTRAINT_TYPE=CAST (PKnUKEY.type_desc AS VARCHAR(30)) , PARENT_TABLE_NAME=CAST (PKnUTable.name AS VARCHAR(30)) , PARENT_COL_NAME=CAST ( PKnUKEYCol.name AS VARCHAR(30)) , PARENT_COL_NAME_DATA_TYPE= oParentColDtl.DATA_TYPE, REFERENCE_TABLE_NAME='' , REFERENCE_COL_NAME='' FROM sys.key_constraints as PKnUKEY INNER JOIN sys.tables as PKnUTable ON PKnUTable.object_id = PKnUKEY.parent_object_id INNER JOIN sys.index_columns as PKnUColIdx ON PKnUColIdx.object_id = PKnUTable.object_id AND PKnUColIdx.index_id = PKnUKEY.unique_index_id INNER JOIN sys.columns as PKnUKEYCol ON PKnUKEYCol.object_id = PKnUTable.object_id AND PKnUKEYCol.column_id = PKnUColIdx.column_id INNER JOIN INFORMATION_SCHEMA.COLUMNS oParentColDtl ON oParentColDtl.TABLE_NAME=PKnUTable.name AND oParentColDtl.COLUMN_NAME=PKnUKEYCol.name UNION ALL SELECT CONSTRAINT_NAME= CAST (oConstraint.name AS VARCHAR(30)) , CONSTRAINT_TYPE='FK', PARENT_TABLE_NAME=CAST (oParent.name AS VARCHAR(30)) , PARENT_COL_NAME=CAST ( oParentCol.name AS VARCHAR(30)) , PARENT_COL_NAME_DATA_TYPE= oParentColDtl.DATA_TYPE, REFERENCE_TABLE_NAME=CAST ( oReference.name AS VARCHAR(30)) , REFERENCE_COL_NAME=CAST (oReferenceCol.name AS VARCHAR(30)) FROM sys.foreign_key_columns FKC INNER JOIN sys.sysobjects oConstraint ON FKC.constraint_object_id=oConstraint.id INNER JOIN sys.sysobjects oParent ON FKC.parent_object_id=oParent.id INNER JOIN sys.all_columns oParentCol ON FKC.parent_object_id=oParentCol.object_id /* ID of the object to which this column belongs.*/ AND FKC.parent_column_id=oParentCol.column_id/* ID of the column. Is unique within the object.Column IDs might not be sequential.*/ INNER JOIN sys.sysobjects oReference ON FKC.referenced_object_id=oReference.id INNER JOIN INFORMATION_SCHEMA.COLUMNS oParentColDtl ON oParentColDtl.TABLE_NAME=oParent.name AND oParentColDtl.COLUMN_NAME=oParentCol.name INNER JOIN sys.all_columns oReferenceCol ON FKC.referenced_object_id=oReferenceCol.object_id /* ID of the object to which this column belongs.*/ AND FKC.referenced_column_id=oReferenceCol.column_id/* ID of the column. Is unique within the object.Column IDs might not be sequential.*/ ) select * from ALL_KEYS_IN_TABLE where PARENT_TABLE_NAME in ('YOUR_TABLE_NAME') or REFERENCE_TABLE_NAME in ('YOUR_TABLE_NAME') ORDER BY PARENT_TABLE_NAME,CONSTRAINT_NAME; 

有关参考请阅读 – http://blogs.msdn.com/b/sqltips/archive/2005/09/16/469136.aspx

Answer 9

哥们，谢啦。

我用它来查找所有表的所有主键。

 SELECT A.Name,Col.Column_Name from INFORMATION_SCHEMA.TABLE_CONSTRAINTS Tab, INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE Col , (select NAME from dbo.sysobjects where xtype='u') AS A WHERE Col.Constraint_Name = Tab.Constraint_Name AND Col.Table_Name = Tab.Table_Name AND Constraint_Type = 'PRIMARY KEY ' AND Col.Table_Name = A.Name 

Answer 10

 SELECT A.TABLE_NAME as [Table_name], A.CONSTRAINT_NAME as [Primary_Key] FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS A, INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE B WHERE CONSTRAINT_TYPE = 'PRIMARY KEY' AND A.CONSTRAINT_NAME = B.CONSTRAINT_NAME 

Answer 11

这一个给你的PK列。

 SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE TABLE_NAME = 'TableName' 

Answer 12

我发现这个很有用，给出了一个列表，列表中有一列逗号分隔的列，然后是一个逗号分隔的列表，其中主键是哪个

 SELECT T.TABLE_SCHEMA, T.TABLE_NAME, STUFF(( SELECT ', ' + C.COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS C WHERE C.TABLE_SCHEMA = T.TABLE_SCHEMA AND T.TABLE_NAME = C.TABLE_NAME FOR XML PATH ('') ), 1, 2, '') AS Columns, STUFF(( SELECT ', ' + C.COLUMN_NAME FROM INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE C INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS TC ON C.TABLE_SCHEMA = TC.TABLE_SCHEMA AND C.TABLE_NAME = TC.TABLE_NAME WHERE C.TABLE_SCHEMA = T.TABLE_SCHEMA AND T.TABLE_NAME = C.TABLE_NAME AND TC.CONSTRAINT_TYPE = 'PRIMARY KEY' FOR XML PATH ('') ), 1, 2, '') AS [Key] FROM INFORMATION_SCHEMA.TABLES T ORDER BY T.TABLE_SCHEMA, T.TABLE_NAME 

Answer 13

下面的查询将列出特定表的 主键 ：

 SELECT DISTINCT CONSTRAINT_NAME AS [Constraint], TABLE_SCHEMA AS [Schema], TABLE_NAME AS TableName FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE TABLE_NAME = 'mytablename' 

Answer 14

 SELECT t.name AS 'table', i.name AS 'index', it.xtype, (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 1 AND k.id = t.id) AS 'column1', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 2 AND k.id = t.id) AS 'column2', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 3 AND k.id = t.id) AS 'column3', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 4 AND k.id = t.id) AS 'column4', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 5 AND k.id = t.id) AS 'column5', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 6 AND k.id = t.id) AS 'column6', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 7 AND k.id = t.id) AS 'column7', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 8 AND k.id = t.id) AS 'column8', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 9 AND k.id = t.id) AS 'column9', (SELECT c.name FROM syscolumns c INNER JOIN sysindexkeys k ON k.indid = i.indid AND c.colid = k.colid AND c.id = t.id AND k.keyno = 10 AND k.id = t.id) AS 'column10', FROM sysobjects t INNER JOIN sysindexes i ON i.id = t.id INNER JOIN sysobjects it ON it.parent_obj = t.id AND it.name = i.name WHERE it.xtype = 'PK' ORDER BY t.name, i.name 

Answer 15

系统存储过程sp_help会为您提供信息。 执行以下语句：

 execute sp_help table_name 

Answer 16

试试这个：

 SELECT CONSTRAINT_CATALOG AS DataBaseName, CONSTRAINT_SCHEMA AS SchemaName, TABLE_NAME AS TableName, CONSTRAINT_Name AS PrimaryKey FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS WHERE CONSTRAINT_TYPE = 'Primary Key' and Table_Name = 'YourTable' 

Answer 17

此版本显示架构，表名和主键的以逗号分隔的有序列表。 Object_Id（）不适用于链接服务器，所以我们按表名进行过滤。

如果没有REPLACE（Si1.Column_Name，''，''），它会在我testing的数据库上显示Column_Name的xml打开和closures标记。 我不知道为什么数据库需要replace'Column_Name'，所以如果有人知道，那么请评论。

 DECLARE @TableName VARCHAR(100) = ''; WITH Sysinfo AS (SELECT Kcu.Table_Name , Kcu.Table_Schema AS Schema_Name , Kcu.Column_Name , Kcu.Ordinal_Position FROM [LinkServer].Information_Schema.Key_Column_Usage Kcu JOIN [LinkServer].Information_Schema.Table_Constraints AS Tc ON Tc.Constraint_Name = Kcu.Constraint_Name WHERE Tc.Constraint_Type = 'Primary Key') SELECT Schema_Name ,Table_Name , STUFF( ( SELECT ', ' , REPLACE(Si1.Column_Name, '', '') FROM Sysinfo Si1 WHERE Si1.Table_Name = Si2.Table_Name ORDER BY Si1.Table_Name , Si1.Ordinal_Position FOR XML PATH('') ), 1, 2, '') AS Primary_Keys FROM Sysinfo Si2 WHERE Table_Name = CASE WHEN @TableName NOT IN( '', 'All') THEN @TableName ELSE Table_Name END GROUP BY Si2.Table_Name, Si2.Schema_Name; 

而使用George的查询的模式也是如此：

 DECLARE @TableName VARCHAR(100) = ''; WITH Sysinfo AS (SELECT S.Name AS Schema_Name , T.Name AS Table_Name , Tc.Name AS Column_Name , Ic.Key_Ordinal AS Ordinal_Position FROM [LinkServer].Sys.Schemas S JOIN [LinkServer].Sys.Tables T ON S.Schema_Id = T.Schema_Id JOIN [LinkServer].Sys.Indexes I ON T.Object_Id = I.Object_Id JOIN [LinkServer].Sys.Index_Columns Ic ON I.Object_Id = Ic.Object_Id AND I.Index_Id = Ic.Index_Id JOIN [LinkServer].Sys.Columns Tc ON Ic.Object_Id = Tc.Object_Id AND Ic.Column_Id = Tc.Column_Id WHERE I.Is_Primary_Key = 1) SELECT Schema_Name ,Table_Name , STUFF( ( SELECT ', ' , REPLACE(Si1.Column_Name, '', '') FROM Sysinfo Si1 WHERE Si1.Table_Name = Si2.Table_Name ORDER BY Si1.Table_Name , Si1.Ordinal_Position FOR XML PATH('') ), 1, 2, '') AS Primary_Keys FROM Sysinfo Si2 WHERE Table_Name = CASE WHEN @TableName NOT IN('', 'All') THEN @TableName ELSE Table_Name END GROUP BY Si2.Table_Name, Si2.Schema_Name; 

Answer 18

Sys.Objects表包含每个用户定义的模式范围对象的行。

像主键或其他创build的约束将是对象 ，表名称将是parent_object

查询sys.Objects并收集必需types的对象ID

 declare @TableName nvarchar(50)='TblInvoice' -- your table name declare @TypeOfKey nvarchar(50)='PK' -- For Primary key SELECT Name FROM sys.objects WHERE type = @TypeOfKey AND parent_object_id = OBJECT_ID (@TableName) 

Answer 19

我正在讲一个我遵循的简单的技术

 SP_HELP table_name 

作为查询运行此代码。 在你想知道主键的table_name的地方提到你的表名。 结果将显示像附加的图像。 希望它会帮助你