简单的问题，如何用 T-SQL 列出表的主键？我知道如何获取表上的索引，但不记得如何获取 PK。
swilliams
问问题
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27 回答
160
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>'
于 2008-09-18T19:29:07.260 回答
36
现在通常建议在 SQL Server 中使用sys.*视图INFORMATION_SCHEMA，所以除非您计划迁移数据库，否则我会使用这些。以下是您将如何处理sys.*视图：
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>');
于 2015-09-10T20:26:24.247 回答
25
这是一个仅使用sys -tables 的解决方案。
它列出了数据库中的所有主键。它为每个主键返回架构、表名、列名和正确的列排序顺序。
如果要获取特定表的主键，则需要对SchemaNameand进行过滤TableName。
恕我直言，这个解决方案非常通用，不使用任何字符串文字，因此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 ;
于 2016-05-16T09:19:34.840 回答
9
在使用 MS SQL Server 时，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'
如果您想要一个特定的表，您还可以过滤 table_name 列。
于 2008-09-18T19:31:08.473 回答
9
这是问题get table primary key using sql query的另一种方法：
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>'
它用于KEY_COLUMN_USAGE确定给定表的约束
然后用于确定每个表是否是主键OBJECTPROPERTY(id, 'IsPrimaryKey')
于 2014-06-04T18:52:27.273 回答
8
我喜欢 INFORMATION_SCHEMA 技术，但我使用的另一个技术是：exec sp_pkeys 'table'
于 2008-10-07T16:06:22.087 回答
4
--这是另一个修改版本，也是相关查询的一个例子
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')
于 2011-09-26T06:36:00.973 回答
4
我正在讲述一个我遵循的简单技术
SP_HELP 'table_name'
将此代码作为查询运行。在您想知道主键的 table_name 位置提及您的表名（不要忘记单引号）。结果将显示为附加图像。希望对你有帮助
于 2017-11-03T08:20:16.527 回答
3
这应该列出所有约束（主键和外键）并在查询末尾放置表名
/* 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;
供参考，请阅读 - http://blogs.msdn.com/b/sqltips/archive/2005/09/16/469136.aspx
于 2013-04-03T20:29:05.783 回答
2
系统存储过程sp_help将为您提供信息。执行以下语句：
execute sp_help table_name
于 2008-09-18T19:23:42.577 回答
2
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
于 2008-09-18T20:06:54.317 回答
2
这一个为您提供了 PK 列。
SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE TABLE_NAME = 'TableName'
于 2015-06-27T00:17:40.773 回答
1
谢了，兄弟们。
稍有变化，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
于 2010-01-20T02:57:10.423 回答
1
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
于 2011-09-26T06:18:23.130 回答
1
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'
于 2016-10-04T11:27:21.480 回答
1
如果您正在寻找自己的 ORM 或从给定的表生成代码，那么这可能是您正在寻找的形式：
declare @table varchar(100) = 'mytable';
with cte as
(
select
tc.CONSTRAINT_SCHEMA
, tc.CONSTRAINT_TYPE
, tc.TABLE_NAME
, ccu.COLUMN_NAME
, IS_NULLABLE
, DATA_TYPE
, CHARACTER_MAXIMUM_LENGTH
, NUMERIC_PRECISION
from
INFORMATION_SCHEMA.TABLE_CONSTRAINTS tc
inner join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE ccu on tc.TABLE_NAME=ccu.TABLE_NAME and tc.TABLE_SCHEMA=ccu.TABLE_SCHEMA
inner join information_schema.COLUMNS c on ccu.COLUMN_NAME=c.COLUMN_NAME and ccu.TABLE_NAME=c.TABLE_NAME and ccu.TABLE_SCHEMA=c.TABLE_SCHEMA
where
tc.table_name=@table
and
ccu.CONSTRAINT_NAME=tc.CONSTRAINT_NAME
union
select TABLE_SCHEMA,'COLUMN', TABLE_NAME, COLUMN_NAME, IS_NULLABLE, DATA_TYPE,CHARACTER_MAXIMUM_LENGTH, NUMERIC_PRECISION from INFORMATION_SCHEMA.COLUMNS where TABLE_NAME=@table
and COLUMN_NAME not in (select COLUMN_NAME from INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE where TABLE_NAME = @table)
)
select
cast(iif(CONSTRAINT_TYPE='PRIMARY KEY',1,0) as bit) PrimaryKey
,cast(iif(CONSTRAINT_TYPE='FOREIGN KEY',1,0) as bit) ForeignKey
,cast(iif(CONSTRAINT_TYPE='COLUMN',1,0) as bit) NotKey
,COLUMN_NAME
,cast(iif(is_nullable='NO',0,1) as bit) IsNullable
, DATA_TYPE
, CHARACTER_MAXIMUM_LENGTH
, NUMERIC_PRECISION
from
cte
order by
case CONSTRAINT_TYPE
when 'PRIMARY KEY' then 1
when 'FOREIGN KEY' then 2
else 3 end
, COLUMN_NAME
结果如下：
<table cellspacing=0 border=1>
<tr>
<td style=min-width:50px>PrimaryKey</td>
<td style=min-width:50px>ForeignKey</td>
<td style=min-width:50px>NotKey</td>
<td style=min-width:50px>COLUMN_NAME</td>
<td style=min-width:50px>IsNullable</td>
<td style=min-width:50px>DATA_TYPE</td>
<td style=min-width:50px>CHARACTER_MAXIMUM_LENGTH</td>
<td style=min-width:50px>NUMERIC_PRECISION</td>
</tr>
<tr>
<td style=min-width:50px>1</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>LectureNoteID</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>int</td>
<td style=min-width:50px>NULL</td>
<td style=min-width:50px>10</td>
</tr>
<tr>
<td style=min-width:50px>0</td>
<td style=min-width:50px>1</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>LectureId</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>int</td>
<td style=min-width:50px>NULL</td>
<td style=min-width:50px>10</td>
</tr>
<tr>
<td style=min-width:50px>0</td>
<td style=min-width:50px>1</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>NoteTypeID</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>int</td>
<td style=min-width:50px>NULL</td>
<td style=min-width:50px>10</td>
</tr>
<tr>
<td style=min-width:50px>0</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>1</td>
<td style=min-width:50px>Body</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>nvarchar</td>
<td style=min-width:50px>-1</td>
<td style=min-width:50px>NULL</td>
</tr>
<tr>
<td style=min-width:50px>0</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>1</td>
<td style=min-width:50px>DisplayOrder</td>
<td style=min-width:50px>0</td>
<td style=min-width:50px>int</td>
<td style=min-width:50px>NULL</td>
<td style=min-width:50px>10</td>
</tr>
</table>
于 2018-05-07T07:47:28.650 回答
1
对于给定 TableName 和 Schema 的主键列的逗号分隔列表：
Select distinct SUBSTRING ( stuff(( select distinct ',' + [COLUMN_NAME]
from INFORMATION_SCHEMA.KEY_COLUMN_USAGE
where OBJECTPROPERTY(OBJECT_ID(CONSTRAINT_SCHEMA + '.' + QUOTENAME(CONSTRAINT_NAME)), 'IsPrimaryKey') = 1
AND TABLE_NAME = 'TableName' AND TABLE_SCHEMA = 'Schema'
order by 1 FOR XML PATH(''), TYPE).value('.', 'NVARCHAR(MAX)'),1,0,'' )
,2,9999)
于 2019-08-26T01:13:17.420 回答
0
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'
于 2008-09-18T19:32:07.363 回答
0
我发现这很有用，给出了一个表列表，其中包含一个逗号分隔的列列表，然后还有一个逗号分隔的主键列表
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
于 2015-06-18T23:55:33.087 回答
0
这个版本显示了模式、表名和一个有序的、逗号分隔的主键列表。Object_Id() 不适用于链接服务器，因此我们按表名过滤。
如果没有 REPLACE(Si1.Column_Name, '', '') 它会在我正在测试的数据库上显示 Column_Name 的 xml 开始和结束标记。我不确定为什么数据库需要替换“Column_Name”，所以如果有人知道，请发表评论。
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;
使用 George 查询的相同模式：
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;
于 2017-03-23T19:26:24.420 回答
0
Sys.Objects 表包含每个用户定义的模式范围对象的行。
像主键或其他创建的约束将是对象，表名将是parent_object
查询 sys.Objects 并收集 Object 的所需类型的 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)
于 2017-08-02T05:52:50.097 回答
0
我可以对下面的原始问题提出一个更准确的简单答案吗
SELECT
KEYS.table_schema, KEYS.table_name, KEYS.column_name, KEYS.ORDINAL_POSITION
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE keys
INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS CONS
ON cons.TABLE_SCHEMA = keys.TABLE_SCHEMA
AND cons.TABLE_NAME = keys.TABLE_NAME
AND cons.CONSTRAINT_NAME = keys.CONSTRAINT_NAME
WHERE cons.CONSTRAINT_TYPE = 'PRIMARY KEY'
笔记：
- 上面的一些答案缺少仅用于主键列的过滤器！
- 我在 CTE 中使用下面来加入更大的列列表，以提供来自源的元数据，以提供 BIML 生成的临时表和 SSIS 代码
于 2018-02-01T22:49:31.467 回答
0
可能最近发布，但希望这将帮助某人通过使用此 t-sql 查询来查看 sql server 中的主键列表：
SELECT schema_name(t.schema_id) AS [schema_name], t.name AS TableName,
COL_NAME(ic.OBJECT_ID,ic.column_id) AS PrimaryKeyColumnName,
i.name AS PrimaryKeyConstraintName
FROM sys.tables t
INNER JOIN sys.indexes AS i on t.object_id=i.object_id
INNER JOIN sys.index_columns AS ic ON i.OBJECT_ID = ic.OBJECT_ID
AND i.index_id = ic.index_id
WHERE OBJECT_NAME(ic.OBJECT_ID) = 'YourTableNameHere'
如果您愿意，可以使用此查询查看所有外键的列表：
SELECT
f.name as ForeignKeyConstraintName
,OBJECT_NAME(f.parent_object_id) AS ReferencingTableName
,COL_NAME(fc.parent_object_id, fc.parent_column_id) AS ReferencingColumnName
,OBJECT_NAME (f.referenced_object_id) AS ReferencedTableName
,COL_NAME(fc.referenced_object_id, fc.referenced_column_id) AS
ReferencedColumnName ,delete_referential_action_desc AS
DeleteReferentialActionDesc ,update_referential_action_desc AS
UpdateReferentialActionDesc
FROM sys.foreign_keys AS f
INNER JOIN sys.foreign_key_columns AS fc
ON f.object_id = fc.constraint_object_id
--WHERE OBJECT_NAME(f.parent_object_id) = 'YourTableNameHere'
--If you want to know referecing table details
WHERE OBJECT_NAME(f.referenced_object_id) = 'YourTableNameHere'
--If you want to know refereced table details
ORDER BY f.name
于 2018-04-19T05:16:34.487 回答
0
我从我的朋友那里找到了这个，如果您要在特定模式下查找所有表的主键，这非常有效。
SELECT tc.constraint_name AS IndexName,tc.table_name AS TableName,tc.table_schema
AS SchemaName,kc.column_name AS COLUMN_NAME
FROM information_schema.table_constraints tc,information_schema.key_column_usage kc
WHERE tc.constraint_type = 'PRIMARY KEY' AND kc.table_name = tc.table_name AND kc.table_schema = tc.table_schema
AND kc.constraint_name = tc.constraint_name AND tc.table_schema='<SCHEMA_NAME>'
于 2018-04-25T05:14:58.590 回答
0
如果需要主键和类型，此查询可能有用：
SELECT L.TABLE_SCHEMA, L.TABLE_NAME, L.COLUMN_NAME, R.TypeName
FROM(
SELECT COLUMN_NAME, TABLE_NAME, TABLE_SCHEMA
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE OBJECTPROPERTY(OBJECT_ID(CONSTRAINT_SCHEMA + '.' + QUOTENAME(CONSTRAINT_NAME)), 'IsPrimaryKey') = 1
)L
LEFT JOIN (
SELECT
OBJECT_NAME(c.OBJECT_ID) TableName ,c.name AS ColumnName ,t.name AS TypeName
FROM sys.columns AS c
JOIN sys.types AS t ON c.user_type_id=t.user_type_id
)R ON L.COLUMN_NAME = R.ColumnName AND L.TABLE_NAME = R.TableName
于 2018-07-07T20:37:10.153 回答
0
可能是最简单的解决方案:)
EXEC sp_pkeys YourTable
于 2021-07-12T13:10:13.430 回答
0
如果您在 Oracle 中需要它，那就太简单了。
SELECT `Constraint_Name`
FROM `All_Constraints`
WHERE `Constraint_Type` = `'P'`
AND `Owner` = `'your schema here';`
于 2021-10-12T15:55:29.747 回答