¿Cómo listar la clave principal de una tabla de SQL Server?
Pregunta sencilla, ¿cómo listar la clave principal de una tabla con T-SQL? Sé cómo obtener índices en una tabla, pero no puedo recordar cómo obtener el PK.
24 answers
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>'
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2014-06-04 18:58:29
Es una práctica generalmente recomendada ahora usar las vistas sys.*
sobre INFORMATION_SCHEMA
en SQL Server, así que a menos que esté planeando migrar bases de datos, las usaría. Así es como lo harías con las vistas sys.*
:
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>');
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2015-09-10 20:26:24
Esta es una solución que utiliza solo sys-tablas.
Enumera todas las claves primarias de la base de datos. Devuelve esquema, nombre de la tabla, nombre de la columna y el orden de ordenación correcto de la columna para cada clave primaria.
Si desea obtener la clave principal de una tabla específica, entonces usted necesita para filtro en SchemaName
y TableName
.
En mi humilde opinión, esta solución es muy genérica y no utiliza ningún literal de cadena, por lo que se ejecutará en cualquier equipo.
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 ;
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2016-05-16 09:19:34
Aquí hay otra manera de la pregunta obtener la clave primaria de la tabla usando la consulta sql :
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>'
Utiliza KEY_COLUMN_USAGE
para determinar las restricciones para una tabla dada
Luego utiliza OBJECTPROPERTY(id, 'IsPrimaryKey')
para determinar si cada uno es una clave primaria
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2018-03-27 19:57:16
Está usando MS SQL Server puede hacer lo siguiente:
--List all tables primary keys
select * from information_schema.table_constraints
where constraint_type = 'Primary Key'
También puede filtrar en la columna table_name si desea una tabla específica.
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2011-11-28 21:40:02
Me gusta la técnica INFORMATION_SCHEMA, pero otra que he usado es: exec sp_pkeys 'table'
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2008-10-07 16:06:22
This Esta es otra versión modificada que también es un ejemplo para la consulta Co-Relacionada
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')
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2011-09-26 06:36:00
Esto debería listar todas las restricciones (clave primaria y Claves Foráneas ) y al final de la consulta poner el nombre de la tabla
/* 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;
Para referencia, por favor lea thru - http://blogs.msdn.com/b/sqltips/archive/2005/09/16/469136.aspx
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2013-04-21 06:08:03
Gracias Guy.
Con una ligera variación lo utilicé para encontrar todas las claves primarias para todas las tablas.
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
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2010-01-20 02:57:10
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
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2012-01-26 09:22:03
Este te da las columnas que son PK.
SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE TABLE_NAME = 'TableName'
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2015-06-27 00:17:40
El procedimiento almacenado en el sistema sp_help
le dará la información. Ejecute la siguiente instrucción:
execute sp_help table_name
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2015-12-04 13:05:06
Encontré esto útil, da una lista de tablas con una lista separada por comas de las columnas y luego también una lista separada por comas de cuáles son la clave primaria
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
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2017-03-23 19:56:19
A continuación la consulta enumerará claves primarias de tabla particular :
SELECT DISTINCT
CONSTRAINT_NAME AS [Constraint],
TABLE_SCHEMA AS [Schema],
TABLE_NAME AS TableName
FROM
INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE
TABLE_NAME = 'mytablename'
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2017-04-05 13:45:48
Estoy diciendo una técnica simple que sigo
SP_HELP 'table_name'
Ejecute este código como query. Mencione el nombre de su tabla en el lugar de nombre_dela tabla para la que desea saber la Clave primaria (no olvide las comillas simples). El resultado se mostrará como imagen adjunta. Espero que te ayude
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2017-12-01 22:26:02
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
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2014-03-08 22:49:05
Pruebe esto:
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'
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2016-01-06 03:36:35
Esta versión muestra el esquema, el nombre de la tabla y una lista ordenada de claves primarias separadas por comas. Object_Id () no funciona para servidores de enlaces, por lo que filtramos por el nombre de la tabla.
Sin el REEMPLAZO(Si1.Column_Name,",") mostraría las etiquetas xml de apertura y cierre para Column_Name en la base de datos en la que estaba probando. No estoy seguro de por qué la base de datos requiere un reemplazo para 'Column_Name', así que si alguien lo sabe, por favor comente.
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;
Y el mismo patrón usando La pregunta de George:
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;
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2017-03-23 19:51:23
Sys.La tabla Objects contiene una fila para cada ámbito de esquema definido por el usuario objeto .
Las restricciones creadas como Clave primaria u otras serán el objeto y Nombre de la tabla será el parent_object
Consulta sys.Objetos y recopilar los Identificadores del Objeto del Tipo Requerido
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)
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2017-08-02 05:52:50
Puedo sugerir una respuesta más precisa y sencilla a la pregunta original a continuación
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'
Notas:
- A algunas de las respuestas anteriores les falta un filtro solo para la clave primaria columnas!
- Estoy usando abajo en un CTE para unirse a una columna más grande listado para proporcionar los metadatos de una fuente para alimentar la generación BIML de tablas de estadificación y código SSIS
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2018-02-01 23:09:32
Podría publicarse últimamente, pero esperamos que esto ayude a alguien a ver la lista de claves primarias en sql server mediante el uso de esta consulta t-sql:
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'
Puede ver la lista de todas las claves foráneas utilizando esta consulta si lo desea:
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
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2018-04-19 05:16:34
Encontré esto de mi amigo, muy efectivo si está buscando todas las claves principales de la tabla bajo un esquema particular.
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>'
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2018-04-25 05:14:58
Si está buscando hacer su propio OR o generar código a partir de una tabla dada, entonces esto podría ser lo que está buscando:
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
Así es como se vería el resultado:
<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>
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/ajaxhispano.com/template/agent.layouts/content.php on line 61
2018-05-07 07:47:28
Si se necesita Clave primaria y tipo, esta consulta puede ser útil:
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
Warning: date(): Invalid date.timezone value 'Europe/Kyiv', we selected the timezone 'UTC' for now. in /var/www/agent_stack/data/www/ajaxhispano.com/template/agent.layouts/content.php on line 61
2018-07-07 20:42:41