Consulta del servidor SQL para obtener la lista de columnas en una tabla junto con los tipos de datos, NOT NULL y restricciones PRIMARY KEY

Necesito escribir una consulta en el servidor SQL para obtener la lista de columnas en una tabla en particular, sus tipos de datos asociados (con longitud) y si no son nulos. Y me las he arreglado para hacer esto.

Pero ahora también necesito obtener, en la misma tabla, contra una columna, TRUEsi esa columna es una clave principal.

¿Cómo hago esto?

Mi salida esperada es:

Column name | Data type | Length | isnull | Pk

Para evitar filas duplicadas para algunas columnas, use user_type_id en lugar de system_type_id.

SELECT 
    c.name 'Column Name',
    t.Name 'Data type',
    c.max_length 'Max Length',
    c.precision ,
    c.scale ,
    c.is_nullable,
    ISNULL(i.is_primary_key, 0) 'Primary Key'
FROM    
    sys.columns c
INNER JOIN 
    sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
    sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
    sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
WHERE
    c.object_id = OBJECT_ID('YourTableName')

Simplemente reemplácelo YourTableNamecon el nombre real de la tabla: funciona para SQL Server 2005 y versiones posteriores.

En caso de que esté utilizando esquemas, reemplace YourTableName por YourSchemaName.YourTableNamedonde YourSchemaNameestá el nombre real del esquema y YourTableNameel nombre real de la tabla.

El procedimiento almacenado sp_columns devuelve información detallada de la tabla.

exec sp_columns MyTable

Puedes usar la consulta:

select COLUMN_NAME, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH, 
       NUMERIC_PRECISION, DATETIME_PRECISION, 
       IS_NULLABLE 
from INFORMATION_SCHEMA.COLUMNS
where TABLE_NAME='TableName'

para obtener todos los metadatos que necesita, excepto la información de Pk.

En SQL 2012 puedes usar:

EXEC sp_describe_first_result_set N'SELECT * FROM [TableName]'

Esto le dará los nombres de columna junto con sus propiedades.

Prueba esto:

select COLUMN_NAME, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH, IS_NULLABLE 
from INFORMATION_SCHEMA.COLUMNS IC
where TABLE_NAME = 'tablename' and COLUMN_NAME = 'columnname'

Para asegurarse de obtener la longitud correcta, deberá considerar los tipos Unicode como un caso especial. Ver el código a continuación.

Para obtener más información, consulte: https://msdn.microsoft.com/en-us/library/ms176106.aspx

SELECT 
   c.name 'Column Name',
   t.name,
   t.name +
   CASE WHEN t.name IN ('char', 'varchar','nchar','nvarchar') THEN '('+

             CASE WHEN c.max_length=-1 THEN 'MAX'

                  ELSE CONVERT(VARCHAR(4),

                               CASE WHEN t.name IN ('nchar','nvarchar')

                               THEN  c.max_length/2 ELSE c.max_length END )

                  END +')'

          WHEN t.name IN ('decimal','numeric')

                  THEN '('+ CONVERT(VARCHAR(4),c.precision)+','

                          + CONVERT(VARCHAR(4),c.Scale)+')'

                  ELSE '' END

   as "DDL name",
   c.max_length 'Max Length in Bytes',
   c.precision ,
   c.scale ,
   c.is_nullable,
   ISNULL(i.is_primary_key, 0) 'Primary Key'
FROM    
   sys.columns c
INNER JOIN 
   sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
   sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
   sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
WHERE
   c.object_id = OBJECT_ID('YourTableName')

Ampliando la respuesta de Alex, puede hacer esto para obtener la restricción PK

Select C.COLUMN_NAME, C.DATA_TYPE, C.CHARACTER_MAXIMUM_LENGTH, C.NUMERIC_PRECISION, C.IS_NULLABLE, TC.CONSTRAINT_NAME
From INFORMATION_SCHEMA.COLUMNS As C
    Left Join INFORMATION_SCHEMA.TABLE_CONSTRAINTS As TC
      On TC.TABLE_SCHEMA = C.TABLE_SCHEMA
          And TC.TABLE_NAME = C.TABLE_NAME
          And TC.CONSTRAINT_TYPE = 'PRIMARY KEY'
Where C.TABLE_NAME = 'Table'

Debo haber pasado por alto que desea una bandera para determinar si la columna dada era parte de la PK en lugar del nombre de la restricción PK. Para eso usarías:

Select C.COLUMN_NAME, C.DATA_TYPE, C.CHARACTER_MAXIMUM_LENGTH
    , C.NUMERIC_PRECISION, C.NUMERIC_SCALE
    , C.IS_NULLABLE
    , Case When Z.CONSTRAINT_NAME Is Null Then 0 Else 1 End As IsPartOfPrimaryKey
From INFORMATION_SCHEMA.COLUMNS As C
    Outer Apply (
                Select CCU.CONSTRAINT_NAME
                From INFORMATION_SCHEMA.TABLE_CONSTRAINTS As TC
                    Join INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE As CCU
                        On CCU.CONSTRAINT_NAME = TC.CONSTRAINT_NAME
                Where TC.TABLE_SCHEMA = C.TABLE_SCHEMA
                    And TC.TABLE_NAME = C.TABLE_NAME
                    And TC.CONSTRAINT_TYPE = 'PRIMARY KEY'
                    And CCU.COLUMN_NAME = C.COLUMN_NAME
                ) As Z
Where C.TABLE_NAME = 'Table'

wite el nombre de la tabla en el editor de consultas seleccione el nombre y presione Alt + F1 y traerá toda la información de la tabla.

SELECT COLUMN_NAME, IS_NULLABLE, DATA_TYPE, CHARACTER_MAXIMUM_LENGTH FROM information_schema.columns WHERE table_name = '<name_of_table_or_view>'

Ejecute SELECT *en la declaración anterior para ver qué información_esquema.columnas devuelve.

Esta pregunta ha sido respondida previamente: https://stackoverflow.com/a/11268456/6169225

Estoy un poco sorprendido de que nadie mencione

sp_help 'mytable'

IF EXISTS (SELECT 1 FROM INFORMATION_SCHEMA.TABLES 
     WHERE TABLE_TYPE = 'BASE TABLE' AND TABLE_NAME = 'Table')
      BEGIN
        SELECT COLS.COLUMN_NAME, COLS.DATA_TYPE, COLS.CHARACTER_MAXIMUM_LENGTH, 
              (SELECT 'Yes' FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS TC JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE KCU
                              ON COLS.TABLE_NAME = TC.TABLE_NAME 
                             AND TC.CONSTRAINT_TYPE = 'PRIMARY KEY'
                             AND KCU.TABLE_NAME = TC.TABLE_NAME
                             AND KCU.CONSTRAINT_NAME = TC.CONSTRAINT_NAME
                             AND KCU.COLUMN_NAME = COLS.COLUMN_NAME) AS KeyX
        FROM INFORMATION_SCHEMA.COLUMNS COLS WHERE TABLE_NAME = 'Table' ORDER BY KeyX DESC, COLUMN_NAME
      END

Lanzando otra respuesta al ring, esto te dará esas columnas y más:

SELECT col.TABLE_CATALOG AS [Database]
     , col.TABLE_SCHEMA AS Owner
     , col.TABLE_NAME AS TableName
     , col.COLUMN_NAME AS ColumnName
     , col.ORDINAL_POSITION AS OrdinalPosition
     , col.COLUMN_DEFAULT AS DefaultSetting
     , col.DATA_TYPE AS DataType
     , col.CHARACTER_MAXIMUM_LENGTH AS MaxLength
     , col.DATETIME_PRECISION AS DatePrecision
     , CAST(CASE col.IS_NULLABLE
                WHEN 'NO' THEN 0
                ELSE 1
            END AS bit)AS IsNullable
     , COLUMNPROPERTY(OBJECT_ID('[' + col.TABLE_SCHEMA + '].[' + col.TABLE_NAME + ']'), col.COLUMN_NAME, 'IsIdentity')AS IsIdentity
     , COLUMNPROPERTY(OBJECT_ID('[' + col.TABLE_SCHEMA + '].[' + col.TABLE_NAME + ']'), col.COLUMN_NAME, 'IsComputed')AS IsComputed
     , CAST(ISNULL(pk.is_primary_key, 0)AS bit)AS IsPrimaryKey
  FROM INFORMATION_SCHEMA.COLUMNS AS col
       LEFT JOIN(SELECT SCHEMA_NAME(o.schema_id)AS TABLE_SCHEMA
                      , o.name AS TABLE_NAME
                      , c.name AS COLUMN_NAME
                      , i.is_primary_key
                   FROM sys.indexes AS i JOIN sys.index_columns AS ic ON i.object_id = ic.object_id
                                                                     AND i.index_id = ic.index_id
                                         JOIN sys.objects AS o ON i.object_id = o.object_id
                                         LEFT JOIN sys.columns AS c ON ic.object_id = c.object_id
                                                                   AND c.column_id = ic.column_id
                  WHERE i.is_primary_key = 1)AS pk ON col.TABLE_NAME = pk.TABLE_NAME
                                                  AND col.TABLE_SCHEMA = pk.TABLE_SCHEMA
                                                  AND col.COLUMN_NAME = pk.COLUMN_NAME
 WHERE col.TABLE_NAME = 'YourTableName'
   AND col.TABLE_SCHEMA = 'dbo'
 ORDER BY col.TABLE_NAME, col.ORDINAL_POSITION;

select
      c.name as [column name], 
      t.name as [type name],
      tbl.name as [table name]
from sys.columns c
         inner join sys.types t 
      on c.system_type_id = t.system_type_id 
         inner join sys.tables tbl
      on c.object_id = tbl.object_id
where
      c.object_id = OBJECT_ID('YourTableName1') 
          and 
      t.name like '%YourSearchDataType%'
union
(select
      c.name as [column name], 
      t.name as [type name],
      tbl.name as [table name]
from sys.columns c
         inner join sys.types t 
      on c.system_type_id = t.system_type_id 
         inner join sys.tables tbl
      on c.object_id = tbl.object_id
where
      c.object_id = OBJECT_ID('YourTableName2') 
          and 
      t.name like '%YourSearchDataType%')
union
(select
      c.name as [column name], 
      t.name as [type name],
      tbl.name as [table name]
from sys.columns c
         inner join sys.types t 
      on c.system_type_id = t.system_type_id 
         inner join sys.tables tbl
      on c.object_id = tbl.object_id
where
      c.object_id = OBJECT_ID('YourTableName3') 
          and 
      t.name like '%YourSearchDataType%')
order by tbl.name

Para buscar qué columna está en qué tabla en función de su tipo de datos de búsqueda para tres tablas diferentes en una base de datos. Esta consulta se puede expandir a 'n' tablas.

Encuentra el resultado combinado para Datatype and Length y es anulable en forma de "NULL" y "Not null". Use la consulta a continuación.

SELECT c.name AS 'Column Name',
       t.name + '(' + cast(c.max_length as varchar(50)) + ')' As 'DataType',
       case 
         WHEN  c.is_nullable = 0 then 'null' else 'not null'
         END AS 'Constraint'
  FROM sys.columns c
  JOIN sys.types t
    ON c.user_type_id = t.user_type_id
 WHERE c.object_id    = Object_id('TableName')

Encontrará el resultado como se muestra a continuación.

Gracias.

SELECT  
   T.NAME AS [TABLE NAME]
   ,C.NAME AS [COLUMN NAME]
   ,P.NAME AS [DATA TYPE]
   ,P.MAX_LENGTH AS [Max_SIZE]
   ,C.[max_length] AS [ActualSizeUsed]
   ,CAST(P.PRECISION AS VARCHAR) +'/'+ CAST(P.SCALE AS VARCHAR) AS [PRECISION/SCALE]
FROM SYS.OBJECTS AS T
JOIN SYS.COLUMNS AS C
    ON T.OBJECT_ID = C.OBJECT_ID
JOIN SYS.TYPES AS P
    ON C.SYSTEM_TYPE_ID = P.SYSTEM_TYPE_ID
    AND C.[user_type_id] = P.[user_type_id]
WHERE T.TYPE_DESC='USER_TABLE'
  AND T.name = 'InventoryStatus'
ORDER BY 2

Consulta: EXEC SP_DESCRIBE_FIRST_RESULT_SET N'SELECT ANNUAL_INCOME FROM [BSLID2C]. [DBO]. [EMPLOYEE] '

NOTA: EN ALGUNA IDE ANTES DE SELECCIONAR N ESTÁ FUNCIONANDO O, EN ALGUNA IDE SIN N ESTÁ FUNCIONANDO

No hay una clave principal aquí, pero esto puede ayudar a otros usuarios que desean tener un nombre de tabla con el nombre del campo y las propiedades básicas del campo.

USE [**YourDB**]
GO
SELECT tbl.name, fld.[Column Name],fld.[Constraint],fld.DataType 
FROM sys.all_objects as tbl left join 
(SELECT c.OBJECT_ID,  c.name AS 'Column Name',
       t.name + '(' + cast(c.max_length as varchar(50)) + ')' As 'DataType',
       case 
         WHEN  c.is_nullable = 0 then 'null' else 'not null'
         END AS 'Constraint'
  FROM sys.columns c
  JOIN sys.types t
    ON c.user_type_id = t.user_type_id
) as fld on tbl.OBJECT_ID = fld.OBJECT_ID
WHERE ( tbl.[type]='U' and tbl.[is_ms_shipped] = 0)
ORDER BY tbl.[name],fld.[Column Name]
GO

Acabo de hacer marc_s "presentación lista":

SELECT 
    c.name 'Column Name',
    t.name 'Data type',
    IIF(t.name = 'nvarchar', c.max_length / 2, c.max_length) 'Max Length',
    c.precision 'Precision',
    c.scale 'Scale',
    IIF(c.is_nullable = 0, 'No', 'Yes') 'Nullable',
    IIF(ISNULL(i.is_primary_key, 0) = 0, 'No', 'Yes') 'Primary Key'
FROM    
    sys.columns c
INNER JOIN 
    sys.types t ON c.user_type_id = t.user_type_id
LEFT OUTER JOIN 
    sys.index_columns ic ON ic.object_id = c.object_id AND ic.column_id = c.column_id
LEFT OUTER JOIN 
    sys.indexes i ON ic.object_id = i.object_id AND ic.index_id = i.index_id
WHERE
    c.object_id = OBJECT_ID('YourTableName')