Ik heb een SQL Server-database en ik wil weten welke kolommen en typen deze heeft. Ik zou dit liever doen via een query in plaats van een GUI zoals Enterprise Manager te gebruiken. Is er een manier om dit te doen?
Antwoord 1, autoriteit 100%
U kunt de sp_columnsopgeslagen procedure gebruiken:
exec sp_columns MyTable
Antwoord 2, autoriteit 37%
Er zijn een paar methoden om metadata over een tabel te verkrijgen:
EXEC sp_help tablename
Retourneert verschillende resultaatsets, die de tabel, de kolommen en beperkingen beschrijven.
De INFORMATION_SCHEMAweergaven geven je de informatie die je wilt, maar helaas moet je de weergaven opvragen en handmatig toevoegen.
Antwoord 3, autoriteit 15%
Voor het geval je de opgeslagen procedure niet wilt gebruiken, hier is een eenvoudige queryversie
select *
from information_schema.columns
where table_name = 'aspnet_Membership'
order by ordinal_position
Antwoord 4, autoriteit 11%
U kunt het volgende gebruiken: sp_help tablename
Voorbeeld: sp_help Customer
OF Sneltoetsen gebruiken
Selectde gewenste tafel en druk op ALT+F1.
Voorbeeld: klant drukt op ALT+F1.
Antwoord 5, autoriteit 8%
Gebruik deze zoekopdracht
Select * From INFORMATION_SCHEMA.COLUMNS Where TABLE_NAME = 'TABLENAME'
Antwoord 6, autoriteit 4%
Gebruik de volgende sql-query; dit werkte voor mijn geval.
select * FROM INFORMATION_SCHEMA.Columns where table_name = 'tablename';
Antwoord 7, autoriteit 4%
Naast de manieren die in andere antwoorden worden getoond, kunt u
. gebruiken
SELECT TOP 0 * FROM table_name
Hierdoor krijgt u de naam van elke kolom zonder resultaten, en wordt vrijwel onmiddellijk voltooid met minimale overhead.
Antwoord 8, autoriteit 3%
Selecteer gewoon een tafel en druk op Alt+F1,
het zal alle informatie over de tabel tonen, zoals kolomnaam, datatype, sleutels enz.
Antwoord 9, autoriteit 2%
Ik schreef een sql*plus DESC(RIBE) zoals select (geeft ook de kolomopmerkingen weer) in t-sql:
USE YourDB
GO
DECLARE @objectName NVARCHAR(128) = 'YourTable';
SELECT
a.[NAME]
,a.[TYPE]
,a.[CHARSET]
,a.[COLLATION]
,a.[NULLABLE]
,a.[DEFAULT]
,b.[COMMENTS]
-- ,a.[ORDINAL_POSITION]
FROM
(
SELECT
COLUMN_NAME AS [NAME]
,CASE DATA_TYPE
WHEN 'char' THEN DATA_TYPE + '(' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + ')'
WHEN 'numeric' THEN DATA_TYPE + '(' + CAST(NUMERIC_PRECISION AS VARCHAR) + ', ' + CAST(NUMERIC_SCALE AS VARCHAR) + ')'
WHEN 'nvarchar' THEN DATA_TYPE + '(' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + ')'
WHEN 'varbinary' THEN DATA_TYPE + '(' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + ')'
WHEN 'varchar' THEN DATA_TYPE + '(' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + ')'
ELSE DATA_TYPE
END AS [TYPE]
,CHARACTER_SET_NAME AS [CHARSET]
,COLLATION_NAME AS [COLLATION]
,IS_NULLABLE AS [NULLABLE]
,COLUMN_DEFAULT AS [DEFAULT]
,ORDINAL_POSITION
FROM
INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = @objectName
) a
FULL JOIN
(
SELECT
CAST(value AS NVARCHAR) AS [COMMENTS]
,CAST(objname AS NVARCHAR) AS [NAME]
FROM
::fn_listextendedproperty ('MS_Description', 'user', 'dbo', 'table', @objectName, 'column', default)
) b
ON a.NAME COLLATE YourCollation = b.NAME COLLATE YourCollation
ORDER BY
a.[ORDINAL_POSITION];
De hierboven genoemde selecte select kan worden gebruikt in een gemarkeerde opgeslagen procedure van het systeem en het kan op een eenvoudige manier vanuit elke database van uw exemplaar worden genoemd:
USE master;
GO
IF OBJECT_ID('sp_desc', 'P') IS NOT NULL
DROP PROCEDURE sp_desc
GO
CREATE PROCEDURE sp_desc (
@tableName nvarchar(128)
) AS
BEGIN
DECLARE @dbName sysname;
DECLARE @schemaName sysname;
DECLARE @objectName sysname;
DECLARE @objectID int;
DECLARE @tmpTableName varchar(100);
DECLARE @sqlCmd nvarchar(4000);
SELECT @dbName = PARSENAME(@tableName, 3);
IF @dbName IS NULL SELECT @dbName = DB_NAME();
SELECT @schemaName = PARSENAME(@tableName, 2);
IF @schemaName IS NULL SELECT @schemaName = SCHEMA_NAME();
SELECT @objectName = PARSENAME(@tableName, 1);
IF @objectName IS NULL
BEGIN
PRINT 'Object is missing from your function call!';
RETURN;
END;
SELECT @objectID = OBJECT_ID(@dbName + '.' + @schemaName + '.' + @objectName);
IF @objectID IS NULL
BEGIN
PRINT 'Object [' + @dbName + '].[' + @schemaName + '].[' + @objectName + '] does not exist!';
RETURN;
END;
SELECT @tmpTableName = '#tmp_DESC_' + CAST(@@SPID AS VARCHAR) + REPLACE(REPLACE(REPLACE(REPLACE(CAST(CONVERT(CHAR, GETDATE(), 121) AS VARCHAR), '-', ''), ' ', ''), ':', ''), '.', '');
--PRINT @tmpTableName;
SET @sqlCmd = '
USE ' + @dbName + '
CREATE TABLE ' + @tmpTableName + ' (
[NAME] nvarchar(128) NOT NULL
,[TYPE] varchar(50)
,[CHARSET] varchar(50)
,[COLLATION] varchar(50)
,[NULLABLE] varchar(3)
,[DEFAULT] nvarchar(4000)
,[COMMENTS] nvarchar(3750));
INSERT INTO ' + @tmpTableName + '
SELECT
a.[NAME]
,a.[TYPE]
,a.[CHARSET]
,a.[COLLATION]
,a.[NULLABLE]
,a.[DEFAULT]
,b.[COMMENTS]
FROM
(
SELECT
COLUMN_NAME AS [NAME]
,CASE DATA_TYPE
WHEN ''char'' THEN DATA_TYPE + ''('' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + '')''
WHEN ''numeric'' THEN DATA_TYPE + ''('' + CAST(NUMERIC_PRECISION AS VARCHAR) + '', '' + CAST(NUMERIC_SCALE AS VARCHAR) + '')''
WHEN ''nvarchar'' THEN DATA_TYPE + ''('' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + '')''
WHEN ''varbinary'' THEN DATA_TYPE + ''('' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + '')''
WHEN ''varchar'' THEN DATA_TYPE + ''('' + CAST(CHARACTER_MAXIMUM_LENGTH AS VARCHAR) + '')''
ELSE DATA_TYPE
END AS [TYPE]
,CHARACTER_SET_NAME AS [CHARSET]
,COLLATION_NAME AS [COLLATION]
,IS_NULLABLE AS [NULLABLE]
,COLUMN_DEFAULT AS [DEFAULT]
,ORDINAL_POSITION
FROM
INFORMATION_SCHEMA.COLUMNS
WHERE
TABLE_NAME = ''' + @objectName + '''
) a
FULL JOIN
(
SELECT
CAST(value AS NVARCHAR) AS [COMMENTS]
,CAST(objname AS NVARCHAR) AS [NAME]
FROM
::fn_listextendedproperty (''MS_Description'', ''user'', ''' + @schemaName + ''', ''table'', ''' + @objectName + ''', ''column'', default)
) b
ON a.NAME COLLATE Hungarian_CI_AS = b.NAME COLLATE Hungarian_CI_AS
ORDER BY
a.[ORDINAL_POSITION];
SELECT * FROM ' + @tmpTableName + ';'
--PRINT @sqlCmd;
EXEC sp_executesql @sqlCmd;
RETURN;
END;
GO
EXEC sys.sp_MS_marksystemobject sp_desc
GO
Om het proceduretype uit te voeren:
EXEC sp_desc 'YourDB.YourSchema.YourTable';
Als u een beschrijving van een object van de huidige database (en schema) wilt, eenvoudig type:
EXEC sp_desc 'YourTable';
Aangezien sp_desc een door het systeem gemarkeerde procedure is, kunt u zelfs het exec-commando verlaten (hoe dan ook niet aanbevolen):
sp_desc 'YourTable';
Antwoord 10, autoriteit 2%
Het SQL Server-equivalent van Oracle’s describe-commando is het opgeslagen proc sp_help
Het describecommando geeft je de informatie over de kolomnamen, typen, lengte, etc.
In SQL Server, laten we zeggen dat u een tabel ‘mytable’ in schema ‘myschema’ in de database ‘mydb’ wilt beschrijven, kunt u het volgende doen:
USE mydb;
exec sp_help 'myschema.mytable';
Antwoord 11, autoriteit 2%
U kunt de sp_help ‘TableName’ gebruiken
Antwoord 12
probeer het:
EXEC [ServerName].[DatabaseName].dbo.sp_columns 'TableName'
en u kunt informatie over de tabelstructuur krijgen, zoals:
TABLE_QUALIFIER, TABLE_OWNER, TABLE_NAME, COLUMN_NAME, DATA_TYPE, TYPE_NAME…
Antwoord 13
Het probleem met die antwoorden is dat je de belangrijkste informatie mist.
Hoewel dit een beetje rommelig is, is dit een snelle versie die ik heb bedacht om ervoor te zorgen dat deze dezelfde informatie bevat als de MySQL Describe-weergave.
Select SC.name AS 'Field', ISC.DATA_TYPE AS 'Type', ISC.CHARACTER_MAXIMUM_LENGTH AS 'Length', SC.IS_NULLABLE AS 'Null', I.is_primary_key AS 'Key', SC.is_identity AS 'Identity'
From sys.columns AS SC
LEFT JOIN sys.index_columns AS IC
ON IC.object_id = OBJECT_ID('dbo.Expenses') AND
IC.column_id = SC.column_id
LEFT JOIN sys.indexes AS I
ON I.object_id = OBJECT_ID('dbo.Expenses') AND
IC.index_id = I.index_id
LEFT JOIN information_schema.columns ISC
ON ISC.TABLE_NAME = 'Expenses'
AND ISC.COLUMN_NAME = SC.name
WHERE SC.object_id = OBJECT_ID('dbo.Expenses')
Antwoord 14
Dit is de code die ik gebruik in de EntityFramework Reverse POCO Generator(beschikbaar hier)
Tabel-SQL:
SELECT c.TABLE_SCHEMA AS SchemaName,
c.TABLE_NAME AS TableName,
t.TABLE_TYPE AS TableType,
c.ORDINAL_POSITION AS Ordinal,
c.COLUMN_NAME AS ColumnName,
CAST(CASE WHEN IS_NULLABLE = 'YES' THEN 1
ELSE 0
END AS BIT) AS IsNullable,
DATA_TYPE AS TypeName,
ISNULL(CHARACTER_MAXIMUM_LENGTH, 0) AS [MaxLength],
CAST(ISNULL(NUMERIC_PRECISION, 0) AS INT) AS [Precision],
ISNULL(COLUMN_DEFAULT, '') AS [Default],
CAST(ISNULL(DATETIME_PRECISION, 0) AS INT) AS DateTimePrecision,
ISNULL(NUMERIC_SCALE, 0) AS Scale,
CAST(COLUMNPROPERTY(OBJECT_ID(QUOTENAME(c.TABLE_SCHEMA) + '.' + QUOTENAME(c.TABLE_NAME)), c.COLUMN_NAME, 'IsIdentity') AS BIT) AS IsIdentity,
CAST(CASE WHEN COLUMNPROPERTY(OBJECT_ID(QUOTENAME(c.TABLE_SCHEMA) + '.' + QUOTENAME(c.TABLE_NAME)), c.COLUMN_NAME, 'IsIdentity') = 1 THEN 1
WHEN COLUMNPROPERTY(OBJECT_ID(QUOTENAME(c.TABLE_SCHEMA) + '.' + QUOTENAME(c.TABLE_NAME)), c.COLUMN_NAME, 'IsComputed') = 1 THEN 1
WHEN DATA_TYPE = 'TIMESTAMP' THEN 1
ELSE 0
END AS BIT) AS IsStoreGenerated,
CAST(CASE WHEN pk.ORDINAL_POSITION IS NULL THEN 0
ELSE 1
END AS BIT) AS PrimaryKey,
ISNULL(pk.ORDINAL_POSITION, 0) PrimaryKeyOrdinal,
CAST(CASE WHEN fk.COLUMN_NAME IS NULL THEN 0
ELSE 1
END AS BIT) AS IsForeignKey
FROM INFORMATION_SCHEMA.COLUMNS c
LEFT OUTER JOIN (SELECT u.TABLE_SCHEMA,
u.TABLE_NAME,
u.COLUMN_NAME,
u.ORDINAL_POSITION
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE u
INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS tc
ON u.TABLE_SCHEMA = tc.CONSTRAINT_SCHEMA
AND u.TABLE_NAME = tc.TABLE_NAME
AND u.CONSTRAINT_NAME = tc.CONSTRAINT_NAME
WHERE CONSTRAINT_TYPE = 'PRIMARY KEY') pk
ON c.TABLE_SCHEMA = pk.TABLE_SCHEMA
AND c.TABLE_NAME = pk.TABLE_NAME
AND c.COLUMN_NAME = pk.COLUMN_NAME
LEFT OUTER JOIN (SELECT DISTINCT
u.TABLE_SCHEMA,
u.TABLE_NAME,
u.COLUMN_NAME
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE u
INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS tc
ON u.TABLE_SCHEMA = tc.CONSTRAINT_SCHEMA
AND u.TABLE_NAME = tc.TABLE_NAME
AND u.CONSTRAINT_NAME = tc.CONSTRAINT_NAME
WHERE CONSTRAINT_TYPE = 'FOREIGN KEY') fk
ON c.TABLE_SCHEMA = fk.TABLE_SCHEMA
AND c.TABLE_NAME = fk.TABLE_NAME
AND c.COLUMN_NAME = fk.COLUMN_NAME
INNER JOIN INFORMATION_SCHEMA.TABLES t
ON c.TABLE_SCHEMA = t.TABLE_SCHEMA
AND c.TABLE_NAME = t.TABLE_NAME
WHERE c.TABLE_NAME NOT IN ('EdmMetadata', '__MigrationHistory')
SQLL SQL:
SELECT FK.name AS FK_Table,
FkCol.name AS FK_Column,
PK.name AS PK_Table,
PkCol.name AS PK_Column,
OBJECT_NAME(f.object_id) AS Constraint_Name,
SCHEMA_NAME(FK.schema_id) AS fkSchema,
SCHEMA_NAME(PK.schema_id) AS pkSchema,
PkCol.name AS primarykey,
k.constraint_column_id AS ORDINAL_POSITION
FROM sys.objects AS PK
INNER JOIN sys.foreign_keys AS f
INNER JOIN sys.foreign_key_columns AS k
ON k.constraint_object_id = f.object_id
INNER JOIN sys.indexes AS i
ON f.referenced_object_id = i.object_id
AND f.key_index_id = i.index_id
ON PK.object_id = f.referenced_object_id
INNER JOIN sys.objects AS FK
ON f.parent_object_id = FK.object_id
INNER JOIN sys.columns AS PkCol
ON f.referenced_object_id = PkCol.object_id
AND k.referenced_column_id = PkCol.column_id
INNER JOIN sys.columns AS FkCol
ON f.parent_object_id = FkCol.object_id
AND k.parent_column_id = FkCol.column_id
ORDER BY FK_Table, FK_Column
Uitgebreide eigenschappen:
SELECT s.name AS [schema],
t.name AS [table],
c.name AS [column],
value AS [property]
FROM sys.extended_properties AS ep
INNER JOIN sys.tables AS t
ON ep.major_id = t.object_id
INNER JOIN sys.schemas AS s
ON s.schema_id = t.schema_id
INNER JOIN sys.columns AS c
ON ep.major_id = c.object_id
AND ep.minor_id = c.column_id
WHERE class = 1
ORDER BY t.name
Antwoord 15
In aanvulling op bovenstaande vragen, als we een tabel in DB hebben zoals db_name.dbo.table_name, kunnen we de volgende stappen gebruiken
-
Verbinden met DB
USE db_name; -
Gebruik
EXEC sp_helpen vergeet niet de tabelnaam in te voeren als'dbo.tablename'als udboals schema.exec sp_help 'dbo.table_name'
Dit zou moeten werken!
Antwoord 16
Ik heb dit geprobeerd en het werkt voor mij
exec sp_help TABLE_NAME
Antwoord 17
gebruik
SELECT COL_LENGTH('tablename', 'colname')
Geen enkele andere oplossing werkte voor mij.
Antwoord 18
Ik hou van dit formaat:
name DataType Collation Constraints PK FK Comment
id int NOT NULL IDENTITY PK Order Line Id
pid int NOT NULL tbl_orders Order Id
itemCode varchar(10) Latin1_General_CI_AS NOT NULL Product Code
Dus ik heb dit gebruikt:
DECLARE @tname varchar(100) = 'yourTableName';
SELECT col.name,
CASE typ.name
WHEN 'nvarchar' THEN 'nvarchar('+CAST((col.max_length / 2) as varchar)+')'
WHEN 'varchar' THEN 'varchar('+CAST(col.max_length as varchar)+')'
WHEN 'char' THEN 'char('+CAST(col.max_length as varchar)+')'
WHEN 'nchar' THEN 'nchar('+CAST((col.max_length / 2) as varchar)+')'
WHEN 'binary' THEN 'binary('+CAST(col.max_length as varchar)+')'
WHEN 'varbinary' THEN 'varbinary('+CAST(col.max_length as varchar)+')'
WHEN 'numeric' THEN 'numeric('+CAST(col.precision as varchar)+(CASE WHEN col.scale = 0 THEN '' ELSE ','+CAST(col.scale as varchar) END) +')'
WHEN 'decimal' THEN 'decimal('+CAST(col.precision as varchar)+(CASE WHEN col.scale = 0 THEN '' ELSE ','+CAST(col.scale as varchar) END) +')'
ELSE typ.name
END DataType,
ISNULL(col.collation_name,'') Collation,
CASE WHEN col.is_nullable = 0 THEN 'NOT NULL ' ELSE '' END + CASE WHEN col.is_identity = 1 THEN 'IDENTITY' ELSE '' END Constraints,
ISNULL((SELECT 'PK'
FROM sys.key_constraints kc INNER JOIN
sys.tables tb ON tb.object_id = kc.parent_object_id INNER JOIN
sys.indexes si ON si.name = kc.name INNER JOIN
sys.index_columns sic ON sic.index_id = si.index_id AND sic.object_id = si.object_id
WHERE kc.type = 'PK'
AND tb.name = @tname
AND sic.column_id = col.column_id),'') PK,
ISNULL((SELECT (SELECT name FROM sys.tables st WHERE st.object_id = fkc.referenced_object_id)
FROM sys.foreign_key_columns fkc INNER JOIN
sys.columns c ON c.column_id = fkc.parent_column_id AND fkc.parent_object_id = c.object_id INNER JOIN
sys.tables t ON t.object_id = c.object_id
WHERE t.name = tab.name
AND c.name = col.name),'') FK,
ISNULL((SELECT value
FROM sys.extended_properties
WHERE major_id = tab.object_id
AND minor_id = col.column_id),'') Comment
FROM sys.columns col INNER JOIN
sys.tables tab ON tab.object_id = col.object_id INNER JOIN
sys.types typ ON typ.system_type_id = col.system_type_id
WHERE tab.name = @tname
AND typ.name != 'sysname'
ORDER BY col.column_id;
Antwoord 19
SELECT C.COLUMN_NAME, C.IS_NULLABLE, C.DATA_TYPE, TC.CONSTRAINT_TYPE, C.COLUMN_DEFAULT
FROM INFORMATION_SCHEMA.COLUMNS AS C
FULL JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE AS CC ON C.COLUMN_NAME = CC.COLUMN_NAME
FULL JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS TC ON CC.CONSTRAINT_NAME = TC.CONSTRAINT_NAME
WHERE C.TABLE_NAME = '<Table Name>';
Antwoord 20
Als u FirstResponderKitvan het Brent Ozar-team gebruikt, kunt u deze query ook uitvoeren :
exec sp_blitzindex @tablename='MyTable'
Het geeft alle informatie over de tafel terug:
- indexen met hun gebruiksstatistieken (lezen, schrijven, vergrendelen, enz.), spatie
gebruikt en andere - ontbrekende indexen
- kolommen
- vreemde sleutels
- statistieken inhoud
Natuurlijk is het geen systeem en niet zo universele stp zoals sp_helpof sp_columns, maar het geeft alle mogelijke informatie over je tabel terug en ik denk dat het de moeite waard is om het te maken op uw omgeving en vermeld het hier.
Antwoord 21
Dubbelklik op de tabelnaam en druk op Alt+F1
Antwoord 22
CREATE PROCEDURE [dbo].[describe]
(
@SearchStr nvarchar(max)
)
AS
BEGIN
SELECT
CONCAT([COLUMN_NAME],' ',[DATA_TYPE],' ',[CHARACTER_MAXIMUM_LENGTH],' ',
(SELECT CASE [IS_NULLABLE] WHEN 'NO' THEN 'NOT NULL' ELSE 'NULL' END),
(SELECT CASE WHEN [COLUMN_DEFAULT] IS NULL THEN '' ELSE CONCAT(' DEFAULT ',[COLUMN_DEFAULT]) END)
) AS DESCRIPTION
FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME LIKE @SearchStr
END
Antwoord 23
De onderstaande query levert vergelijkbare uitvoer op als de functie info()in python, Pandas-bibliotheek.
USE [Database_Name]
IF OBJECT_ID('tempdo.dob.#primary_key', 'U') IS NOT NULL DROP TABLE #primary_key
SELECT
CONS_T.TABLE_CATALOG,
CONS_T.TABLE_SCHEMA,
CONS_T.TABLE_NAME,
CONS_C.COLUMN_NAME,
CONS_T.CONSTRAINT_TYPE,
CONS_T.CONSTRAINT_NAME
INTO #primary_key
FROM
INFORMATION_SCHEMA.TABLE_CONSTRAINTS AS CONS_T
JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE AS CONS_C ON CONS_C.CONSTRAINT_NAME= CONS_T.CONSTRAINT_NAME
SELECT
SMA.name AS [Schema Name],
ST.name AS [Table Name],
SC.column_id AS [Column Order],
SC.name AS [Column Name],
PKT.CONSTRAINT_TYPE,
PKT.CONSTRAINT_NAME,
SC.system_type_id,
STP.name AS [Data Type],
SC.max_length,
SC.precision,
SC.scale,
SC.is_nullable,
SC.is_masked
FROM sys.tables AS ST
JOIN sys.schemas AS SMA ON SMA.schema_id = ST.schema_id
JOIN sys.columns AS SC ON SC.object_id = ST.object_id
JOIN sys.types AS STP ON STP.system_type_id = SC.system_type_id
LEFT JOIN #primary_key AS PKT ON PKT.TABLE_SCHEMA = SMA.name
AND PKT.TABLE_NAME = ST.name
AND PKT.COLUMN_NAME = SC.name
ORDER BY ST.name ASC, SMA.name ASC
Antwoord 24
selecteer * uit sysobjects waar name=’TABLENAME’