Bir SQL Server tablosunun birincil anahtarını nasıl listelersiniz?

Basit soru, T-SQL ile bir tablonun birincil anahtarını nasıl listelersiniz? Bir tabloda indeksleri nasıl elde edeceğimi biliyorum, ancak PK'yi nasıl elde edeceğimi hatırlayamıyorum.

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>'

sys.*Görünümleri şimdi INFORMATION_SCHEMASQL Server'da kullanmak genel olarak tavsiye edilir , bu nedenle veritabanlarını taşımayı planlamıyorsanız, bunları kullanırım. sys.*Görünümlerle bunu şu şekilde yaparsınız :

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>');

Bu, yalnızca sys tablolarını kullanan bir çözümdür .

Veritabanındaki tüm birincil anahtarları listeler. Her birincil anahtar için şema, tablo adı, sütun adı ve doğru sütun sıralama düzenini döndürür .

Belirli bir tablo için birincil anahtarı almak istiyorsanız, o zaman SchemaNameve üzerinde filtrelemeniz gerekir TableName.

IMHO, bu çözüm çok geneldir ve herhangi bir dize değişmezi kullanmaz, bu nedenle herhangi bir makinede çalışacaktır.

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 ;

İşte sorudan , sql sorgusu kullanarak tablo birincil anahtarını almanın başka bir yolu :

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>'

Bu kullanan KEY_COLUMN_USAGEbir tablo için kısıtlamaları belirlemek için
Sonra kullanan her bir birincil anahtar olup olmadığını belirlemek içinOBJECTPROPERTY(id, 'IsPrimaryKey')

INFORMATION_SCHEMA tekniğini beğendim, ancak kullandığım başka bir teknik: exec sp_pkeys 'table'

MS SQL Server kullanıyor, aşağıdakileri yapabilirsiniz:

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

Belirli bir tablo istiyorsanız tablo_adı sütununu da filtreleyebilirsiniz.

- Bu, aynı zamanda Co-Related Query için bir örnek olan başka bir Değiştirilmiş Sürümdür

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')

Bu, tüm kısıtlamaları (birincil Anahtar ve Yabancı Anahtarlar) listelemeli ve sorgunun sonuna tablo adını koymalıdır

/* 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;

Referans için lütfen okuyun - http://blogs.msdn.com/b/sqltips/archive/2005/09/16/469136.aspx

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

Teşekkürler Guy.

Küçük bir varyasyonla, tüm tablolar için tüm birincil anahtarları bulmak için kullandım.

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

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

Bu size PK olan sütunları verir.

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

Depolanan sistem yordamı sp_helpsize bilgi verecektir. Aşağıdaki ifadeyi yürütün:

execute sp_help table_name

Aşağıdaki sorgu , belirli bir tablonun birincil anahtarlarını listeleyecektir :

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

İzlediğim basit bir Teknik anlatıyorum

SP_HELP 'table_name'

bu kodu sorgu olarak çalıştırın. Birincil Anahtarı öğrenmek istediğiniz tablo_adı yerine tablo adınızı belirtin (tek tırnak işaretlerini unutmayın). Sonuç ekli Resim gibi görünecektir. Umarım sana yardımcı olur

Belirli bir TabloAdı ve Şema için birincil anahtar sütunlarının virgülle ayrılmış listesi için:

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) 

Bunu bir deneyin:

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'

Bu sürüm şemayı, tablo adını ve sıralı, virgülle ayrılmış birincil anahtarlar listesini görüntüler. Object_Id () bağlantı sunucuları için çalışmaz, bu nedenle tablo adına göre filtre uygularız.

REPLACE (Si1.Column_Name, '', '') olmadan üzerinde test ettiğim veritabanındaki Sütun_Adı için xml açma ve kapama etiketlerini gösterecekti. Veritabanının neden 'Sütun_Adı' için bir değişiklik gerektirdiğinden emin değilim, bu yüzden birisi biliyorsa lütfen yorum yapın.

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;

Ve George'un sorgusunu kullanan aynı model:

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;

Bunu yararlı buldum, sütunların virgülle ayrı bir listesini içeren bir tablo listesi ve ardından hangilerinin birincil anahtar olduğuna ilişkin virgülle ayrı bir liste veriyor

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

Sys.Objects Tablosu, her kullanıcı tanımlı, şema kapsamlı nesne için satır içerir.

Birincil Anahtar veya diğerleri gibi yaratılan kısıtlamalar nesne olacak ve Tablo adı parent_object olacaktır.

Sys.Objects sorgulayın ve Nesnenin Gerekli Türdeki Kimliklerini toplayın

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)

Aşağıdaki orijinal soruya daha doğru ve basit bir cevap önerebilir miyim?

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'

Notlar:

1. Yukarıdaki yanıtlardan bazılarında yalnızca birincil anahtar sütunları için bir filtre eksik!
2. BIML oluşturma tabloları ve SSIS kodunu beslemek için bir kaynaktan meta verileri sağlamak üzere daha büyük bir sütun listesine katılmak için aşağıdaki CTE'yi kullanıyorum

Son zamanlarda gönderilebilir, ancak umarım bu, birisinin bu t-sql sorgusunu kullanarak sql sunucusundaki birincil anahtar listesini görmesine yardımcı olur:

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'

İsterseniz bu sorguyu kullanarak tüm yabancı anahtarların listesini görebilirsiniz:

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

Bunu arkadaşımdan buldum, tablonun belirli bir şema altındaki tüm birincil anahtarlarını arıyorsanız çok etkili.

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>'

Kendi ORM'nizi yapmak veya belirli bir tablodan kod oluşturmak istiyorsanız, aradığınız biçim bu olabilir:

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

Sonuç şöyle görünecektir:

				<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>
				

Birincil Anahtar ve tür gerekiyorsa, bu sorgu yararlı olabilir:

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