web-dev-qa-db-ja.com

SQL Serverテーブルの主キーをどのようにリストしますか?

簡単な質問です。T-SQLでテーブルの主キーをどのようにリストしますか?テーブルのインデックスを取得する方法は知っていますが、PKを取得する方法を思い出せません。

87
swilliams
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>'
131
Guy Starbuck

現在、SQL Serverでsys.*よりもINFORMATION_SCHEMAビューを使用することをお勧めします。したがって、データベースの移行を計画していない限り、これらを使用します。 sys.*ビューを使用してこれを行う方法は次のとおりです。

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>');
25
Dave Zych

これは、sys-tablesのみを使用するソリューションです。

データベース内のすべての主キーをリストします。 スキーマ、テーブル名、列名および各主キーに対して正しい列のソート順を返します。

特定のテーブルの主キーを取得する場合は、SchemaNameおよびTableNameでフィルタリングする必要があります。

私見、このソリューションは非常に汎用的であり、文字列リテラルを使用しないため、どのマシンでも実行できます。

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 ;
18
SQLGeorge

質問 SQLクエリを使用してテーブルの主キーを取得 から別の方法を次に示します。

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>'

KEY_COLUMN_USAGE を使用して、特定のテーブルの制約を決定します
次に OBJECTPROPERTY(id, 'IsPrimaryKey') を使用して、それぞれが主キーかどうかを判断します

7
KyleMit

INFORMATION_SCHEMA手法は好きですが、私が使用した別の手法は次のとおりです。execsp_pkeys 'table'

6
user12861

MS SQL Serverを使用している場合、次のことができます。

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

特定のテーブルが必要な場合は、table_name列でフィルタリングすることもできます。

6
Dwight T

-これは、共同関連クエリの例でもある別の修正バージョンです

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')
4

これはすべての制約(プライマリキーと外部キー)をリストし、クエリの最後にテーブル名を入れます

/* 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;

参照するには、以下をお読みください- http://blogs.msdn.com/b/sqltips/archive/2005/09/16/469136.aspx

3
dekdev
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
1

私が従う簡単なテクニックを伝えています

SP_HELP 'table_name'

このコードをクエリとして実行します。主キーを知りたいtable_nameの場所にテーブル名を記載します(一重引用符を忘れないでください)。結果は添付画像のように表示されます。それがあなたを助けることを願っています

enter image description here

1
Bha15

以下のクエリは主キー of 特定のテーブルをリストします:

SELECT DISTINCT
    CONSTRAINT_NAME AS [Constraint],
    TABLE_SCHEMA AS [Schema],
    TABLE_NAME AS TableName
FROM
    INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE
    TABLE_NAME = 'mytablename'
1
Anjan Kant

みんなありがとう。

わずかなバリエーションを加えて、すべてのテーブルのすべての主キーを見つけるために使用しました。

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
1
MartinC

これにより、PKである列が得られます。

SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE WHERE TABLE_NAME = 'TableName'
1
Tanner Ornelas

システムストアドプロシージャsp_helpが情報を提供します。次のステートメントを実行します。

execute sp_help table_name
1
boes

特定のスキーマですべてのテーブルの主キーを探している場合、非常に効果的であることが友人からわかった

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>'
0
WEshruth

このバージョンでは、スキーマ、テーブル名、および順序付けられたカンマ区切りの主キーのリストが表示されます。 Object_Id()はリンクサーバーでは機能しないため、テーブル名でフィルタリングします。

REPLACE(Si1.Column_Name、 ''、 '')がなければ、テスト中のデータベースのColumn_Nameのxml開始タグと終了タグが表示されます。データベースが 'Column_Name'の置換を必要とした理由がわからないため、誰かが知っている場合はコメントしてください。

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;

そして、ジョージのクエリを使用した同じパターン:

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;
0
Soenhay

これを試してみてください:

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'
0
Austin Salonen

以下の元の質問に対するより正確で簡単な答えを提案できますか

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'

ノート:

  1. 上記の回答の中には、主キー列だけのフィルターがありません!
  2. CTEで以下を使用して、より大きな列リストに参加し、ソースからのメタデータを提供して、ステージングテーブルとSSISコードのBIML生成をフィードします。
0
Saxman

最近投稿される可能性がありますが、このt-sqlクエリを使用することで、誰かがsqlサーバーのプライマリキーリストを確認できるようになります。

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'

必要に応じて、このクエリを使用して、すべての外部キーのリストを表示できます。

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
0
Humayoun_Kabir
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
0
Liban01

独自のORMを実行したり、特定のテーブルからコードを生成したりする場合は、次のようになります。

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

結果は次のようになります。

                            <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>
                                
0
TruthSeeker

特定のTableNameおよびSchemaの主キー列のコンマ区切りリストの場合:

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) 
0
Allan F

主キーとタイプが必要な場合、このクエリは便利です。

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
0
Hamed Nikzad

私はこれが便利だと思いました、列のコンマ区切りリストと、主キーであるコンマ区切りリストを含むテーブルのリストを提供します

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
0
Pricey

Sys.Objectsテーブルには、ユーザー定義のスキーマスコープの各オブジェクトの行が含まれます。

主キーなどのように作成された制約はobjectになり、テーブル名はparent_objectになります

Sys.Objectsを照会し、オブジェクトの必須タイプのIDを収集します

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)
0
UJS