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SQLサーバーに存在する場合にのみ外部キー制約を削除するにはどうすればよいですか？

次のコードを使用してテーブルが存在する場合、テーブルを削除できますが、制約で同じことを行う方法がわかりません。

IF EXISTS(SELECT 1 FROM sys.objects WHERE OBJECT_ID = OBJECT_ID(N'TableName') AND type = (N'U')) DROP TABLE TableName
go

また、このコードを使用して制約を追加します。

ALTER TABLE [dbo].[TableName] 
  WITH CHECK ADD CONSTRAINT [FK_TableName_TableName2] FOREIGN KEY([FK_Name])
    REFERENCES [dbo].[TableName2] ([ID])
go
sqlsql-serversql-server-2005tsql
208
solrevdev

より簡単なソリューションは、Eric Isaacsの回答で提供されています。ただし、任意のテーブルの制約を見つけます。特定のテーブルの外部キー制約を対象とする場合、これを使用します。

IF EXISTS (SELECT * 
  FROM sys.foreign_keys 
   WHERE object_id = OBJECT_ID(N'dbo.FK_TableName_TableName2')
   AND parent_object_id = OBJECT_ID(N'dbo.TableName')
)
  ALTER TABLE [dbo.TableName] DROP CONSTRAINT [FK_TableName_TableName2]
295
James L

これは、現在提案されているソリューションよりもはるかに簡単です。

IF (OBJECT_ID('dbo.FK_ConstraintName', 'F') IS NOT NULL)
BEGIN
    ALTER TABLE dbo.TableName DROP CONSTRAINT FK_ConstraintName
END

別のタイプの制約を削除する必要がある場合、これらは2番目のパラメーター位置でOBJECT_ID（）関数に渡す適切なコードです。

C = CHECK constraint
D = DEFAULT (constraint or stand-alone)
F = FOREIGN KEY constraint
PK = PRIMARY KEY constraint
UQ = UNIQUE constraint

2番目のパラメーターなしでOBJECT_IDを使用することもできます。

タイプの完全なリスト こちら ：

オブジェクトタイプ：

AF = Aggregate function (CLR)
C = CHECK constraint
D = DEFAULT (constraint or stand-alone)
F = FOREIGN KEY constraint
FN = SQL scalar function
FS = Assembly (CLR) scalar-function
FT = Assembly (CLR) table-valued function
IF = SQL inline table-valued function
IT = Internal table
P = SQL Stored Procedure
PC = Assembly (CLR) stored-procedure
PG = Plan guide
PK = PRIMARY KEY constraint
R = Rule (old-style, stand-alone)
RF = Replication-filter-procedure
S = System base table
SN = Synonym
SO = Sequence object

適用対象：SQL Server 2012からSQL Server 2014（

SQ = Service queue
TA = Assembly (CLR) DML trigger
TF = SQL table-valued-function
TR = SQL DML trigger
TT = Table type
U = Table (user-defined)
UQ = UNIQUE constraint
V = View
X = Extended stored procedure
281
Eric Isaacs

SQL Server 2016では、DROP IF EXISTSを使用できます。

CREATE TABLE t(id int primary key, 
               parentid int
                    constraint tpartnt foreign key references t(id))
GO
ALTER TABLE t
DROP CONSTRAINT IF EXISTS tpartnt
GO
DROP TABLE IF EXISTS t

http://blogs.msdn.com/b/sqlserverstorageengine/archive/2015/11/03/drop-if-exists-new-thing-in-sql-server-2016.aspx を参照してください

23
Jovan MSFT
IF (OBJECT_ID('DF_Constraint') IS NOT NULL)
BEGIN
    ALTER TABLE [dbo].[tableName]
    DROP CONSTRAINT DF_Constraint
END
15
DevDave

実際の制約の名前を知っていれば、ジェームズの答えはうまく機能します。トリッキーなことは、レガシーおよびその他の現実世界のシナリオでは、know制約が何と呼ばれるかということです。

これが当てはまる場合、重複する制約を作成するリスクがあるため、回避するには以下を使用できます。

create function fnGetForeignKeyName
(
    @ParentTableName nvarchar(255), 
    @ParentColumnName nvarchar(255),
    @ReferencedTableName nvarchar(255),
    @ReferencedColumnName nvarchar(255)
)
returns nvarchar(255)
as
begin 
    declare @name nvarchar(255)

    select @name = fk.name  from sys.foreign_key_columns fc
    join sys.columns pc on pc.column_id = parent_column_id and parent_object_id = pc.object_id
    join sys.columns rc on rc.column_id = referenced_column_id and referenced_object_id = rc.object_id 
    join sys.objects po on po.object_id = pc.object_id
    join sys.objects ro on ro.object_id = rc.object_id 
    join sys.foreign_keys fk on fk.object_id = fc.constraint_object_id
    where 
        po.object_id = object_id(@ParentTableName) and 
        ro.object_id = object_id(@ReferencedTableName) and
        pc.name = @ParentColumnName and 
        rc.name = @ReferencedColumnName

    return @name
end

go

declare @name nvarchar(255)
declare @sql nvarchar(4000)
-- hunt for the constraint name on 'Badges.BadgeReasonTypeId' table refs the 'BadgeReasonTypes.Id'
select @name = dbo.fnGetForeignKeyName('dbo.Badges', 'BadgeReasonTypeId', 'dbo.BadgeReasonTypes', 'Id')
-- if we find it, the name will not be null
if @name is not null 
begin 
    set @sql = 'alter table Badges drop constraint ' + replace(@name,']', ']]')
    exec (@sql)
end
14
Sam Saffron
ALTER TABLE [dbo].[TableName]
    DROP CONSTRAINT FK_TableName_TableName2
6
Mitch Wheat
Declare @FKeyRemoveQuery NVarchar(max)

IF EXISTS(SELECT 1 FROM sys.foreign_keys WHERE parent_object_id = OBJECT_ID(N'dbo.TableName'))

BEGIN
    SELECT @FKeyRemoveQuery='ALTER TABLE dbo.TableName DROP CONSTRAINT [' + LTRIM(RTRIM([name])) + ']'   
    FROM sys.foreign_keys
    WHERE parent_object_id = OBJECT_ID(N'dbo.TableName')

    EXECUTE Sp_executesql @FKeyRemoveQuery 

END
3
Rajalingam

これはあなたに役立つと思います...

    DECLARE @ConstraintName nvarchar(200)
SELECT 
    @ConstraintName = KCU.CONSTRAINT_NAME
FROM INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS RC 
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE AS KCU
    ON KCU.CONSTRAINT_CATALOG = RC.CONSTRAINT_CATALOG  
    AND KCU.CONSTRAINT_SCHEMA = RC.CONSTRAINT_SCHEMA 
    AND KCU.CONSTRAINT_NAME = RC.CONSTRAINT_NAME
WHERE
    KCU.TABLE_NAME = 'TABLE_NAME' AND
    KCU.COLUMN_NAME = 'TABLE_COLUMN_NAME'
IF @ConstraintName IS NOT NULL EXEC('alter table TABLE_NAME drop  CONSTRAINT ' + @ConstraintName)

特定のテーブルと列に基づいて外部キー制約を削除します。

1
Samir Savasani

これらのクエリを使用して、テーブルのすべてのFKを見つけることができます。

Declare @SchemaName VarChar(200) = 'Schema Name'
Declare @TableName VarChar(200) = 'Table name'

-- Find FK in This table.
SELECT 
    'IF  EXISTS (SELECT * FROM sys.foreign_keys WHERE object_id = OBJECT_ID(N''' + 
      '[' + OBJECT_SCHEMA_NAME(FK.parent_object_id) + '].[' + FK.name + ']' 
      + ''') AND parent_object_id = OBJECT_ID(N''' + 
      '[' + OBJECT_SCHEMA_NAME(FK.parent_object_id) + '].[' 
      + OBJECT_NAME(FK.parent_object_id) + ']' + ''')) ' +

    'ALTER TABLE ' +  OBJECT_SCHEMA_NAME(FK.parent_object_id) +
    '.[' + OBJECT_NAME(FK.parent_object_id) + 
    '] DROP CONSTRAINT ' + FK.name
    , S.name , O.name, OBJECT_NAME(FK.parent_object_id)
FROM sys.foreign_keys AS FK
INNER JOIN Sys.objects As O 
  ON (O.object_id = FK.parent_object_id )
INNER JOIN SYS.schemas AS S 
  ON (O.schema_id = S.schema_id)  
WHERE 
      O.name = @TableName
      And S.name = @SchemaName


-- Find the FKs in the tables in which this table is used
  SELECT 
    ' IF  EXISTS (SELECT * FROM sys.foreign_keys WHERE object_id = OBJECT_ID(N''' + 
      '[' + OBJECT_SCHEMA_NAME(FK.parent_object_id) + '].[' + FK.name + ']' 
      + ''') AND parent_object_id = OBJECT_ID(N''' + 
      '[' + OBJECT_SCHEMA_NAME(FK.parent_object_id) + '].[' 
      + OBJECT_NAME(FK.parent_object_id) + ']' + ''')) ' +

    ' ALTER TABLE ' +  OBJECT_SCHEMA_NAME(FK.parent_object_id) +
    '.[' + OBJECT_NAME(FK.parent_object_id) + 
    '] DROP CONSTRAINT ' + FK.name
    , S.name , O.name, OBJECT_NAME(FK.parent_object_id)
FROM sys.foreign_keys AS FK
INNER JOIN Sys.objects As O 
  ON (O.object_id = FK.referenced_object_id )
INNER JOIN SYS.schemas AS S 
  ON (O.schema_id = S.schema_id)  
WHERE 
      O.name = @TableName
      And S.name = @SchemaName
0
Ardalan Shahgholi

この質問に対する受け入れられた答えは、私にとってはうまくいかないようです。私はわずかに異なる方法で同じことを達成しました：

IF (select object_id from sys.foreign_keys where [name] = 'FK_TableName_TableName2') IS NOT NULL
BEGIN
    ALTER TABLE dbo.TableName DROP CONSTRAINT FK_TableName_TableName2
END
0
Melbourne Developer