Sql server SQL Server使用时态表删除特殊架构下的所有表

我正在尝试删除带有时态表的数据库方案。 通过谷歌搜索找到的非现有脚本支持时态表

有人已经这样做了吗

该方案上有许多时态表，其中包含许多具有依赖关系的约束。因此，当我试图放弃该方案时，它会抱怨依赖性。 基本上，我在寻找一个存储过程或一些可以遍历所有DB对象并逐个删除的东西

创建示例表的脚本

USE [master];
GO

CREATE DATABASE [TestDb];
GO

USE [TestDb];
GO

CREATE SCHEMA [TestScheme];
GO

SET ANSI_NULLS ON;
GO
SET QUOTED_IDENTIFIER ON;
GO

CREATE TABLE [TestScheme].[Country]
(
    [CountryCode] [char](2) NOT NULL, 
    [Country] [varchar](60) NOT NULL, 
    [ValidFrom] [datetime2](2) GENERATED ALWAYS AS ROW START NOT NULL, 
    [ValidTo] [datetime2](2) GENERATED ALWAYS AS ROW END NOT NULL, 

    CONSTRAINT [PK_TestScheme_Country_CountryCode] 
        PRIMARY KEY CLUSTERED([CountryCode] ASC) 
                WITH (PAD_INDEX = ON, STATISTICS_NORECOMPUTE = OFF, 
                      IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, 
                      ALLOW_PAGE_LOCKS = ON, FILLFACTOR = 95) ON [PRIMARY], 
    PERIOD FOR SYSTEM_TIME([ValidFrom], [ValidTo])
) ON [PRIMARY] 
     WITH (SYSTEM_VERSIONING = ON (HISTORY_TABLE = [TestScheme].[CountryHistory]));
GO

SET ANSI_NULLS ON;
GO
SET QUOTED_IDENTIFIER ON;
GO

CREATE TABLE [TestScheme].[Address]
(
     [AddressId] [int] IDENTITY(1, 1) NOT NULL, 
     [City] [varchar](100) NOT NULL, 
     [CountryCode] [char](2) NOT NULL, 
     [ValidFrom] [datetime2](7) GENERATED ALWAYS AS ROW START NOT NULL, 
     [ValidTo] [datetime2](7) GENERATED ALWAYS AS ROW END NOT NULL, 

     CONSTRAINT [PK_TestScheme_Address_AddressId] 
         PRIMARY KEY CLUSTERED([AddressId] ASC) 
                 WITH (PAD_INDEX = ON, STATISTICS_NORECOMPUTE = OFF, 
                       IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, 
                       ALLOW_PAGE_LOCKS = ON, FILLFACTOR = 100) ON [PRIMARY], 
     PERIOD FOR SYSTEM_TIME([ValidFrom], [ValidTo])
)
ON [PRIMARY] 
   WITH (SYSTEM_VERSIONING = ON(HISTORY_TABLE = [TestScheme].[AddressHistory]));
GO

ALTER TABLE [TestScheme].[Address] WITH CHECK
    ADD CONSTRAINT [FK_TestScheme_CountryCode] 
        FOREIGN KEY([CountryCode]) REFERENCES [TestScheme].[Country]([CountryCode]);
GO

ALTER TABLE [TestScheme].[Address] CHECK CONSTRAINT [FK_TestScheme_CountryCode];
GO

查询删除方案：

USE [TestDb];
GO

DROP SCHEMA [TestScheme];
GO

要删除表的查询：

USE [TestDb]
GO

ALTER TABLE [TestScheme].[Country] SET (SYSTEM_VERSIONING = OFF)
GO

IF EXISTS (SELECT * FROM sys.objects 
           WHERE object_id = OBJECT_ID(N'[TestScheme].[Country]') AND type in (N'U'))
    DROP TABLE [TestScheme].[Country]
GO

IF EXISTS (SELECT * FROM sys.objects 
           WHERE object_id = OBJECT_ID(N'[TestScheme].[CountryHistory]') AND type in (N'U'))
    DROP TABLE [TestScheme].[CountryHistory]
GO

所以问题是有很多DB对象，我真的不想创建一个巨大的脚本来逐个删除

---

谢谢

谢谢大家，下面是我创建的脚本，它对我很有用

USE TestDb;
GO
DECLARE @SchemeName varchar(50)= 'TestScheme';
DECLARE @DatabaseName varchar(50)= 'TestDb';
DECLARE @sql nvarchar(max)= '';

/*Removing versioning on temporal tables*/
WITH selectedTables
     AS (SELECT concat('[', @DatabaseName, '].[', @SchemeName, '].[', name, ']') AS TableName
         FROM SYS.TABLES    WHERE history_table_id IS NOT NULL AND  SCHEMA_NAME(schema_id) = @SchemeName)
     SELECT @sql = COALESCE(@sql, N'') + 'ALTER TABLE   ' + TableName + '   SET ( SYSTEM_VERSIONING = OFF  );'
     FROM selectedTables;

SELECT @sql;
EXEC sp_executesql @sql;

/*Remove constraints*/
SET @sql = N'';
SELECT @sql = COALESCE(@sql, N'') + N'ALTER TABLE ' + QUOTENAME(s.name) + N'.' + QUOTENAME(t.name) + N' DROP CONSTRAINT ' + QUOTENAME(c.name) + ';'
FROM SYS.OBJECTS AS c   INNER JOIN  SYS.TABLES AS t ON c.parent_object_id = t.[object_id]
     INNER JOIN SYS.SCHEMAS AS s    ON t.[schema_id] = s.[schema_id]
WHERE c.[type] IN( 'D', 'C', 'F', 'PK', 'UQ' ) AND  s.name = @SchemeName
ORDER BY c.[type];

SELECT @sql;
EXEC sp_executesql @sql;

/*Delete Tables*/
SET @sql = N'';
SELECT @sql =   COALESCE(@sql, N'') + N'DROP TABLE ['+@SchemeName+'].' + QUOTENAME(TABLE_NAME) + N';' + CHAR(13)
FROM INFORMATION_SCHEMA.TABLES  WHERE TABLE_SCHEMA = @SchemeName    AND TABLE_TYPE = 'BASE TABLE';

SELECT @sql
EXEC sp_executesql @sql;

/*Drop scheme*/
SET @sql = N'';
SELECT @sql =   COALESCE(@sql, N'') + N'DROP SCHEMA IF EXISTS ' + @SchemeName + ';' + CHAR(13);

SELECT @sql
EXEC sp_executesql @sql;
GO

---

再次感谢

请添加一个您的确切意思的示例-您不能删除已启用系统版本控制的表。希望此说明足够，我现在已经在创建脚本，如果工作正常，将上载它，谢谢！如果存在xyz，您完全可以简化代码以删除一个表来使用

删除表

并避免所有那些签入

sys.objects

。。