我希望这是关于如何使用SQL语句检查SQL Server 2000/2005中是否存在表的最终讨论。
这里有两种可能的方法。哪一种是做这件事的标准/最佳方式?
第一个方法:
IF EXISTS (SELECT 1
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_TYPE='BASE TABLE'
AND TABLE_NAME='mytablename')
SELECT 1 AS res ELSE SELECT 0 AS res;
第二种方式:
IF OBJECT_ID (N'mytablename', N'U') IS NOT NULL
SELECT 1 AS res ELSE SELECT 0 AS res;
MySQL提供简单的
SHOW TABLES LIKE '%tablename%';
声明。我在找类似的东西。
当前回答
您可以使用下面的代码
IF (OBJECT_ID('TableName') IS NOT NULL )
BEGIN
PRINT 'Table Exists'
END
ELSE
BEGIN
PRINT 'Table NOT Exists'
END
Or
IF (EXISTS (SELECT * FROM sys.tables WHERE [name] = 'TableName'))
BEGIN
PRINT 'Table Exists'
END
ELSE
BEGIN
PRINT 'Table NOT Exists'
END
其他回答
您可以使用下面的代码
IF (OBJECT_ID('TableName') IS NOT NULL )
BEGIN
PRINT 'Table Exists'
END
ELSE
BEGIN
PRINT 'Table NOT Exists'
END
Or
IF (EXISTS (SELECT * FROM sys.tables WHERE [name] = 'TableName'))
BEGIN
PRINT 'Table Exists'
END
ELSE
BEGIN
PRINT 'Table NOT Exists'
END
我以创建一个视图为例。
因为ALTER/CREATE命令不能在BEGIN/END块中。在创建之前，您需要测试是否存在并删除它
IF Object_ID('TestView') IS NOT NULL
DROP VIEW TestView
GO
CREATE VIEW TestView
as
. . .
GO
如果您担心权限丢失，您也可以编写GRANT语句，并在最后重新运行它们。
你可以把create/alter包装成一个字符串，然后执行EXEC——对于大的视图来说，这可能会很难看
DECLARE @SQL as varchar(4000)
-- set to body of view
SET @SQL = 'SELECT X, Y, Z FROM TABLE'
IF Object_ID('TestView') IS NULL
SET @SQL = 'CREATE VIEW TestView AS ' + @SQL
ELSE
SET @SQL = 'ALTER VIEW TestView AS ' + @SQL
IF OBJECT_ID('mytablename') IS NOT NULL
还有一个选项可以检查表是否跨数据库存在
IF EXISTS(SELECT 1 FROM [change-to-your-database].SYS.TABLES WHERE NAME = 'change-to-your-table-name')
BEGIN
-- do whatever you want
END
select name from SysObjects where xType='U' and name like '%xxx%' order by name
