-- 连接数据库GOUSEYour_Database;GO-- 查询表是否存在GOIFEXISTS(SELECT*FROMsys.objectsWHEREobject_id=OBJECT_ID(N'[dbo].[Your_Table]')ANDtypein(N'U'))BEGINSELECT'表已存在'ASResult;ENDELSEBEGINSELECT'表不存在'ASResult;ENDGO-- 关闭连
select count(*) into c from user_tables where table_name = upper('continent'); if c = 1 then execute immediate 'drop table continent'; end if; end; 这两个脚本都运行良好,但我的老板想要IF EXIT类的东西。任何人都可以帮助我。在这种情况下如何使用 IF EXIT ?
Public Function ExistTable(TName As String, YHcnn As ADODB.Connection) As Boolean Dim cat As New ADOX.Catalog Dim tbl As ADOX.Table Dim Ret As Boolean Ret = False Set cat.ActiveConnection = YHcnn For Each tbl In cat.Tables If tbl.Type = "TABLE" And tbl.Name = TName Then Ret = T...
这个IF EXISTS语句将返回表存在与否的结果。 如果表存在,查询结果将是:Table exists. 如果表不存在,查询结果将是:Table does not exist. 此外,我们还可以在IF EXISTS中执行其他SQL语句。 IF EXISTS (SELECT * FROM information_schema.tables WHERE table_schema = 'test_database' AND table_name = 'test_tabl...
if exists(select * from syscolumns where id=object_id('table1') and name='name') 7、判断数据库是否存在 if exists( select * from master.dbo.sysdatabases where dbid=db_ID( 'scbjdb' ) ) drop database scbjdb else print 'no exist scbjdb'...
if object_id(’tempdb..#临时表名’) is not null drop table #临时表名 if object_id(’tempdb..#临时表名’) is not null drop table #临时表名 5 判断视图是否存在 Sql代码 --SQL Server 2000 IF EXISTS (SELECT * FROM sysviews WHERE object_id = ’[dbo].[视图名]’ ...
IF EXISTS (SELECT 1 FROM SYSCOLUMNS WHERE TBL_NAME = 'your_table_name' AND TBL_TYPE = 'TABLE')BEGIN PRINT 'Table exists';END ELSE BEGIN PRINT 'Table does not exist';END ```上述SQL语句通过IF EXISTS语句判断名为your_table_name的表是否存在，如果存在则输出'Table exists'，否则输出'Table ...
SET is_it_there = does_table_exist ( 'religion', 'choices' ); if is_it_there = false then SET function_succeeded = true; ELSE SET function_succeeded = false; END IF; ELSE SET function_succeeded = false; END IF; select function_succeeded as function_succeeded ; END| ...
if object_id('tb_table') is not null print 'exist' else print'not exist' 如上,可用object_id()来快速达到相同的目的,tb_table就是我将要创建的资源的名称,所以要先判断当前数据库中不存在相同的资源object_id()可接受两个参数,第一个如上所示,代表资源的名称,上面的就是表的名字,但往往我们要说明我...
$exist = 0; while($i < $num_tables) { $tablename = mysql_tablename($tables, $i); if ($tablename=='table_search') $exist=1; $i++; } if ($exist==0) { // something to do if the table does not exist } else { // something to do if the table exist }...