SQL Server 2005及以后版本：

IF EXISTS(SELECT 1 FROM sys.columns 
          WHERE Name = N'columnName'
          AND Object_ID = Object_ID(N'schemaName.tableName'))
BEGIN
    -- Column Exists
END

马丁·史密斯的版本更短：

IF COL_LENGTH('schemaName.tableName', 'columnName') IS NOT NULL
BEGIN
    -- Column Exists
END

有几种方法可以检查列的存在。我强烈建议使用INFORMATION_SCHEMA.COLUMNS，因为它是为了与用户通信而创建的。考虑下表：

 sys.objects
 sys.columns

甚至可以使用一些其他访问方法来检查系统目录。

此外，无需使用SELECT*，只需通过NULL值测试即可

IF EXISTS(
           SELECT NULL 
           FROM INFORMATION_SCHEMA.COLUMNS
           WHERE
             TABLE_NAME = 'myTableName'
             AND COLUMN_NAME = 'myColumnName'
         ) 

IF NOT EXISTS(SELECT NULL
              FROM  INFORMATION_SCHEMA.COLUMNS
              WHERE table_name = 'TableName'
                    AND table_schema = 'SchemaName'
                    AND column_name = 'ColumnName') BEGIN

  ALTER TABLE [SchemaName].[TableName] ADD [ColumnName] int(1) NOT NULL default '0';

END;

还有一种变化。。。

SELECT 
  Count(*) AS existFlag 
FROM 
  sys.columns 
WHERE 
  [name] = N 'ColumnName' 
  AND [object_id] = OBJECT_ID(N 'TableName')

试试这个。。。

IF NOT EXISTS(
  SELECT TOP 1 1
  FROM INFORMATION_SCHEMA.COLUMNS
  WHERE 
    [TABLE_NAME] = 'Employees'
    AND [COLUMN_NAME] = 'EmployeeID')
BEGIN
  ALTER TABLE [Employees]
    ADD [EmployeeID] INT NULL
END

select distinct object_name(sc.id)
from syscolumns sc,sysobjects so  
where sc.name like '%col_name%' and so.type='U'