How to check if a column exists in a SQL Server table?

Answer: SQL Server 2005 onwards:

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

How to check if a column exists in a SQL Server table?

A more concise version

IF COL_LENGTH('table_name','column_name') IS NULL
BEGIN
/* Column does not exist or caller does not have permission to view the object */
END

The point about permissions on viewing metadata applies to all answers not just this one.

Note that the first parameter table name to COL_LENGTH can be in one, two, or three part name format as required.

An example referencing a table in a different database is

COL_LENGTH('AdventureWorks2012.HumanResources.Department','ModifiedDate')

One difference with this answer compared to using the metadata views is that metadata functions such as COL_LENGTH always only return data about committed changes irrespective of the isolation level in effect.

Answer #3:

Tweak the below to suit your specific requirements:

if not exists (select
                     column_name
               from
                     INFORMATION_SCHEMA.columns
               where
                     table_name = 'MyTable'
                     and column_name = 'MyColumn')
    alter table MyTable add MyColumn int

Edit to deal with edit to question: That should work – take a careful look over your code for stupid mistakes; are you querying INFORMATION_SCHEMA on the same database as your insert is being applied to for example? Do you have a typo in your table/column name in either statement?

Answer #4:

Try this…

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

Answer #5:

For the people who are checking the column existence before dropping it.

From SQL Server 2016 you can use new DIE statements instead of big IF wrappers

ALTER TABLE Table_name DROP COLUMN IF EXISTS Column_name

Answer #6:

You can use the information schema system views to find out pretty much anything about the tables you’re interested in:

SELECT *
  FROM INFORMATION_SCHEMA.COLUMNS
 WHERE TABLE_NAME = 'yourTableName'
 ORDER BY ORDINAL_POSITION

You can also interrogate views, stored procedures and pretty much anything about the database using the Information_schema views.

Answer #7:

Try something like:

CREATE FUNCTION ColumnExists(@TableName varchar(100), @ColumnName varchar(100))
RETURNS varchar(1) AS
BEGIN
DECLARE @Result varchar(1);
IF EXISTS (SELECT * FROM INFORMATION_SCHEMA.Columns WHERE TABLE_NAME = @TableName AND COLUMN_NAME = @ColumnName)
BEGIN
    SET @Result = 'T'
END
ELSE
BEGIN
    SET @Result = 'F'
END
RETURN @Result;
END
GO

GRANT EXECUTE ON  [ColumnExists] TO [whoever]
GO

Then use it like this:

IF ColumnExists('xxx', 'yyyy') = 'F'
BEGIN
  ALTER TABLE xxx
  ADD yyyyy varChar(10) NOT NULL
END
GO

It should work on both SQL Server 2000 & SQL Server 2005. Not sure about SQL Server 2008, but don’t see why not.

Answer #8:

First check if the table/column(id/name) combination exists in dbo.syscolumns (an internal SQL Server table that contains field definitions), and if not issue the appropriate ALTER TABLE query to add it. For example:

IF NOT EXISTS ( SELECT  *
            FROM    syscolumns
            WHERE   id = OBJECT_ID('Client')
                    AND name = 'Name' ) 
ALTER TABLE Client
ADD Name VARCHAR(64) NULL

Answer #9:

A good friend and colleague of mine showed me how you can also use an IF block with SQL functions OBJECT_ID and COLUMNPROPERTY in SQL SERVER 2005+ to check for a column. You can use something similar to the following:

IF (OBJECT_ID(N'[dbo].[myTable]') IS NOT NULL AND
    COLUMNPROPERTY( OBJECT_ID(N'[dbo].[myTable]'), 'ThisColumnDoesNotExist', 'ColumnId') IS NULL)
BEGIN
    SELECT 'Column does not exist -- You can add TSQL to add the column here'
END

Hope you learned something from this post. The primary source of this article is StackOverflow.

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