How to get list of all tables that has identity columns

Question

I would like to learn how to fetch list of all tables that has identity columns from a MS SQL database.

Answer 1

The script below will do:

SELECT a.name as TableName,
  CASE WHEN b.name IS NULL
    THEN 'No Identity Column'
    ELSE b.name
  END as IdentityColumnName
FROM sys.tables a
  LEFT JOIN sys.identity_columns b on a.object_id = b.object_id 

Answer 2

SELECT 
  [schema] = s.name,
  [table] = t.name
FROM sys.schemas AS s
INNER JOIN sys.tables AS t
  ON s.[schema_id] = t.[schema_id]
WHERE EXISTS 
(
  SELECT 1 FROM sys.identity_columns
    WHERE [object_id] = t.[object_id]
);

Answer 3

Select OBJECT_NAME(object_Id) Rrom sys.identity_columns where is_identity = 1;

Answer 4

      select COLUMN_NAME, TABLE_NAME
      from INFORMATION_SCHEMA.COLUMNS
       where TABLE_SCHEMA = 'dbo'
       and COLUMNPROPERTY(object_id(TABLE_NAME), COLUMN_NAME, 'IsIdentity') = 1
       order by TABLE_NAME

Answer 5

I like this approach because it uses a join instead of a WHERE EXISTS or a call to COLUMNPROPERTY. Note that the group by is only necessary if you a) have tables with more than one IDENTITY column and b) don't want duplicate results:

SELECT 
    SchemaName = s.name,
    TableName = t.name
FROM
    sys.schemas AS s
    INNER JOIN sys.tables AS t ON s.schema_id = t.schema_id
    INNER JOIN sys.columns AS c ON t.object_id = c.object_id
    INNER JOIN sys.identity_columns AS ic on c.object_id = ic.object_id AND c.column_id = ic.column_id
GROUP BY
    s.name,
    t.name
ORDER BY
    s.name,
    t.name;