How to get all index columns including filtered columns?
if object_id('dbo.tTestIndex') is not null
drop table dbo.tTestIndex;
create table dbo.tTestIndex (a int, b int, c int);
create index X_tTestIndex on dbo.tTestIndex (a) include (b) where c > 0;
select i.name, i.filter_definition, c.name, ic.is_included_column
from sys.indexes i
inner join sys.index_columns ic on i.object_id = ic.object_id and i.index_id = ic.index_id
left join sys.columns c on ic.object_id = c.object_id and ic.column_id = c.column_id
where i.object_id = object_id('dbo.tTestIndex');
if object_id('dbo.tTestIndex') is not null
drop table dbo.tTestIndex;
This example gives two rows instead of three.
index_name | filter_definition | column_name | is_included_column
-------------+-------------------+-------------+-------------------
X_tTestIndex | ([c]>(0)) | a | 0
X_tTestIndex | ([c]>(0)) | b | 1
You can use the catalog view sys.sql_expression_dependencies to find out the columns on which your filter relies:
SELECT i.name, i.filter_definition, c.name, is_included_column = 0
FROM sys.indexes AS i
INNER JOIN sys.sql_expression_dependencies AS d
ON d.referencing_id = i.[object_id]
AND d.referencing_minor_id = i.index_id
AND d.referencing_class_desc = 'INDEX'
INNER JOIN sys.columns AS c
ON c.[object_id] = d.referenced_id
AND c.column_id = d.referenced_minor_id
WHERE i.[object_id] = OBJECT_ID(N'dbo.tTestIndex');
You can then just UNION
this to your original query to get all the columns.