Question

How do I get a list of all index & index columns in SQL Server 2005+? The closest I could get is:

select s.name, t.name, i.name, c.name from sys.tables t inner join sys.schemas s on t.schema_id = s.schema_id inner join sys.indexes i on i.object_id = t.object_id inner join sys.index_columns ic on ic.object_id = t.object_id inner join sys.columns c on c.object_id = t.object_id and         ic.column_id = c.column_id  where i.index_id > 0      and i.type in (1, 2) -- clustered & nonclustered only  and i.is_primary_key = 0 -- do not include PK indexes  and i.is_unique_constraint = 0 -- do not include UQ  and i.is_disabled = 0  and i.is_hypothetical = 0  and ic.key_ordinal > 0  order by ic.key_ordinal 

Which is not exactly what I want.
What I want is, to list all user-defined indexes, (which means no indexes which support unique constraints & primary keys) with all columns (ordered by how do they appear in index definition) plus as much metadata as possible.

Answer 1

There are two "sys" catalog views you can consult: sys.indexes and sys.index_columns.

Those will give you just about any info you could possibly want about indices and their columns.

EDIT: This query’s getting pretty close to what you’re looking for:

SELECT       TableName = t.name,      IndexName = ind.name,      IndexId = ind.index_id,      ColumnId = ic.index_column_id,      ColumnName = col.name,      ind.*,      ic.*,      col.*  FROM       sys.indexes ind  INNER JOIN       sys.index_columns ic ON  ind.object_id = ic.object_id and ind.index_id = ic.index_id  INNER JOIN       sys.columns col ON ic.object_id = col.object_id and ic.column_id = col.column_id  INNER JOIN       sys.tables t ON ind.object_id = t.object_id  WHERE       ind.is_primary_key = 0       AND ind.is_unique = 0       AND ind.is_unique_constraint = 0       AND t.is_ms_shipped = 0  ORDER BY       t.name, ind.name, ind.index_id, ic.is_included_column, ic.key_ordinal;