"Markus Schiltknecht" <[EMAIL PROTECTED]> writes: > CREATE INDEX x ON test(a, b, c); > > That's why I'd say, the first column of an index would have to be equal to all > of the columns used in the partitioning key.
That's certainly the simple case. It would be nice to be able to create an index like that and have the system automatically recognize that the leading column is identical to the partition key of (a) and therefore build indexes on each partition on (b,c). However there are also cases such as where you have a=0..99 in one partition and a=100..199 in partition two, etc. It could still automatically build indexes on (a,b,c) on each partition and somehow note that the unique constraint is guaranteed across the whole partitioned table. -- Gregory Stark EnterpriseDB http://www.enterprisedb.com ---------------------------(end of broadcast)--------------------------- TIP 6: explain analyze is your friend
