I think the correct way is SELECT equipo_attr[1] AS year, equip_attr[2] AS make FROM vehicle_tb; On Mon, 12 Jun 2000, Bernie Huang wrote: > Hi, I have a following query: > > "select equip_attr[1], equip_attr[2] > as year, make > from vehicle_tb;" > > which gives me the following error: > > "ERROR: Attribute 'make' not found" > > Is there something wrong with my "as" usage? Thanks > > > - Bernie >
- [SQL] Problem regarding 'select...as...' Bernie Huang
- [SQL] Re: [PHP-DB] Problem regarding 'select...as...' Vince LaMonica
- Jesus Aneiros