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Re: [PHP-DB] Problem regarding 'select...as...'

From: Vince LaMonica <Vince(dot)LaMonica(at)quest(dot)com>
To: Bernie Huang <bernie(dot)huang(at)ec(dot)gc(dot)ca>
Cc: PHP Database <php-db(at)lists(dot)php(dot)net>, PGSQL-SQL <pgsql-sql(at)postgresql(dot)org>
Subject: Re: [PHP-DB] Problem regarding 'select...as...'
Date: 2000-06-12 22:40:27
Message-ID: Pine.WNT.4.21.0006121539160.256-100000@vlamonica.quest.com (view raw or flat)
Thread:
Lists: pgsql-sql
On 2000-06-12, Bernie Huang stated:

} "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

Try: 
	"select equip_attr[1] as year, equip_attr[2] as make"
instead.

HTH,

/vjl/


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