am Wed, dem 21.02.2007, um 19:21:09 +0100 mailte Sergio Andreozzi folgendes: > Dear all, > > given a column which type is for instance varchar(20)[], is there any SQL > command that let me generate the list of distinct scalar values? > > > e.g.: > col1 > row 1: (aaa, bb, c) > row 2: (dddd, eeee) > row 3: (aaa, eeee) > > the query should return: > > aaa > bb > c > dddd > eeee
Okay, next solution: test=*# select * from a; c ------------- {aaa,bb,c} {dddd,eeee} {aaa,eeee} (3 rows) test=*# select distinct c[s] from a, generate_series(1,3)s where c[s] is not null; c ------ aaa bb c dddd eeee (5 rows) You need to know the greatest upper dimension of the array, in this case 3, for the generate_series - function. Andreas -- Andreas Kretschmer Kontakt: Heynitz: 035242/47150, D1: 0160/7141639 (mehr: -> Header) GnuPG-ID: 0x3FFF606C, privat 0x7F4584DA http://wwwkeys.de.pgp.net ---------------------------(end of broadcast)--------------------------- TIP 9: In versions below 8.0, the planner will ignore your desire to choose an index scan if your joining column's datatypes do not match