Good Morning Laurenz,=0D=0A=0D=0AThat worked perfectly.=0D=0A=0D=0AThank Yo=
u=0D=0A=0D=0AJeff=0D=0A=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=
=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=5F=0D=0A=
Jeffrey Schade=20=0D=0ASystems Consultant, Technology Engineering=0D=0A=0D=0A=
ISO=0D=0A545 Washington Boulevard=0D=0AJersey City, NJ 07310=0D=0AVoice: (2=
01) 469-3738=0D=0A=A0 FAX: (201) 748-1500=A0=A0=0D=0Ajschade@iso.com=0D=0A=0D=
=0A-----Original Message-----=0D=0AFrom: Albe Laurenz [mailto:laurenz.albe@=
wien.gv.at]=20=0D=0ASent: Friday, February 22, 2013 8:31 AM=0D=0ATo: Schade=
, Jeffrey; pgsql-general@postgresql.org=0D=0ASubject: RE: Redefining a colu=
mn within a view=0D=0A=0D=0AJeffrey Schade wrote:=0D=0A> We have a table wh=
ich contains a 3 byte column with datatype CHAR(3)=20=0D=0A> which we want =
to redefine within the view as a CHAR(1) column and a=20=0D=0A> CHAR(2) col=
umn. When I code the SUBSTR function the resulting column=20=0D=0A> datatyp=
e is TEXT. I would like to see the CHAR datatype, is there anything I can d=
o to set the proper datatype. The sample view is below:=0D=0A>=20=0D=0A> CR=
EATE OR REPLACE VIEW schema.jeff=5Fview AS SELECT col1 ,=0D=0A> =
col2,=0D=0A> Substr(col3,1,1) as col3=5Fpart1,=0D=0A>=
Substr(col3,2) as col3=5Fpart2,=0D=0A> =
col4=0D=0A> FROM schema.jeff=5Ftable;=0D=0A=0D=0ATry:=0D=0A[...]=0D=0A =
Substr(col3,1,1)::char(1) as col3=5Fpart1,=0D=0A =
Substr(col3,2)::char(2) as col3=5Fpart2,=0D=0A[...]=0D=0A=0D=0AYours=
,=0D=0ALaurenz Albe=0D=0A=0D=0AThis email is intended for the recipient onl=
y. If you are not the intended recipient please disregard, and do not use =
the information for any purpose.=0D=0A
