do something like:
from DCOracle import *
conn=Connect( "%s/%s@%s" % (user,pw,srv) )
h=conn.prepare("insert into table_x (x,y,z) values :p1, p2:, :p3")
long_str = 'a'*8000
h.execute(1,2,long_str)
assuming column z is declared as LONG
hth
peter.
On Fri, 17 Nov 2000, Dorothea Kuehn wrote:
:
Hi,
If I try to insert an ca.5000 char string with DCOracle(1.3.2) into a LONG
column in Oracle (8i) I keep getting ORA-1704 'string literal too
long'-Errors.
Ok, I read here
http://osi.oracle.com/~tkyte/Misc/LargeStrings.html
that it will work (a) with bind variables or (b) that I have to use
O
