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declare
lowerl number:= 1;
upperl number:= 3;
num varchar2(10);
begin
for i into lowerl..upperl
loop
num:=num||to_char(lowerl);
if i=3 then upperl:=5;
end loop;
message(num);
What will be the output ?

Question Posted / guest

Answer Posted / guest

result will be 111

Is This Answer Correct ?    6 Yes 0 No



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