Unable to understand about the NVL

Question

i want to know the space used by the following column , it may be possible that one of these column will bw null, for taht purpose i m using NVL , but i m unable to understand that where i will put null, to get the exact space
i have three queries here , let me know which will give me correct result ,because all these three queries gives me different result
select (
nvl(sum(length(mis_skill_lookup)),0) +
nvl(sum(length(mis_skill_selected)),0) +
nvl(sum(length(mis_place_selected)),0) +
nvl(sum(length(mis_agent_selected)),0) +
nvl(sum(length(mis_dblookup_checkpoint)),0) +
nvl(sum(length(mis_call_duration)),0) +
nvl(sum(length(mis_survey_result)),0) +
nvl(sum(length(mis_survey_id)),0)
     / 1024) / 1024 a
from mis_data where mis_dblookup_checkpoint <> 0 and mis_call_duration <> 0;
in the second and third query i have put nvl only in one column , just for an example, i will latter on put it into all the column
select ((sum(length(mis_skill_lookup)) +
sum(length(mis_skill_selected)) +
sum(length(mis_place_selected)) +
sum(length(mis_agent_selected)) +
sum(length(mis_dblookup_checkpoint)) +
sum(length(mis_call_duration)) +
sum (length (nvl(mis_survey_result,0))) +
sum(length(mis_survey_id))) / 1024) / 1024 a
from mis_data where mis_dblookup_checkpoint <> 0 and mis_call_duration <> 0;
select ((sum(length(mis_skill_lookup)) +
sum(length(mis_skill_selected)) +
sum(length(mis_place_selected)) +
sum(length(mis_agent_selected)) +
sum(length(mis_dblookup_checkpoint)) +
sum(length(mis_call_duration)) +
nvl(sum(nvl(length('||mis_survey_result||'),0)),0) +
sum(length(mis_survey_id))) / 1024) / 1024 a
from mis_data where mis_dblookup_checkpoint <> 0 and mis_call_duration <> 0;

Answer 1

Hi,
assuming all your columns have datatype varchar2 none
of your select statements will give you a correct
result. You will have to put NVL inside the SUM, so
the statements are reading:
sum ( length ( nvl ( <column_name>, 0 ) ) )
then you will get the right result.
Regards Markus ... btw: please change your username
to something human readableThis will never give correct answer
use
sum ( nvl( length( <column_name> ),0 ) )
eg:
sql>select * from test;
STR 
a 
     --null here  
     ---null here
b 
sql>
select sum(length(nvl(str,0))) sm1,sum(nvl(length(str),0)) sm2,nvl(sum(length(str)),0) sm3
from test;
SM1 SM2 SM3 
4  2  2