1. Display employee name and salary as

'Above 1500’ if salary >1500 (using Decode)
‘Below 1500’ if salary<1500
‘Exactly 1500’ If salary=1500


2. Write a query to to find out number of days in that month for a given date ?

select to_char(lastday(to_date('02-01-2009'), sysdate from dual

select to_char(lastday(to_date('02-01-2009'), sysdate from dual.

Correct me if I am wrong

better to use case, that is more effective

Hi,
it is some what complex to get your solution using Decode Function..
For your requirement better use case statement..it is advisable..
Anyhow check it out this..

select ename,
sum(decode(greatest(SALARY,29000),least(SALARY,30000),1,0))
from employee
group by ename;

It displays list of ename who r having salary in the range of 29000-30000.

For ur second Query,,
Try this..

SELECT TO_CHAR(LAST_DAY('01-jan-2008'),'dd')
from dual;

thanks alot

SELECT TO_CHAR(LAST_DAY('01-jan-2008'),'dd')
from dual;

sorry, and the above query is working

any way thanks for giving support
your second query is corect

Hi,

If you need to use the decode function then first you have to create a formula that will evaluate to a single number for each one of your ranges


The following code may give you some idea,though I am not sure of the code because I havent run it before posting it to you.

SELECT emp_name,
decode(trunc (( salary + 1500) / 1500), 1, "BELOW",
2, "EXACT",
"ABOVE") as SALARY
FROM employees;


For the second question,I think the solution provided by the other members is sufficient.

Select ename,
DECODE (SUBSTR (sal - 1500, 1, 1),
'-', 'Below',
'0', 'Exact',
'Above'
)
From scott.emp

__________________
Sireesha

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