sql code to convert varchar colum to int - sql

Question

I am stuck on converting a varchar column schedule containing the following data 0, 1, 2, 3,4,8,9,10,11,12,15,16,17,18,19 to INT. I know, please don't ask why this schedule column was not created as INT initially, long story.
So I tried this, but it doesn't work. and give me an error:
select CAST(schedule AS int) from shift_test:
the query should check if the numbers representing days are found in the schedule filed using the sql code below
select empid, case when ((DateDiff(hour,'01-01-2014 07:00' , '01-01-2014 00:00')/ 24 )% 15) in ( CAST(schedule AS int))
then 'A' else '*' end as shift_A from Shift_test
After executing i get this error.
Conversion failed when converting the varchar value to int.
Any help will be appriciated

Answer 1

Use ISNUMERIC() test if you are using version 2008 or 2008R2. In SQL SERVER 2012 you can use TRY_CAST() function, which checks if the data conversion is allowed for given literal.
Mock up code for SQL Server 2008/R2:
Select col1, col2,
case
when <condition> and isnumeric(col2) then cast(col2 as int)
else <do whatever...>
end
as converted_col2
from <yourtable>;
For SQL Server 2012:
Select col1, col2,
case
when <condition> then try_cast(col2 as int)
else <do whatever...>
end
as converted_col2
from <yourtable>;
Example with SQl Server 2008
declare #T table (empid int, schedule varchar(2)) ;
insert into #T(empid, schedule) values (1, '1');
insert into #T(empid, schedule) values (2, '2');
insert into #T(empid, schedule) values (3, '03');
insert into #T(empid, schedule) values (4, '4');
insert into #T(empid, schedule) values (5, '05');
insert into #T(empid, schedule) values (6, 'A');
select empid,
case
when ISNUMERIC(schedule) = 1
and ((DateDiff(hour,'01-01-2014 07:00' , '10-01-2014 00:00')/ 24 )% 15)
in ( CAST(schedule AS int)) then 'A'
else '*'
end
as shift_A
from #T;

Answer 2

# Binaya Ive added my code for more help.
The schedule column contain (0, 1, 2, 3,4,8,9,10,11,12,15,16,17,18,19) which is varchar.
i want to output A if after this calculation ((DateDiff(hour,'01-01-2014 07:00',startdate)/ 24 )% 15) the result is found in the schedule column else if after the calculation and the result is not found output *
;with Shift_runover (shift_code,schedule,endd,startdate)
-- Start at the beginning of shift.
as
(select shift_code,schedule,Cast(end_date as DateTime) as endd,Cast(start_date as DateTime)as startdate from dbo.Shift_test
union all
-- Add hours up to the desired end date.
select shift_code,schedule,endd,DateAdd(hour, 1,startdate)from Shift_runover where startdate<=endd),
Extendedsamples as
(
-- Calculate the number of days since the beginning of the first shift on 1/1/2014.
select shift_code,schedule,startdate,DateDiff(hour,'01-01-2014 07:00',startdate)/ 24 as Days from Shift_runover ),
Shifts as
(
-- the schedule column contain (0, 1, 2, 3,4,8,9,10,11,12,15,16,17,18,19) which is varchar.
-- i want to output A if ((DateDiff(hour,'01-01-2014 07:00',startdate)/ 24 )% 15) is found in the schedule colume
select *,
case when (DateDiff(hour,'01-01-2014 07:00',startdate)/ 24 )% 15 in(schedule)
then 'A' else '*' end as shift_A
from ExtendedSamples
)
select *
from Shifts
option ( maxrecursion 0 )