Create a scheduled job using MSSQL Agent

Question

It's my first time working with SQL Agent and i don't have much experience with it!
I need to create a scheduled job that checks few columns of a table everyday and updates one of them based on today's date!

Project table's columns:

CREATE TABLE [dbo].[Project](
[projectID] [int] IDENTITY(1,1) NOT NULL,
[prID] [nvarchar](50) NULL,
[projectName] [nvarchar](20) NOT NULL,
[startDate] [datetime] NOT NULL,
[dueDate] [datetime] NOT NULL,
[progress] [int] NULL,
[status] [bit] NULL,

I need to update the status column everyday based on the progress that the project has this now using the startDate column and today's date. basically something like:

ExpectedProgress = (todayDate-startDate)*(100/(dueDate-StartDate)) Now by comparing the progress column and the ExpectedProgress i can set the value of status column true (on time) or false (late)

Thanks in Advance!

Answer 1

You need to create a stored procedure having the above mentioned query, and then add that stored procedure as a step for the job and schedule it. Your required query should be like:

Update TableName Set ExpectedProgress = datediff(dd,startDate,todaydate)*(100/daediff(dd,StartDate,duedate))
Update TableName set status=(case when progress>= ExpectedProgress then 'true' else 'false' end.

For creating a job here are the steps: http://technet.microsoft.com/en-us/library/ms190268.aspx

Answer 2

You could use a computed column that would give you the expected progress each time you query it. Might not be as good for performance as its recalculated each time it's read but for a reasonable number of rows and for a simple function, like you have in your question, it should perform well.