Question

We have a requirement in project to store all the revisions(Change History) for the entities in the database. Currently we have 2 designed proposals for this:

e.g. for ‘Employee’ Entity

Design 1:

-- Holds Employee Entity 'Employees (EmployeeId, FirstName, LastName, DepartmentId, .., ..)'  -- Holds the Employee Revisions in Xml. The RevisionXML will contain -- all data of that particular EmployeeId 'EmployeeHistories (EmployeeId, DateModified, RevisionXML)' 

Design 2:

-- Holds Employee Entity 'Employees (EmployeeId, FirstName, LastName, DepartmentId, .., ..)'  -- In this approach we have basically duplicated all the fields on Employees  -- in the EmployeeHistories and storing the revision data. 'EmployeeHistories (EmployeeId, RevisionId, DateModified, FirstName,        LastName, DepartmentId, .., ..)' 

Is there any other way of doing this thing?

The problem with the ‘Design 1’ is that we have to parse XML each time when you need to access data. This will slow the process and also add some limitations like we cannot add joins on the revisions data fields.

And the problem with the ‘Design 2’ is that we have to duplicate each and every field on all entities (We have around 70-80 entities for which we want to maintain revisions).

Answer 1

1. Do not put it all in one table with an IsCurrent discriminator attribute. This just causes problems down the line, requires surrogate keys and all sorts of other problems.
2. Design 2 does have problems with schema changes. If you change the Employees table you have to change the EmployeeHistories table and all the related sprocs that go with it. Potentially doubles you schema change effort.
3. Design 1 works well and if done properly does not cost much in terms of a performance hit. You could use an xml schema and even indexes to get over possible performance problems. Your comment about parsing the xml is valid but you could easily create a view using xquery – which you can include in queries and join to. Something like this…

CREATE VIEW EmployeeHistory AS , FirstName, , DepartmentId  SELECT EmployeeId, RevisionXML.value('(/employee/FirstName)[1]', 'varchar(50)') AS FirstName,    RevisionXML.value('(/employee/LastName)[1]', 'varchar(100)') AS LastName,    RevisionXML.value('(/employee/DepartmentId)[1]', 'integer') AS DepartmentId,  FROM EmployeeHistories