Org chart

[22strider]

Hello Friends

I have a table that looks like as given under

Employee Manager
E100
E105 E100
E110 E105
E115 E110

As you can see that E100 is manage of E105; E105 is the manager of E110 and E110 is the manager of E115. Is there any way (SQL script may be) for getting the organization chart from this table? Basically I want to display who is directly or indirectly reporting to E100.
Having is third degree of reporting (E110 is also kind of under E100) is throwing me off.

Thanks for you help
Rajesh

 

[Trevor G]

Why don't you use Word or PowerPoint with the Organisational Chart?

 

[22strider]

Well Trever there are other reports that will be generated once I figure out how to display all the org reports. I know I can use self join to display one level of reports. It's the second and third level that I need to figure out.
Thanks for your response

 

[alansidman]

I did a google search "Create Org Chart in MS Access" and got several hits. This one looked to be the easiest, while many required using the information in Access and integrating it into another program which may be a good option for you.

http://www.ehow.com/how_5843962_create-chart-ms-access-2007.html

Alan

 

[22strider]

Hello Alan

Thanks for your message. Probably I was not clear in my initial post. I am looking for query result that would display the chain of command.
For example if I want to display all the employees that direcly or indirectly report to E100. Result of my query should be list of employees E105, E110 and E115.
Even though E110 and E115 are not directly reporting to E100. But they are reporting to manager(s) that are reporting to E100.

Thanks
Rajesh

 

[jackd]

The guru on SQL and hierarchies is Joe Celko
look here for your sample

http://www.ibase.ru/devinfo/DBMSTrees/sqltrees.html

 

[jackd]

Here are some other references

There are other examples out there (not as intimidating perhaps)

http://www.sqlteam.com/article/more-trees-hierarchies-in-sql

http://www.mvps.org/access/queries/qry0023.htm

 

[22strider]

That is one brilliant algorithm; thanks for sharing it.