Select Inverse Function

[talkiechick]

Is there a select Inverse function in MS Access? I have a query that generates a list of sales people, some of the sales people do not show up on the list because they do not have any current deals.

I want a function or query of some sort that will connect the the original query and generate the sales people who have been excluded.

Does anyone have any ideas how i might accomplish this?

 

[SOS]

You use an OUTER join in the query.

 

[talkiechick]

Do I need a full outer join? Could i use just a right or left outer join? Also is there an easier way to accomplish this without an outer join?, I only need one field to show up.

 

[SOS]

Access doesn't have Full Outer Joins. What you need is to have an outer join from the table with the sales information to the table with the sales people. Left or right it really makes no difference as long as you have the arrow going towards the sales people table. When it is a LEFT join, that means the first table referenced is the one that you are joining from and a Right one is the second you are joining from. In other words, think of the graphical interface - Query by Example grid. When you add two tables in, one is on the left and one is on the right. If you are wanting all records from the first table and those that match on the second then it is a Left Join. If you were wanting all records from the second table and only those that matched those then you would have a Right Join.

Does that help?

 

[talkiechick]

Ok, understood and implemented... However, It shows all the sales people, but I am trying to exclude the ones that are equal to the query. So only show sales people with no deals.

 

[SOS]

Okay, this may be hard to describe, but you basically also include the field (without displaying) that is linked on and then use IS NULL for the criteria for that one.

 

[talkiechick]

Smeghead,

Thank you soooo Much... worked like a charm.

 

[SOS]

No problem, glad you understood what I was saying. I wasn't sure I could understand what I was saying.