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Grouping and active or non active problem

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Fadius

IS-IT--Management
Jul 25, 2001
139
US
I have a database that contains multiple sites. The report I am working on is groups by site, then by person. The problem I am having is this:

A person may be linked to multiple sites, but inactive at 1 site and active at another site. The person shows up at both sites. I have thus far been unable to have the person not appear on the report for the site they are inactive at.

I have tried adding a third group based on Active status, and the person shows up twice, once as active and once as inactive for that site.

Any help would be greatly appreciated on this.



 
If there is a code for "Active/Inactive", why not use that in your record selection formula:

{table.status} = "Active"

-LB
 
I have tried that. It does not work for whatever reason and will will show them because they are active at another site.

This is why I have posted here as I am perplexed. I am wondering if a while printing records may work for this, but sure exactly how or where to place this.
 
Please explain more fully the results you are receiving after adding the selection criterion and in what way it is not working. It might help if you show some sample results.

-LB
 
Well, if a person is linked to multiple sites, and is inactive at one of site, they still show up at the site in whcih they are inactive for.

Example:

Minneapolis
----John Doe Active

St. Paul
----John Doe Inactive




Because John Doe is linked to both sites, he will show and be displayed under both locations. Thsi happens even if I use record selection table.active=true
 
I think you must be using the wrong criterion. You should right click on the {table.active} field->browse field and report back with the values you see there.

-LB
 
0.00 or 1.00

I have tried the criteron with true and 1.00 as this is a check box true/false field in the database.

When I browse the data in the field, I see both 0.00 and 1.00
 
If 1 = Active, then change that in your record selection formula:

{table.status} = 1

Is it a string? If so, then use:

{table.status} = "1.00"

-LB
 
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