Pi chart across 2 tables

[UserInterface]

I spent the whole day on this yesterday and I just can't seem to get my head around it.

I was to create a PI graph that shows if a feild is empty or not, but the information is in 2 different tables (i have another post about this which is closed, it solved a diferent issue so created a new one as to not confuse).

"Workstation_Fields.UDF_CHAR1" has the field I want to report on but only contains a row if it has data, so I wish to compair it to "Systeminfo.WORKSTATIONID" which has a row for every device, but I can't seem to work out how to do it. 

There is a relationship between the 2 tables, but I can still only report on one table at a time. I'm sure im missing something simple as I have no problem doing this in SMS.

Behind the black box is just each hostname in my envro, and you can see that by joining the 2 tables i end up with Null values where the field is missing in the second table.

Below is my relationships so everyone can know what the hell I am talking about 🙂

 

[UserInterface · Accepted solution]

I knew there had to be a simple soultion!!

I just had to add a data source with my original SQL Query, then I simply used the same formular to run my report off the correct field.

[v-jiascu-msft]

Hi @UserInterface,

Since the two tables have relationship, just drag these two columns into the visual. But the NULL values could be a problem, they won't show up in the visual. Is this your issue?

Best Regards,

Dale

[UserInterface]

Yes. I was creating a new field that had a Found/NotFound value, and I am just after a percent complete. 

Like this:

This image is off a defferent field that I did as a test, it worked because there was a line item regardless of wether there was a value or not. Unfortunatly when it use the field that i actually need (not my test one) it is in another table that only contains a line item when there is a value.

Basicly, old one was like this 

WorkstationID, Comp, 123

WorkstationID, Comp2, 456

WorkstationID, Comp3, Null

WorkstationID, Comp4, 789

New one is like this. (No WorkstationID for the Null value)

WorkstationID, 123

WorkstationID, 456

WorkstationID, 789

So I need to use the relationship to look at every Workstation ID, and if it doesn't have a match in other table report it as null..

Otherwise i get this 

I fell like its something simple, i just can't crack it for some reason..

[UserInterface]

After reading this site i thought it was going to be as simple as this

Test = EVALUATE NATURALLEFTOUTERJOIN ( SystemInfo, Workstation_Fields )

 But it doesn't recognise this command