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Unclear definitions: Inventoried - last scanned - unused...

Question asked by TomB. on Oct 21, 2019
Latest reply on Oct 22, 2019 by Samuel

Hello everyone!

 

I'm pretty new to Snow and haven't even yet had my introduction course which is planned for the near future.

However, there is a necessity for me to clarify a few points and I can't wait until then.

 

Can anyone help me figure out the answers to the below questions? I haven't found what I'm looking for on the forums just yet:

 

 

[A]-->Could you please stipulate the difference between an "inventoried computer" and a "last scanned..." computer?

[B]-->And does an inventoried computer always have an Snow Agent installed?

 

Context: I have two reports open: 1)-All Computers- and 2)-Physical and virtual servers per datacenter- 1) should show all inventoried Computers -aka any machine which has an agent and sends (during the daily scan) information. 2) should show all servers (physical and virtual) divided under their respectively named clusters. In report 2) there is a Column Filter called "Inventoried" with "Yes, no or blank" as input possibilities.

 

[C]-->Are all computers mentioned under "no" uninventoried because they do not have an agent? If that's the case how do we still see them listed in that report?

 

[D]--> How is it possible to have a server listed in 1)-All Computers- and simultaneously "Not inventoried" in 2)-Physical and virtual servers per datacenter-?

 

[E] The Unused Applications Report will sometime show applications that aren't really "unused" and are actually background running apps or services that will be "wrongly" listed. An example of that is the Snow Inventory Agent itself which appears as an "unused application" despite it obviously not being an applications no longer in use and flagged to be uninstalled.

How can I determine what is an actual unused application I can uninstall and what is a service/background application I shouldn't touch?

 

I hope the above is clear enough and that someone more knowledgeable that I am will be able to help out!

 

Thanks in advance!

Kind regards,

 

Tom

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