Splunk Search

sum total unique users in time chart with appendcols

kmccowen
Path Finder

I need to sum the PMBI users and ADF Users to get total user count. Any suggestions?

index=gateway host=sc58lgwap* ApplicationID=GATWY | timechart span=1d dc(RequestorUserID) as "PMBI_Users" | append [search index=ctap host=sc58* sourcetype=gateway userid="userId: *" | timechart span=1d dc(userid) as "ADF_Users"]

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1 Solution

somesoni2
Revered Legend

Try this

index=gateway host=sc58lgwap* ApplicationID=GATWY | timechart span=1d dc(RequestorUserID) as "PMBI_Users" | append [search index=ctap host=sc58* sourcetype=gateway userid="userId: *" | timechart span=1d dc(userid) as "ADF_Users"] | timechart span=1d values(*) as * | addtotals 

View solution in original post

0 Karma

woodcock
Esteemed Legend

Like this:

(index=gateway host=sc58lgwap* ApplicationID=GATWY) OR (index=ctap host=sc58* sourcetype=gateway userid="userId: *") | eval CombinedUser=coalesce(RequestorUserID, userid) | timechart span=1d dc(RequestorUserID) AS PMBI_Users dc(userid) AS ADF_Users dc(CombinedUser) AS Combined_Users

Note: YOU MUST NOT add together dc values to get a combined dc; the combined must be recalculated from the raw data.

0 Karma

somesoni2
Revered Legend

Try this

index=gateway host=sc58lgwap* ApplicationID=GATWY | timechart span=1d dc(RequestorUserID) as "PMBI_Users" | append [search index=ctap host=sc58* sourcetype=gateway userid="userId: *" | timechart span=1d dc(userid) as "ADF_Users"] | timechart span=1d values(*) as * | addtotals 
0 Karma

kmccowen
Path Finder

Awesome! What if I want to remove duplicate users if they were in both apps? Would I table and then dedup?

0 Karma

kmccowen
Path Finder

Meaning if my user id was logged in to both environments in a given day, the total unique users would be 1. Did you already answer this? I'm not finding it.

0 Karma

somesoni2
Revered Legend

@woodcock's answer will give you 3 fields, PMBI_Users ADF_Users and Combined_Users(includes removal of duplicates)

0 Karma

kmccowen
Path Finder

sounds good, thanks Woodcock!

0 Karma

woodcock
Esteemed Legend

The wrong answer was Accepted.

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woodcock
Esteemed Legend

See my answer.

0 Karma
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