How to compare a common field between two indexes and list all values present in one index that are not in the other index?

[tp92222]

Hi,

I have two indexes:
index="abc"

index="dummy"

Now both indexes have one common field ID.

I want to compare index dummy with index abc and list all IDs which are present in index abc, but not in index dummy

eg

index=abc contains

id
2
4
6
8
10

index=dummy contains
1
2
3
4
5
6
7
10

desired result
8

Thanks

[somesoni2 · Accepted solution]

Try this

Updated wrong function used in where.

index=abc OR index=dummy | stats values(index) as index by ID | where mvcount(index)=1 AND index="abc"

Update2
For dbquery

index=abc | table ID | eval From="abc" | append [| dbquery ...your query | table ID | eval From="dbquery" ]  | stats values(From) as From by ID | where mvcount(From)=1 AND From="abc"

[tp92222]

It worked thank you,

but it only works if both data is coming form index

now ,i am fetching abc data directly form database using dbquery and dummy is same (Index data)

can u suggest query for above condition

[tp92222]

|savedsearch "abc"|table ID "IP Address"|eval From="abc"| append [|savedsearch "dummy"|table ID "IP Address"|eval From="dummy"]|stats values(From) as From by ID|where mvcount(From)=1 AND From="abc"

above query doesnt worked for saved search plz suggest modifications
thank you,

[somesoni2]

What's the error do you get? Do the queries work individually?

[somesoni2]

Try the updated answer

[sundareshr]

Try
index=abc OR index=dummy | chart count over id by index | where dummy=0