The labels for the two sides of the division need to match exactly.
If they match 1:1 except for additional labels, then you can use
xxx / on (foo,bar) yyy # foo,bar are the matching labels
or
xxx / ignoring (baz,qux) zzz # baz,qux are the labels to ignore
If they match N:1 then you need to use group_left or group_right.
If you show the results of the two halves of the query separately then we
can be more specific. That is:
sum(kafka_consumergroup_lag{cluster=~"$cluster",consumergroup=~"$consumergroup",topic=~"$topic"})
by (consumergroup, topic)
count(up{job="prometheus.scrape.kafka_exporter"})
On Sunday 26 May 2024 at 08:28:10 UTC+1 Sameer Modak wrote:
> I tried the same i m not getting any data post adding below
>
> sum(kafka_consumergroup_lag{cluster=~"$cluster",consumergroup=~
> "$consumergroup",topic=~"$topic"}) by (consumergroup, topic) / count(up{
> job="prometheus.scrape.kafka_exporter"})
>
> On Saturday, May 25, 2024 at 11:53:44 AM UTC+5:30 Ben Kochie wrote:
>
>> You can use the `up` metric
>>
>> sum(...)
>> /
>> count(up{job="kafka"})
>>
>> On Fri, May 24, 2024 at 5:53 PM Sameer Modak <[email protected]>
>> wrote:
>>
>>> Hello Team,
>>>
>>> I want to know the no of instance data sending to prometheus. How do i
>>> formulate the query .
>>>
>>>
>>> Basically i have below working query but issues is we have 6 instances
>>> hence its summing value of all instances. Instead we just need value from
>>> one instance.
>>> sum(kafka_consumergroup_lag{cluster=~"$cluster",consumergroup=~
>>> "$consumergroup",topic=~"$topic"})by (consumergroup, topic)
>>> I was thinking to divide it / 6 but it has to be variabalise on runtime
>>> if 3 exporters are running then it value/3 to get exact value.
>>>
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>>>
>>
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