Sorry, ignore that. You want one if it's present, *or* the other. On Wednesday, 3 May 2023 at 18:16:01 UTC+1 Brian Candler wrote:
> Another approach is to turn it around using the "unless" operator. This > will *only* give timeseries from the vector on the LHS, and will suppress > any which have matching label sets on the RHS. > > sum by (nodename) (....) unless on (nodename) machine_cpu_cores > > [assuming that machine_cpu_cores has a "nodename" label] > > On Wednesday, 3 May 2023 at 17:18:37 UTC+1 Brian Candler wrote: > >> I think you need to describe: >> * what you actually see >> * what you would like to see instead >> * the results of each of the subexpressions (i.e. left and right of "or") >> in the PromQL browser >> >> Then it should be clearer how to combine them to achieve the result you >> want. >> >> For more info on how the "or" operator works, see: >> https://prometheus.io/docs/prometheus/latest/querying/operators/#logical-set-binary-operators >> >> The most important thing to note is that it's not a boolean: it's a >> union. An expression like "a or b" has a vector of values for "a" and a >> vector of values for "b". The result combines both timeseries a and b into >> the result set. However, if there are any metrics from "a" and "b" which >> match *exactly* the same set of label values, then only the "a" one will be >> included in the result set. >> >> You can restrict the set of labels used for matching using >> "on(labels...)" or "ignoring(labels...)" >> >> In your example: the subexpressions "sum by (nodename) (...)" will only >> have a single label {nodename="XXX"}, whilst the subexpression >> "machine_cpu_cores" very likely has more labels than that (job, instance >> etc) and may not have a "nodename" label at all. Since the labels of the >> LHS and RHS of "or" don't match, both sides are included in the result. >> >> On Wednesday, 3 May 2023 at 15:19:09 UTC+1 Anuj Kumar wrote: >> >>> HI All, >>> >>> I am using the below query but getting output for the two queries . I >>> need to get the output for each query. Any help would be appreciated. >>> >>> machine_cpu_cores or sum by(nodename) >>> (irate(node_cpu_seconds_total{mode="idle"}[5m]) * on(instance) >>> group_left(nodename) node_uname_info) or sum by(nodename) >>> (irate(node_cpu_seconds_total{mode!="idle"}[5m]) * on(instance) >>> group_left(nodename) node_uname_info) >>> >>> Thanks, >>> Anuj Kumar >>> >> -- You received this message because you are subscribed to the Google Groups "Prometheus Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/prometheus-users/5dd1ed7a-3f77-4df9-bd48-151937d5ecb6n%40googlegroups.com.
