regex, remove everything before first underscore

Question

i have this string grafana-stack_alloy and require everything after the first underscore alloy.

I need the result as a group $1. Tried this one, but it failed. (?:)?([^_]+)*$.

Can anyone help me to solve this problem?

I am testing this:

        rule {
        action        = "replace"
        source_labels = [
            "__meta_docker_container_label_com_docker_swarm_service_name",
        ]
        regex        = "^(?:;*)?([^;]+).*$" work but wrong
        //regex      = "[^_]+.$" ----> not work
        //regex      = "([^_]+)$" ----> not work
        replacement  = argument.namespace.value + "/$1"
        target_label = "job"
    }

It is about the Grafana Agent with relabel-regex: Grafana prometheus.relabel which use Google's RE2 as its regex engine.

Thanks in advance

(?:) looks weird. Isn't (?:foo) a non-capturing version of (foo) in Perl regexes? But that would make (?:) a zero-length match... What's the intent here? — ilkkachu
– ilkkachu, Commented Dec 23, 2024 at 11:17

user10489 · Accepted Answer · 2024-12-23 13:16:51Z

1

These will get everything after the last underscore.

_([^_]+)$

or

^.*_([^_]+)$

If you want everything after the first underscore:

^[^_]*_(.+)$

or change the first * to a + if you don't want leading underscores.

If you want to also eliminate the second underscore in a double underscore but get everything after that, try

^[^_]*_+(.+)$

If you want a more specific match, maybe you want

grafana-stack_([^_]+)$

If none of these do what you want, you'll need to give more of the surrounding context than the string you gave.

edited Dec 23, 2024 at 13:16

answered Dec 23, 2024 at 13:09

user10489

11k1 gold badge15 silver badges37 bronze badges

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Gilles Quénot · Accepted Answer · 2024-12-23 11:32:28Z

0

As far as Grafana use Golang regexes flavor, you need to modify the regex to:

([^_])*$

See online demo explanations

Official doc mention Golang too:

expression needs to be a Go RE2 regex string.

edited Dec 23, 2024 at 11:32

answered Dec 23, 2024 at 10:28

Gilles Quénot

36.8k7 gold badges76 silver badges97 bronze badges

Note that this would match even if there is no _ and also matches the empty string. Not a problem with the OP's input string but might be worth mentioning.

terdon
– terdon ♦

2024-12-23 12:50:50 +00:00
Commented Dec 23, 2024 at 12:50
Also, this gives everything after the last underscore, he wanted everything after the first underscore...although there's no difference for the sample string.

user10489
– user10489

2024-12-23 13:12:17 +00:00
Commented Dec 23, 2024 at 13:12

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Stéphane Chazelas · Accepted Answer · 2024-12-23 21:25:22Z

As the "replacement" is:

replacement  = argument.namespace.value + "/$1"

That implies that what is matched by the regexp is replaced with that.

In grafana-stack_alloy, your ([^_]+)$ regexp will correctly match the alloy part after _ and capture it into $1, but because the whole regexp matches only that, after replacement, you'll get:

grafana-stack_<argument.namespace.value>/alloy

If what you want the result to be after replacement is only <argument.namespace.value>/alloy, then you need the regex to match the whole input and out of that capture the part after _.

As RE2 support perl's *? non-greedy repetition operator, that can be just:

^.*?_(.*)$

The anchors (^, $) are not even needed as that regexp is guaranteed to match the whole input as long as it contains at least one _ (and not match at all with or without anchor if there's no _).

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