Best way to get only domain from a URL using command with Go template package

Sensu Go
1

Hello, I have a small problem and need your help.
I used several URL in my check and now need here from only the domain:
Here I found the following solution:

But this is in Jinja and I can not get it to run in sensu.

# lables
#   url: https://api.server.local.cloud:8443/test/health?details=true

here i need as result only:

api.server.local.cloud:8443

i the past i use a check like this

command: check_http -H {{ .labels.url }} -C {{ .labels.cert_check_warn | default "35" }},{{ .labels.cert_check_crit | default "15" }} --ssl --sni -f follow -vvv

to this here as example

command: check_http -H {{ .labels.url.rpartition("//")[-1].partition("/")[0] }} -C {{ .labels.cert_check_warn | default "35" }},{{ .labels.cert_check_crit | default "15" }} --ssl --sni -f follow -vvv

i only find less information and examples in the sensu docs:

2

Hi there,

A quick note–when posting code examples, please make sure you surround it with triple backticks like so:

```
code goes here
```

This helps make your snippets easier to read for the folks in the community who are seeing your post.

Re:

But this is in Jinja and I can not get it to run in sensu.

There is no support in Sensu Go for Jinja2 templating. It uses Go templates, and Jinja2 (AFAIK) is a Python templating tool.

Can you elaborate as to what exactly you’re trying to accomplish by just extracting the URL without the HTTP scheme?

i only find less information and examples in the sensu docs:

We certainly try to cover the most common use cases our users and customers run into–if this is something that you’d like to see more examples of, we’d welcome an issue on our Docs repo.

3

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