'How do I get the External IP of a Kubernetes service as a raw value?
I am running an application with GKE. It works fine but I can not figure out how to get the external IP of the service in a machine readable format.
So i am searching a gcloud or kubectl command that gives me only the external IP or a url of the format http://192.168.0.2:80 so that I can cut out the IP.
Solution 1:[1]
You can use the jsonpath output type to get the data directly without needing the additional jq to process the json:
kubectl get services --namespace ingress-nginx nginx-ingress-controller --output jsonpath='{.status.loadBalancer.ingress[0].ip}'
(Be sure to replace the namespace and service name, respectively, with yours.)
Solution 2:[2]
In my case 'kubectl get services' returns array of items, but not just one service.
So then such jsonpath works fine to me:
kubectl get services -l component=controller,app=nginx-ingress -o jsonpath="{.items[0].status.loadBalancer.ingress[0].ip}"
Solution 3:[3]
The answers above do not provide the output the user asked. The correct command would be:
kubectl -n $namespace get svc $ingressServiceName -o json | jq -r .status.loadBalancer.ingress[].hostname
Solution 4:[4]
...and yet another way... This will list all the "load-balancer" services
kubectl get services --all-namespaces -o json | jq -r '.items[] | { name: .metadata.name, ns: .metadata.namespace, ip: .status.loadBalancer?|.ingress[]?|.ip }'
Depending on the networkPlugin used by your cluster services/pods may be exposed directly on external-ip. But this will also find an Ingress controllers run in the cluster.
Solution 5:[5]
All previous solutions don't work any more for me (on GCP).
To get the IP:
kubectl get ingress <YOUR_INGRESS_NAME> -o jsonpath="{.status.loadBalancer.ingress[0].ip}"
To get the host-name:
kubectl get ingress <YOUR_INGRESS_NAME> -o jsonpath="{.spec.rules[0].host}"
Solution 6:[6]
To get the external-ip on GCP i can use:
kubectl get services --namespace=<your-namespace> -o jsonpath="{.items[0].status.loadBalancer.ingress[0].ip}"
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | kenny |
| Solution 2 | Gadget |
| Solution 3 | Juan Buhagiar |
| Solution 4 | AndyPook |
| Solution 5 | |
| Solution 6 | MC Vermeulen |
