我想使用kubectl的LabelSelector输出"kubectl get pod"排除key的结果

Question

我正在下面的 URL 学习 Kubernetes 标签选择器。

https://kubernetes.io/docs/concepts/overview/working-with-objects/labels/

基于集合的要求

Set-based label requirements allow filtering keys according to a set of values. Three kinds of operators are supported: in, notin and exists (only the key identifier). For example:

! partition

The fourth(This) example selects all resources without a label with key partition; no values are checked.

我的假设是，如果我指定“!test”，所有pods都会被输出，我认为指定“!app”时没有任何输出。

我想按照此处描述的示例排除键和输出，但没有获得预期的结果。

My @ Pc: ~ / Understanding-K8s / chap04 $ kubectl get pod -l! app
My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l !app
-bash: !app: event not found
My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l "!app"
-bash: !app: event not found
My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l "!test"
-bash: !test: event not found
My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l !test
-bash: !test: event not found

执行--show-labels命令的结果如下

My @ Pc: ~ / Understanding-K8s / chap04 $ kubectl get pod --show-labels
NAME          READY   STATUS    RESTARTS   AGE   LABELS
nginx-pod-a   1/1     Running   0          89m   app=photo-view,env=stage
nginx-pod-b   1/1     Running   0          89m   app=imagetrain,env=test
nginx-pod-c   1/1     Running   0          89m   app=prediction,env=test
nginx-pod-d   1/1     Running   0          89m   app=photo-view,env=stage
nginx-pod-e   1/1     Running   0          89m   app=imagetrain,env=stage
nginx-pod-f   1/1     Running   0          89m   app=photo-view,env=prod

如何排除 app 和 env 等预期密钥？

tyzbit给了我答案，我也试过了。评论里没办法很好的描述结果，所以我会在这里补充。

My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l '!app'
No resources found in default namespace.
My@Pc:~/Understanding-K8s/chap04$ kubectl get pod -l '!test'
NAME          READY   STATUS    RESTARTS   AGE
nginx-pod-a   1/1     Running   0          162m
nginx-pod-b   1/1     Running   0          162m
nginx-pod-c   1/1     Running   0          162m
nginx-pod-d   1/1     Running   0          162m
nginx-pod-e   1/1     Running   0          162m
nginx-pod-f   1/1     Running   0          162m

Girdhar Singh Rathore 教我 AWK 和 jsonpath。

(可能是我格式写错了...)

AWK

My@Pc:~$ kubectl get pods --show-labels | awk  ' !~/app/ {print ;}'
NAME          READY   STATUS    RESTARTS   AGE   LABELS

My@Pc:~$ kubectl get pods --show-labels | awk  ' !~/test/ {print ;}'
NAME          READY   STATUS    RESTARTS   AGE   LABELS
nginx-pod-a   1/1     Running   0          18m   app=photo-view,env=stage
nginx-pod-d   1/1     Running   0          18m   app=photo-view,env=stage
nginx-pod-e   1/1     Running   0          18m   app=imagetrain,env=stage
nginx-pod-f   1/1     Running   0          18m   app=photo-view,env=prod

json路径

My@Pc:~$ kubectl get pods -o jsonpath='{range .items[*]} {.metadata.name} {.metadata.labels} {"\n"} {end}' | awk ' !~/app/ {print }'

My@Pc:~$ kubectl get pods -o jsonpath='{range .items[*]} {.metadata.name} {.metadata.labels} {"\n"} {end}' | awk ' !~/test/ {print }'
nginx-pod-a
nginx-pod-d
nginx-pod-e
nginx-pod-f

Answer 1

! 正在被您的 shell 解释，请使用单引号来防止这种情况。正确的语法是：

kubectl get pods -l '!app'

Answer 2

您可以使用awk

kubectl get pods --show-labels | awk  ' !~/app/ {print ;}'

或-o jsonpath

kubectl get pods -o jsonpath='{range .items[*]} {.metadata.name} {.metadata.labels} {"\n"} {end}' | awk ' !~/app/ {print }'

或选择器

kubectl get pods -l '!app'

我想使用kubectl的LabelSelector输出"kubectl get pod"排除key的结果

I want to output the result of "kubectl get pod" excluding the key using LabelSelector of kubectl

kubernetes

kubectl

基于集合的要求