当我运行命令kubectl get pods | grep "apisix"
时,我得到以下数据
apisix-dev-78549978b7-pvh2v 1/1 Running 6 (4m19s ago) 8m14s
apisix-dev-dashboard-646df79bf-mwkpc 1/1 Running 6 (4m35s ago) 8m12s
apisix-dev-etcd-0 1/1 Running 0 8m12s
apisix-dev-etcd-1 1/1 Running 0 8m11s
apisix-dev-etcd-2 0/1 CrashLoopBackOff 4 (24s ago) 8m11s
apisix-dev-ingress-controller-58f7887759-28cm9 1/1 Running 0 8m11s
apisix-dev-ingress-controller-6cc65c7cb5-k6dx2 0/1 Init:0/1 0 8m9s
是否有办法删除所有包含单词apisix
的pod,而不是在kubectl delete命令中提及每个pod名称?
2条答案
按热度按时间mnemlml81#
您可以运行一个简单的命令:
g0czyy6m2#
选择(然后可能删除)一组Pod的常用方法是通过
selector
标签。例如,假设存在值为
apisix
的正确标签app
: