Let’s say I do this:
1, Create a snapshot from a running debian base image (+install apache) and commit it to
2, Then I (+install mysql) and commit it to:
3, Now if I try to delete version2 with:
docker rmi testdocker/version2 I get: Conflict, cannot delete c7a70abff108 because the container 0b6cc7178a3f is using it, use -f to force
What is exactly using version2 when the machines are stopped and that is the latest snapshot??? Although if I ignore this warning and use the -f option it will remove version2 without damaging version1. I just want to make sure 100% understanding what’s going on here to avoid further data corruptions.
Also the version1 should not depend on any image, since a base image was pulled in for creating it (therefore it should be a clone of that debian base image). Is this assumption correct? I did some experiments and regarding how I do it 1->2->3 1->2 1->3 I can delete any of the snapshots with the (-f) without affecting the other two.