Consider this list of blobs (or any storage data):
backup-2018-08-29-0000.archive
backup-2018-08-29-0100.archive
backup-2018-08-29-0200.archive
backup-2018-08-29-0300.archive
backup-2018-08-29-0400.archive
backup-2018-08-29-0500.archive
backup-2018-08-29-0600.archive
backup-2018-08-29-0700.archive
backup-2018-08-29-0800.archive
backup-2018-08-29-0900.archive
backup-2018-08-29-1000.archive
backup-2018-08-29-1100.archive
backup-2018-08-29-1200.archive
backup-2018-08-29-1300.archive
backup-2018-08-29-1400.archive
backup-2018-08-29-1500.archive
backup-2018-08-29-1600.archive
backup-2018-08-29-1700.archive
backup-2018-08-29-1800.archive
backup-2018-08-29-1900.archive
backup-2018-08-29-2000.archive
backup-2018-08-29-2100.archive
backup-2018-08-29-2200.archive
backup-2018-08-29-2300.archive
I wish to delete all files except one. So my initial idea is to use --pattern flag.
--pattern
The pattern used for globbing files or blobs in the source. The supported patterns are '*', '?', '[seq]', and '[!seq]'.
But I can not find info about how '*', '?', '[seq]', and '[!seq]' works.
In the command below, what pattern will seize all files excluding backup-2018-08-29-0000.archive
?
$ az storage blob delete-batch --source mycontainer --pattern <pattern>
Update
Additional issue is that I have about 10000 backups collected in more than one year. Using non-batch operations seems non practical.