On 2010-11-06, at 8:24, Bob Ball <[email protected]> wrote: > I am emptying a set of OST so that I can reformat the underlying RAID-6 > more efficiently. Two questions: > 1. Is there a quick way to tell if the OST is really empty? lfs_find > takes many hours to run.
If you mount the OST as type ldiskfs and look in the O/0/d* directories (capital-O, zero) there should be a few hundred zero-length objects owned by root. > 2. When I reformat, I want it to retain the same ID so as to not make > "holes" in the list. From the following information, am I correct to > assume that the id is 24? If not, how do I determine the correct ID to > use when we re-create the file system? If you still have the existing OST, the easiest way to do this is to save the files last_rcvd, O/0/LAST_ID, and CONFIGS/*, and copy them into the reformatted OST. > /dev/sdj 3.5T 3.1T 222G 94% /mnt/ost51 > 10 UP obdfilter umt3-OST0018 umt3-OST0018_UUID 547 > umt3-OST0018_UUID 3.4T 3.0T 221.1G 88% > /lustre/umt3[OST:24] > 20 IN osc umt3-OST0018-osc umt3-mdtlov_UUID 5 The OST index is indeed 24 (18 hex). As for /dev/sdj, it is hard to know from the above info. If you run "e2label /dev/sdj" the filesystem label should match the OST name umt3-OST0018. Cheers, Andreas _______________________________________________ Lustre-discuss mailing list [email protected] http://lists.lustre.org/mailman/listinfo/lustre-discuss
