I want to shrink an ext4 filesystem to make room for a new partition and came across the resize2fs program. The command looks like this:
resize2fs -p /dev/mapper/ExistingExt4 $size
How should I determine $size if I want to substract exactly 15 GiB from the current ext4 filesystem? Can I use the output of df somehow?
Answers:
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Method 1
You should not use df because it shows the size as reported by the filesystem (in this case, ext4).
Use the dumpe2fs -h /dev/mapper/ExistingExt4 command to find out the real size of the partition. The -h option makes dumpe2fs show super block info without a lot other unnecessary details. From the output, you need the block count and block size.
... Block count: 19506168 Reserved block count: 975308 Free blocks: 13750966 Free inodes: 4263842 First block: 0 Block size: 4096 ...
Multiplicating these values will give the partition size in bytes.
The above numbers happen to be a perfect multiple of 1024, so we can calculate the result in KiB:
$ python -c 'print 19506168.0 * 4096 / 1024' 78024672.0
Since you want to shrink the partition by 15 GiB (which is 15 MiB times 1 KiB):
$ python -c 'print 19506168.0 * 4096 / 1024 - 15 * 1024 * 1024' 62296032.0
As resize2fs accepts several kinds of suffixes, one of them being K for “1024 bytes”, the command for shrinking the partition to 62296032 KiB becomes:
resize2fs -p /dev/mapper/ExistingExt4 62296032K
Without unit, the number will be interpreted as a multiple of the filesystem’s blocksize (4096 in this case). See man resize2fs(8)
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