This question was posed on a forum:

I have a customer who has a computer, 2 SATA disk (striped in RAID config. Windows won’t load. Diag reports bad hard drive. When I disconnect one, it kills the stripe and the computer appears to not have a hard drive at all. Seems kind of silly to have it set this way as it increases the risk of failure. Other than putting each hard drive in another computer, I’d like to determine which of the disk are bad.

Also, not quite sure how to attack data recovery as they are a stripe set and plugging in to a SATA to USB does not appear to be a valid method. If I put a third hard drive in as a boot drive, do i have to reconfig the stripe set and if i do, will it kill the data.

I have reassembled two RAID-0 “striped” drives to a single larger drive by hand before. It’s actually a programmatically simple operation, but you require a lot of low-level knowledge and some guesswork to do it. The specific pair I had appeared to store the metadata somewhere other than the start of the disk, and I was able to discover through a hex editor that the drive was on a 64KB stripe size. I also spotted which drive had a partition table and which didn’t, because that’s only on the first drive which contains the first stripe.

At a Linux command prompt, with the two RAID-0 disks (that were NOT being detected properly by the Linux LVM2 “FakeRAID” algorithms, by the way) and a disk of twice their size connected, I wrote a very simple script that looked something like this (sda/sdb as RAID-0, sdc as the destination disk, and this might work under ash or similar as well).

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#/bin/bash

; X=sda position, Y=sdb position, Z=sdc position, STRIPE=stripe size
X=0; Y=0; Z=0; STRIPE=65536

; Retrieve the size of a RAID-0 disk so we can terminate at the end
SIZE=$(cat /proc/partitions | grep ‘sda$’ | awk ‘{print $3}’)
; Divide size by stripe, including any tail blocks (expr truncates)
SIZE=$(( SIZE + STRIPE – 1 ))
SIZE=$(expr $SIZE / $STRIPE ))
while [ “$Z” -lt “$SIZE” ]
do
dd if=/dev/sda of=/dev/sdc seek=$Z; skip=$X bs=$STRIPE count=1
Z=$(( Z + 1 ))
dd if=/dev/sdb of=/dev/sdc seek=$Z; skip=$Y bs=$STRIPE count=1
Z=$(( Z + 1 ))
X=$(( X + 1 ))
Y=$(( Y + 1 ))
done

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Note that all it does is load 64K at a time from each disk and save it to the third disk in sequential order. This is untested, and requires modification to suit your scenario, and is only being written here as an example. It does not fail if a ‘dd’ command fails, so it will work okay for data recovery; you will lose any stripe that contains a bad block, though, and the algorithm could be improved to use dd_rescue (if you have it) or to copy smaller units in a stripe so that only a partial stripe loss occurs on bad data.