mjhjmtu

The following list represented the tokens between sdb and sd$MAX (while MAX range could be c-z)

For example:

MAX=z
list=$(eval echo sdb..$MAX)
echo $list

strings are: ( example )

sdb sdc sdd sde sdf sdg sdh sdi sdj sdk sdl sdm sdn sdo sdp sdq sdr sds sdt sdu sdv sdw sdx sdy sdz

my question how to shift the list each loop to the next string

example:

for i in disk1 disk2 disk3 disk4 ..................
do
echo $i <...syntax...>
done

expected output

disk1 sdb
disk2 sdc
disk3 sdd
disk4 sde
.
.
.

It is better to use array variables:

#!/bin/bash

MAX=d
eval list=( $(echo sdb..$MAX))
eval disks=( $(echo disk1..$#list[@]) )

for((i=0;i<=$#list[@];i++));do
 echo "$disks[i] $list[i]"
done

On execution:

$ ./script
disk1 sdb
disk2 sdc
disk3 sdd

Adapt as required.

You could use indirection to loop over the array's indices rather than its values directly e.g.

Given

$ list=(sda..f)
$ echo "$list[@]"
sda sdb sdc sdd sde sdf

then

$ for i in "$!list[@]"; do printf '%s %sn' "disk$i" "$list[$i]"; done
disk0 sda
disk1 sdb
disk2 sdc
disk3 sdd
disk4 sde
disk5 sdf

If you really want to do it your way, then you will need to extract a usable integer index from the elements of the strings that you're looping over e.g. by removing the leading string disk using parameter substitution:

for d in disk1 disk2 disk3 disk4; do printf '%s %sn' "$d" "$list[$d##disk]"; done
disk1 sdb
disk2 sdc
disk3 sdd
disk4 sde

You may build a counter for disk column to feet the list & replace strings for use as export

-bash-4.4$ MAX=z
-bash-4.4$ list=$(eval echo sdb..$MAX)
-bash-4.4$ echo $list
sdb sdc sdd sde sdf sdg sdh sdi sdj sdk sdl sdm sdn sdo sdp sdq sdr sds sdt sdu sdv sdw sdx sdy sdz
-bash-4.4$ disk=1 ; for i in $(echo $list) ; do echo $i | sed 's/sd./disk'$disk't &/' ; disk=$(($disk +1)) ; done
disk1 sdb
disk2 sdc
disk3 sdd
disk4 sde
disk5 sdf
disk6 sdg
disk7 sdh
disk8 sdi
disk9 sdj
disk10 sdk
disk11 sdl
disk12 sdm
disk13 sdn
disk14 sdo
disk15 sdp
disk16 sdq
disk17 sdr
disk18 sds
disk19 sdt
disk20 sdu
disk21 sdv
disk22 sdw
disk23 sdx
disk24 sdy
disk25 sdz
-bash-4.4$ 

be careful here I use a define variable disk to count & a t (tabulation) as separator; you may adapt to feet your needs if you prefer a space or something else.

if you wanna get this output in a file add a redirection of output
& of course in a script you may want to use code as this (without ; )

disk=1 
for i in $(echo $list) 
do 
 echo $i | sed 's/sd./disk'$disk't &/' ; disk=$(($disk +1))
done > file

this is the easy way to understand it.
A better method is to use the bash substitution instead of sed
see bash method $!list[@] & either printf command or $parameter/pattern/string substituion advanced functions ...

Bash has two simple and straightforward built-in features that
together will solve your problem: set and shift.

set $list will assign each element of $list to one of bash's
positional parameters.

shift renames all positional parameter $n+1 to $n, effectively
deleting $1 and "shifting" all positional parameters forward.

So, in your example, your statement echo $i <...syntax...> would be
echo $i " " $1; shift.

BTW, personally, I would use bash's printf instead of echo (See
elsewhere for trouble echo can give you): printf "%s %sn" $i $1; shift