Bash Wildcards – How to Glob with Numerical Order

You can use the l parameter expansion flag to pad a number on the left.

i=0; for x in *; do ((++i)); mv -- $x new/${(l:6::0:)i}; done

There is a relatively simple way to do this with only POSIX features: start numbering at 1000001 (for 6 digits) instead of 1, and strip off the leading 1. It is less straightforward but a few characters shorter.

i=1000000; for x in *; do i=$((i+1)); mv -- "$x" new/${i#1}; done

If you want to take advantage of zmv, you can use an arithmetic expression that increments i inside the replacement text.

i=0; zmv '*' '${(l:6::0:)$((++i))}'
i=1000000; zmv '*' '${$((++i))#1}'

Add the o glob qualifier if you need to sort the files in a different order. With zmv, you need to pass the -Q flag when the pattern contains glob qualifiers.

I don't know of a way to change the globbing order, but there's an easy workaround for your case:

grep squiggle /var/log/messages-* /var/log/messages

i.e. don't match the messages files in your glob pattern, and add it to the end of grep's argument list.