Separate a list of files by keyword group, then sort each group by reverse date - bash

Question

I use this:
for f in $( ls -tr ${repSource}*.txt );
...to loop on a list of files, sorted oldest to newest.
I want to add another sort "filter": filenames that don't start with "abc" come first, whatever their timestamp.
So if the files are "abc.txt", "def.txt" and "ghi.txt", then "abc.txt" must be last, and the other two come before in the list (sorted by reverse date).

Answer 1

You don't want to parse the output of the ls command. The construct in your question is a classic bash pitfall and parsing ls is well known to be problematic.
You can instead use stat to get the time of each file that you find in a for loop on a pattern expansion.
#!/usr/bin/env bash
uname_s=$(uname -s)
stamp() {
case "$uname_s" in
Linux) stat -c '%Y' "$1" ;;
Darwin|*BSD) stat -f '%m' "$1" ;;
esac
}
inputfiles=( "$#" )
for file in "${inputfiles[#]}"; do
n=$(stamp "$file")
while [ -n "${forward[$n]}" ]; do
((n++)) # introduce fudge to avoid timestamp collissions
done
forward[$n]="$file"
done
declare -p inputfiles
for n in "${!forward[#]}"; do
reverse[$(( 2**32 - $n ))]="${forward[$n]}"
done
declare -p forward reverse
This example script takes a list of files as command line options (which can be a glob), then uses declare -p to show you the original list, the forward-sorted list, and a reverse-sorted list.
The case statement in the stamp function makes it portable between Linux, OS X (Darwin), FreeBSD, NetBSD, etc. since I don't know what operating system you're using. (If it's something less common like Solaris, HP/UX, etc, then the stat command may not be available or useful and this solution might not work.)
Once you have a sorted (non-associative) array in bash, you can process the files one by one with constructs like:
for file in "${forward[#]}"; do
# something to $file
done
or
for file in "${reverse[#]}"; do
# something to $file
done
And you can trust that since non-associative bash arrays are always numerically ordered, you'll get the files in their date order.
And of course, you've got the dates themselves as indexes, if you want. :)

Answer 2

The files:
ls -log
total 12
-rw-rw-r-- 1 4 Apr 8 11:15 abc.txt
-rw-rw-r-- 1 4 Apr 8 11:15 def.txt
-rw-rw-r-- 1 4 Apr 8 11:16 ghi.txt
This seems to work, but there must be a better way:
ls -log --time-style +%s | tail -n +2 | \
pee \
'grep -v abc.txt | sort -k4 -rn' \
'grep abc.txt | sort -k4 -rn' | \
cut -d " " -f 5
Output:
ghi.txt
def.txt
abc.txt
Note: the unseemly named util pee is from debian's moreutils package, it's a "like tee for pipes".
Now to make a loop of it:
# usage: foo filename # filter above sort code by filename
foo() { ls -log --time-style +%s | tail -n +2 | \
pee 'grep -v '"$1"' | sort -k4 -rn' \
'grep '"$1"' | sort -k4 -rn' | \
cut -d " " -f 5 ; }
for f in $( foo abc.txt )