mjhjmtu

I have a directory of files. There is a line in each file that says:

# order: N

where N is an integer number. I want to list all files in the directory (or even include them in wrapper script) according to that N number. Is this possible from a bash command-line?

With GNU grep, and assuming file names don't contain colon or newline characters:

$ ls
bar baz foo freeble quux
$ cat ./*
# order: 3
# order: 2
# order: 1
# order: 4
# order: 5
$ grep -m1 -EH '^# order: [0-9]+$' ./* | sort -n -k3 | cut -d: -f1
foo
baz
bar
freeble
quux

With single GNU awk process:

awk 'BEGIN PROCINFO["sorted_in"]="@val_num_asc" 
 /order: [0-9]+/ a[FILENAME]=$NF; nextfile 
 END for(i in a) print i ' ./*

With zsh, you can define a glob sorting order based on the content of those lines with:

byOrder() REPLY=$(grep '^# order:' < $REPLY)

and then use it for example with:

printf '%sn' *(.no+byOrder)

or

sorted_file_list=(*(.no+byOrder))

(also adding a . to the glob qualifier to only consider regular files (not directories, fifos, symlinks...)).