不使用-S选项对ls -alh的输出值进行排序

Question

I have a script that recursively goes through directories and appends the result of running ls -alh --block-size=KB | grep ^\\- ls -alh --block-size=KB | grep ^\\- to a file. I then need to sort the resulting file by decreasing file size in the same way that using the -S option would if it was used at the point where ls was called.

Answer 1

The many issues of trying to parse ls are covered well in Why you shouldn't parse the output of ls and Fixing Unix/Linux/POSIX Filenames for an idea of what others have tried before you.

Some additional reasons that your approach will not work reliably:

1. If you recurse over a device boundary, some versions of ls may add a column to show the new device ID and throw off your sorting and parsing;
2. You are using sed to remove the kB / mB / gB magnitude of the output of ls -h . That will sort a 2 byte file, 2 kilobyte file and 2 megabyte file all together as the same size.
3. The output of ls changes when you pipe it or display at the terminal also changing the logic of the parsing / sorting.

The solution is to use a glob and sort based on an added column to the output of ls .

We can use dd to create a list of test files of some known sizes:

dd if=/dev/zero of=A  bs=2  count=1
dd if=/dev/zero of=B  bs=1024  count=2
dd if=/dev/zero of=C  bs=1024  count=3
dd if=/dev/zero of=D  bs=1024  count=150
dd if=/dev/zero of=E  bs=1024  count=2000

Resulting in:

$ ls -lh *
-rw-r--r--  1 andrew  wheel     2B Jan  8 20:52 A
-rw-r--r--  1 andrew  wheel   2.0K Jan  8 20:52 B
-rw-r--r--  1 andrew  wheel   3.0K Jan  8 20:52 C
-rw-r--r--  1 andrew  wheel   150K Jan  8 20:52 D
-rw-r--r--  1 andrew  wheel   2.0M Jan  8 20:52 E

If you sort the output of ls by the -S switch:

$ ls -lhS *
-rw-r--r--  1 andrew  wheel   2.0M Jan  8 20:52 E
-rw-r--r--  1 andrew  wheel   150K Jan  8 20:52 D
-rw-r--r--  1 andrew  wheel   3.0K Jan  8 20:52 C
-rw-r--r--  1 andrew  wheel   2.0K Jan  8 20:52 B
-rw-r--r--  1 andrew  wheel     2B Jan  8 20:52 A

You approach would remove the M K or B in column five and then sort on that. That would result in A, B and E sorting together.

---

(It is possible to crudely sort the output of ls like so:

$ ls -al | grep ^\- | sort -nrk 5
-rw-r--r--   1 andrew  wheel  2048000 Jan  8 20:52 E
-rw-r--r--   1 andrew  wheel   153600 Jan  8 20:52 D
-rw-r--r--   1 andrew  wheel     3072 Jan  8 20:52 C
-rw-r--r--   1 andrew  wheel     2048 Jan  8 20:52 B
-rw-r--r--   1 andrew  wheel        2 Jan  8 20:52 A

but that does not produce the output of -h that you have...)

---

The correct way is to do this is to use a Decorate / Sort / Undecorate pattern with a glob.

for fn in *; do
    [ -f "$fn" ] || continue
    c1=$(($(wc -c < "$fn")))
    c2=$(ls -alh "$fn")
    printf "%s\t%s\n" "$c1" "$c2"
done | sort -nrk 1 | cut -f 2   

Result:

-rw-r--r--  1 andrew  wheel   2.0M Jan  8 20:52 E
-rw-r--r--  1 andrew  wheel   150K Jan  8 20:52 D
-rw-r--r--  1 andrew  wheel   3.0K Jan  8 20:52 C
-rw-r--r--  1 andrew  wheel   2.0K Jan  8 20:52 B
-rw-r--r--  1 andrew  wheel     2B Jan  8 20:52 A

Which is the same as using ls -lhS

If you are recursing a file tree and writing to a file, the general methodology is the same.

Answer 2

My solution which is good enough for my purposes, though the accepted answer is much better:

sed 's/kB//' files.tmp > files1.tmp #remove first instance of "kB" from each line
sed 's/ \+/ /g' files1.tmp > files2.tmp #replace all multiple spaces with single space
sort -k 5n,5 files2.tmp | tac > files3.tmp #sort by numeric file size and reverse

This only works due to giving the --block-size=KB option to ls .