Reputation: 13
This is my ls
command and post-processing:
ls -l $pwd | tail -n +2 | cut -c1-10,50-999999 | sed 's/./& /g' |
sed 's/\(.\{7\}\)/& /g' | sed 's/\(.\{30\}\)/&/g'
This is the output:
- r w x r - - r - - a d d . o l d
I want to remove all the spaces within the filename, so I can end up with something like this: (keep in mind the space pattern is kept at the permissions)
- r w x r - - r - - add.old
Upvotes: 1
Views: 3849
Reputation: 753735
Writing bullet-proof code to parse ls
output is tricky if the file names are not reasonably clean. With the caveat that your file names must not contain newlines and should avoid other control characters (but more or less anything else should be OK) you can just about risk parsing the output from ls
. If your file names are limited to the portable file name character set ([-a-zA-Z0-9_.]
), you should be fine. But be aware that not everyone will be as disciplined as you are with their file names, so your scripts can fail suddenly if someone creates a name with unexpected characters. Note that leading dashes in file names can wreak havoc; a name such as --version
will make the average GNU utility behave unexpectedly, for example.
You can use cut -c1-10,50-
to avoid typing all those 9's. You can combine your 3 sed
commands into one using -e
in front of each expression.
With a little bit of sed
trickery, you can deal with the spacing more easily. On my Mac, the correct column for the start of the file name is column 54, not 50:
$ ls -l |
> sed -e '1d;12q' | # Only 11 file names listed
> cut -c 1-10,54- |
> sed -e 's/\(.\)\(...\)\(...\)\(...\)/\1\2 \3 \4 @@/' \
> -e h -e 's/@@.*/@@/' -e 's/[^@]/& /g' -e G -e 's/\n//' \
> -e 's/@@[^@]*@@//'
- r w - r - - r - - 2da.c
- r w - r - - r - - 3dalloc1.c
- r w - r - - r - - 3dalloc2.c
- r w - r - - r - - 3dalloc3.c
- r w - r - - r - - 3dalloc4.c
- r w - r - - r - - 4 @@ Testing
- r w - r - - r - - @@ Testing
d r w x r - x r - x Floppy
d r w x r - x r - x IQ
d r w x r - x r - x Primes
d r w x r - x r - x SHA-256
$
What does the sed
command do?
@@
at the end of the permissions.@@
and everything after it with just @@
.@
character.@@
markers and everything between that isn't an @
.This leaves @@
markers in the file names untouched — witness the two names with spaces in them that contain @@
.
For the record, the 11 lines passed to cut
and then sed
were:
-rw-r--r-- 1 jleffler staff 2362 Mar 6 19:48 2da.c
-rw-r--r-- 1 jleffler staff 1638 Mar 6 19:48 3dalloc1.c
-rw-r--r-- 1 jleffler staff 2870 Mar 6 19:48 3dalloc2.c
-rw-r--r-- 1 jleffler staff 2968 Mar 6 19:48 3dalloc3.c
-rw-r--r-- 1 jleffler staff 2096 Mar 6 19:48 3dalloc4.c
-rw-r--r-- 1 jleffler staff 0 Mar 21 16:46 4 @@ Testing
-rw-r--r-- 1 jleffler staff 0 Mar 21 16:47 @@ Testing
drwxr-xr-x 4 jleffler staff 136 Mar 9 23:03 Floppy
drwxr-xr-x 8 jleffler staff 272 Mar 6 19:48 IQ
drwxr-xr-x 33 jleffler staff 1122 Mar 6 19:48 Primes
drwxr-xr-x 13 jleffler staff 442 Mar 6 19:48 SHA-256
Upvotes: 0
Reputation: 36346
You simply don't. There's a whole universe of articles out there detailing why you should not be parsing ls
output, but use combinations of tools like your shell's (most likely very comprehensive) file name globbing, find
, and stat.
for example:
for name in * ;do echo $(stat -c '%A' "$name"):$name ; done
EDIT: stat
gives you a lot of formats to help you achieve your desired output, and now that this gives you an unambiguous output (still, this can go wrong with things like newlines in file names), you can just use sed
on the stat
output in isolation. See man stat
.
Upvotes: 4