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My goal is deceptively simple (at least to me). I wish to take the output of ls -l or ls -lh and select just one field.
I am looking for this to be as bulletproof as possible, by which I mean, assume that filenames can have a variable number of spaces, not everything in the field has the same length, etc.
Bonus points for having a script that will take the name of the the field (or even just a field number), and then return the contents of the field.
I want to turn
into:
Try ls -l | awk '{print $7}'.
awk selects columns so it's perfect for this task.
ls. The find command is tremendously helpful.
Jul 18, 2012 at 10:02
ls -l --block-size=M?
Jul 4, 2019 at 18:08
ls command is not POSIX standard, so a script that depends on ls could break on different systems or even over time. See the answer by @ormaaj on this page for more details. One way to do what you ask is with a Bash loop. Here is an example. for FILENAME in *; do FILESIZE=$(( $( stat --format=%s "${FILENAME}" ) / 1024 )); echo "${FILENAME}" ${FILESIZE}; done. Of course, instead of echo, you'd use your processing.
Jul 5, 2019 at 7:30
... | xargs -d '\n' du -m which works just as well.
Jul 5, 2019 at 16:13
Never parse ls. Use GNU find. Or if portability isn't important, stat(1).
find . -maxdepth 1 -printf '%Td\n'
For reading data other than lists of filenames line-by-line and splitting into fields, see: BashFAQ/001
There are no methods to reliably read a newline-delimited list of filenames that make sense under most circumstances.
You may fetch the specific column in shell like:
ls -al | while read perm bsize user group size month day time file; do echo $day; done
or awk as shown in @Corey answer, cut -c44-45 would also work after adjustment (since ls has fixed columns) , or whatever else, however the main problem is that it won't be reliable and bulletproof (e.g. on Unix it may be $6, not $7, and it changes depending on arguments) making it not machine-friendly therefore it is not recommended to parse ls command at all.
The best is to use different available commands such as find or stat, which can provide relevant options to format the output as you need. For example:
$ stat -c "%x %n" *
2016-04-10 04:53:07.000000000 +0100 001.txt
2016-04-10 05:08:42.000000000 +0100 7c1c.txt
To return column of only days of modifications, try this example:
stat -c "%x" * | while read ymd; do date --date="$ymd" "+%d"; done
It's worth to note that GNU stat could have different options to BSD stat, so it still won't be bulletproof across different operating systems.
stat . date however is setup incorrectly. You should add --date=$ymd , since date by itself will print the current day, but the purpose here is to convert date format of the file
Apr 10, 2016 at 4:37
ls -la | cut -c32-33 command in total honesty is just completely unreliable, not only because of the possible pitfalls with the filenames, but simply because it depends on the usernames' length and on the files' size. The stat -c "%x" *.* command looks ok, but by using *.* you're restricing it only to filenames containing a dot. I guess the intention was to catch also hidden files; in that case you should enable globbing for dotfiles beforehand and use * instead of *.*: shopt -s dotglob; stat -c "%x" * | [...].
--date="$ymd" to the date command, otherwise it will print the current day of the month.