I know that there are many answers to this question online. However, I would like to know if this alternate solution would work:
ls -lt `find . -name "*.jpg" -print | head -10`
I'm aware of course that this will only give me the first 10 results. The reason I'm asking is because I'm not sure whether the ls is executing separately for each result of find or not. Thanks
In your solution:
the ls will be executed after the find is evaluated
it is likely that find will yield too many results for ls to process, in which case you might want to look at the xargs command
This should work better:
find . -type f -print0 | xargs -0 stat -f"%m %Sm %N" | sort -rn
The three parts of the command to this:
find all files and print their path
use xargs to process the (long) list of files and print out the modification unixtime, human readable time, and filename for each file
sort the resulting list in reverse numerical order
The main trick is to add the numerical unixtime when the files were last modified to the beginning of the lines, and then sort them.
Related
I want to use the GNU find command to find files based on a pattern, and then have them displayed in order of the most recently modified file to the least recently modified.
I understand this:
find / -type f -name '*.md'
but then what would be added to sort the files from the most recently modified to the least?
find can't sort files, so you can instead output the modification time plus filename, sort on modification time, then remove the modification time again:
find . -type f -name '*.md' -printf '%T# %p\0' | # Print time+name
sort -rnz | # Sort numerically, descending
cut -z -d ' ' -f 2- | # Remove time
tr '\0' '\n' # Optional: make human readable
This uses \0-separated entries to avoid problems with any kind of filenames. You can pass this directly and safely to a number of tools, but here it instead pipes to tr to show the file list as individual lines.
find <dir> -name "*.mz" -printf "%Ts - %h/%f\n" | sort -rn
Print the modified time in epoch format (%Ts) as well as the directories (%h) and file name (%f). Pipe this through to sort -rn to sort in reversed number order.
Pipe the output of find to xargs and ls:
find / -type f -name '*.md' | xargs ls -1t
I'm trying to count the total lines in the files within a directory. To do this I am trying to use a combination of find and wc. However, when I run find . -exec wc -l {}\;, I recieve the error find: missing argument to -exec. I can't see any apparent issues, any ideas?
You simply need a space between {} and \;
find . -exec wc -l {} \;
Note that if there are any sub-directories from the current location, wc will generate an error message for each of them that looks something like that:
wc: ./subdir: Is a directory
To avoid that problem, you may want to tell find to restrict the search to files :
find . -type f -exec wc -l {} \;
Another note: good idea using the -exec option . Too many times people pipe commands together thinking to get the same result, for instance here it would be :
find . -type f | xargs wc -l
The problem with piping commands in such a manner is that it breaks if any files has spaces in it. For instance here if a file name was "a b" , wc would receive "a" and then "b" separately and you would obviously get 2 error messages: a: no such file and b: no such file.
Unless you know for a fact that your file names never have any spaces in them (or non-printable characters), if you do need to pipe commands together, you need to tell all the tools you are piping together to use the NULL character (\0) as a separator instead of a space. So the previous command would become:
find . -type f -print0 | xargs -0 wc -l
With version 4.0 or later of bash, you don't need your find command at all:
shopt -s globstar
wc -l **/*
There's no simple way to skip directories, which as pointed out by Gui Rava you might want to do, unless you can differentiate files and directories by name alone. For example, maybe directories never have . in their name, while all the files have at least one extension:
wc -l **/*.*
I have recently run into a problem.
I used a utility to move all my music files into directories based on tags. This left a LOT of almost empty folders. The folders, in general, contain a thumbs.db file or some sort of image for album art. The mp3s have the correct album art in their new directories, so the old ones are okay to delete.
Basically, I need to find any directories within D:/Music/ that:
-Do not have any subdirectories
-Do not contain any mp3 files
And then delete them.
I figured this would be easier to do in a shell script or bash script or whatever else linux/unix world than in Windows 8.1 (HAHA).
Any suggestions? I'm not very experienced writing scripts like this.
This should get you started
find /music -mindepth 1 -type d |
while read dt
do
find "$dt" -mindepth 1 -type d | read && continue
find "$dt" -iname '*.mp3' -type f | read && continue
echo DELETE $dt
done
Here's the short story...
find . -name '*.mp3' -o -type d -printf '%h\n' | sort | uniq > non-empty-dirs.tmp
find . -type d -print | sort | uniq > all-dirs.tmp
comm -23 all-dirs.tmp non-empty-dirs.tmp > dirs-to-be-deleted.tmp
less dirs-to-be-deleted.tmp
cat dirs-to-be-deleted.tmp | xargs rm -rf
Note that you might have to run all the commands a few times (depending on your repository's directory depth) before you're done deleting all recursive empty directories...
And the long story goes...
You can approach this problem from two basic perspective: either you find all directories, then iterate over each of them, check if it contain any mp3 file or any subdirectory, if not, mark that directory for deletion. It will works, but on large very large repositories, you might expect a significant run time.
Another approach, which is in my sense much more interesting, is to build a list of directories NOT to be deleted, and subtract that list from the list of all directories. Let's work the second strategy, one step at a time...
First of all, to find the path of all directories that contains mp3 files, you can simply do:
find . -name '*.mp3' -printf '%h\n' | sort | uniq
This means "find any file ending with .mp3, then print the path to it's parent directory".
Now, I could certainly name at least ten different approaches to find directories that contains at least one subdirectory, but keeping the same strategy as above, we can easily get...
find . -type d -printf '%h\n' | sort | uniq
What this means is: "Find any directory, then print the path to it's parent."
Both of these queries can be combined in a single invocation, producing a single list containing the paths of all directories NOT to be deleted.. Let's redirect that list to a temporary file.
find . -name '*.mp3' -o -type d -printf '%h\n' | sort | uniq > non-empty-dirs.tmp
Let's similarly produce a file containing the paths of all directories, no matter if they are empty or not.
find . -type d -print | sort | uniq > all-dirs.tmp
So there, we have, on one side, the complete list of all directories, and on the other, the list of directories not to be deleted. What now? There are tons of strategies, but here's a very simple one:
comm -23 all-dirs.tmp non-empty-dirs.tmp > dirs-to-be-deleted.tmp
Once you have that, well, review it, and if you are satisfied, then pipe it through xargs to rm to actually delete the directories.
cat dirs-to-be-deleted.tmp | xargs rm -rf
I'd like to construct a Linux command to list all files (with their full paths) within a specific directory (and subdirectories) ordered by access time.
ls can order by access time, but doesn't give the full path. find gives the full path, but the only control you have over the access time is to specify a range with -atime N (accessed at least 24*N hours ago), which isn't what I want.
Is there a way to order by access time and get the full path at once? I could just write a script, but it seems there should be a way to do this with the standard Linux programs.
find . -type f -exec ls -l {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
This will find all files, and sort them by date and then time. You can then use awk or cut to extract the dates and files name from the ls -l output
you could try:
ls -l $(find /foo/bar -type f )
you can add other options (e.g. -t for sorting) to ls command to achieve your goal.
also you could add your searching criteria to find cmd
find . -type f | xargs ls -ldt should do the trick as long as there's not so many files that you hit the command like argument limit and spawn 2 instances of ls.
pwd | xargs -I % find % -type f
find . -type f -exec ls -l --full-time {} \; 2> /dev/null | sort -t' ' -k +6,6 -k +7,7
Alex's answer did not work for me since I had files older than one year and the sorting got messed up. The above adds the --full-time parameter which nuetralizes the date/time values and makes them sortable regardless of how old they are.
I need to search a file in unix which starts with "catalina"
find ... what to be used effectively -name, -exec ? Whats the expression
Also I need to show few files at a time, then show some more. There are huge set of log files in there. I know there is some expression, but forgot...
find /path/to/search/in -name 'catalina*'
Use iname to match case-insensitively.
To not be overwhelmed with a long list of files, filter through less (append |less). You can also use more instead of less.
If catalina is the file name, then use
find -name 'catalina*'
If catalina is the first word contained in the file, then use
find -type f | xargs head -v -n 1 | grep -B 1 -A 1 -e '^catalina'