I have this folders:
2014-09-01-00:00:01
2014-09-01-01:00:01
2014-09-01-02:00:01
2014-09-01-03:00:01
2014-09-01-04:00:01
(There are many more folders)
I write this names to an array. (folders=("2014-09-01-00:00:01" "2014-09-01-01:00:01"...))
How can I get the folder with the newest date? (Not based on the creation/modified date)
ls is the first thing to think about, but parsing ls is evil. Hence, I would use find for this:
find /your/path -mindepth 1 -maxdepth 1 -type d | sort -rn
It looks for all the directories in /your/path (not sub-directories) and sorts them numerically.
The first one will be the newer. Adding | head -1 we get just that one.
Given the convenient date/time format you're using, you can simply sort lexicographically, using sort:
ls -d PARENT_DIR/*/ | sort | tail -1
you can use below command
ll -rt | tail -n 1
Actually it will give you the latest file or directory in your parent directory.
But as you added in your question that your parent directory contains only directories. You can simply use above command.
bash filename pattern expansion gives you alphabetically sorted results, so:
folders=(*/)
echo "${folders[-1]}"
Related
I'm writing a bash script that should output the first 10 heaviest files in a directory and all subfolders (the directory is passed to the script as an argument).
And for this I use the following command:
sudo du -haS ../../src/ | sort -hr
, but its output contains folder sizes, and I only need files. Help!
Why using du at all? You could do a
ls -S1AF
This will list all entries in the current directoriy, sorted descending by size. It will also include the names of the subdirectories, but they will be at the end (because the size of a directory entry is always zero), and you can recognize them because they have a slash at the end.
To exclude those directories and pick the first 10 lines, you can do a
ls -S1AF | head -n 10 | grep -v '/$'
UPDATE:
If your directory contains not only subdirectories, but also files of length zero, some of those empty files might not be shown in the output, as pointed out in the comment by F.Hauri. If this is an issue for your application, I suggest to exchange the order and do a
ls -S1AF | grep -v '/$' | head -n 10
instead.
Would you please try the following:
dir="../../src/"
sudo find "$dir" -type f -printf "%s\t%p\n" | sort -nr | head -n 10 | cut -f2-
find "$dir" -type f searches $dir for files recursively.
The -printf "%s\t%p\n" option tells find to print the filesize
and the filename delimited by a tab character.
The final cut -f2- in the pipeline prints the 2nd and the following
columns, dropping the filesize column only.
It will work with the filenames which contain special characters such as a whitespace except for
a newline character.
I have two directories with the same list of files. I need to compare all the files present in both the directories using the diff command. Is there a simple command line option to do it, or do I have to write a shell script to get the file listing and then iterate through them?
You can use the diff command for that:
diff -bur folder1/ folder2/
This will output a recursive diff that ignore spaces, with a unified context:
b flag means ignoring whitespace
u flag means a unified context (3 lines before and after)
r flag means recursive
If you are only interested to see the files that differ, you may use:
diff -qr dir_one dir_two | sort
Option "q" will only show the files that differ but not the content that differ, and "sort" will arrange the output alphabetically.
Diff has an option -r which is meant to do just that.
diff -r dir1 dir2
diff can not only compare two files, it can, by using the -r option, walk entire directory trees, recursively checking differences between subdirectories and files that occur at comparable points in each tree.
$ man diff
...
-r --recursive
Recursively compare any subdirectories found.
...
Another nice option is the über-diff-tool diffoscope:
$ diffoscope a b
It can also emit diffs as JSON, html, markdown, ...
If you specifically don't want to compare contents of files and only check which one are not present in both of the directories, you can compare lists of files, generated by another command.
diff <(find DIR1 -printf '%P\n' | sort) <(find DIR2 -printf '%P\n' | sort) | grep '^[<>]'
-printf '%P\n' tells find to not prefix output paths with the root directory.
I've also added sort to make sure the order of files will be the same in both calls of find.
The grep at the end removes information about identical input lines.
If it's GNU diff then you should just be able to point it at the two directories and use the -r option.
Otherwise, try using
for i in $(\ls -d ./dir1/*); do diff ${i} dir2; done
N.B. As pointed out by Dennis in the comments section, you don't actually need to do the command substitution on the ls. I've been doing this for so long that I'm pretty much doing this on autopilot and substituting the command I need to get my list of files for comparison.
Also I forgot to add that I do '\ls' to temporarily disable my alias of ls to GNU ls so that I lose the colour formatting info from the listing returned by GNU ls.
When working with git/svn or multiple git/svn instances on disk this has been one of the most useful things for me over the past 5-10 years, that somebody might find useful:
diff -burN /path/to/directory1 /path/to/directory2 | grep +++
or:
git diff /path/to/directory1 | grep +++
It gives you a snapshot of the different files that were touched without having to "less" or "more" the output. Then you just diff on the individual files.
In practice the question often arises together with some constraints. In that case following solution template may come in handy.
cd dir1
find . \( -name '*.txt' -o -iname '*.md' \) | xargs -i diff -u '{}' 'dir2/{}'
Here is a script to show differences between files in two folders. It works recursively. Change dir1 and dir2.
(search() { for i in $1/*; do [ -f "$i" ] && (diff "$1/${i##*/}" "$2/${i##*/}" || echo "files: $1/${i##*/} $2/${i##*/}"); [ -d "$i" ] && search "$1/${i##*/}" "$2/${i##*/}"; done }; search "dir1" "dir2" )
Try this:
diff -rq /path/to/folder1 /path/to/folder2
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.
How can I print in the Unix shell the number of directories in a tree which contain other directories?
I haven't found a solution yet with commands like find or ls.
You can use find command: find . -type d -not -empty
That will print every subdirectory that is not empty. You can control how deep you want the search with -maxdepth.
To print the number, you can use wc -l.
find . -type d -not -empty | wc -l
If you generate a list of all the directories under a particular directory, and then remove the last component from the name, you have a list of the directories containing subdirectories, but there are likely to be repeats in that list. So, you need to post-process the list, yielding (as a first approximation):
find ${base:-.} -type d |
sed 's%/[^/]*$%%' |
sort -u
Find all the directories under the directory or directories listed in variable $base, defaulting to the current directory, and print their names. The code assumes you don't have directories with a newline in the name. If you do, there are fixes, but the best fix is to rename the directory. The sed command removes the last slash and everything after it. The sort eliminates duplicate entries. What's left is the list of directories containing subdirectories.
Well, more or less. There's the degenerate case to consider: the top-level directories in the list will be listed regardless of whether they have sub-directories or not. Fixing that is a bit harder. You need to eliminate any lines of output that exactly match the directories specified to find before removing trailing material. So, you need something like:
{
printf '\\#^%s$#d\n' ${base:-.}
echo 's%/[^/]*$%%'
} > sed.script
find ${base:-.} -type d |
sed -f sed.script |
sort -u
rm -f sed.script
The \\#^%s$#d assumes you don't use # in directory names. If you do use it, then you need to find a character you don't use in names (maybe Control-A) and use that in place of the #. If you could face absolutely any character, then you'll need to do more work escaping some obscure character, such as Control-A, when it appears in a directory name.
There's a problem still: using a fixed name like sed.script for a temporary file name is bad (for multiple reasons — such as two people trying to run the script at the same time in the same directory, though it can also be a security risk), so use mktemp to create a temporary file name:
tmp=$(mktemp ${TMPDIR:-/tmp}/dircnt.XXXXXX)
trap "rm -f $tmp; exit 1" 0 1 2 3 13 15
{
printf '\\#^%s$#d\n' ${base:-.}
echo 's%/[^/]*$%%'
} > $tmp
find ${base:-.} -type d |
sed -f $tmp |
sort -u
rm -f $tmp
trap 0
This deals with the most common signals (HUP, INT, QUIT, PIPE, TERM) and removes the temporary file even if one of those arrives.
Clearly, if you want to simply count the number of directories, you can pipe the output from the commands above through wc -l to get the count.
ls -1d */*/. | cut -d / -f1 | uniq
Is there any method in Linux to calculate the number of files in a directory (that is, immediate children) in O(1) (independently of the number of files) without having to list the directory first? If not O(1), is there a reasonably efficient way?
I'm searching for an alternative to ls | wc -l.
readdir is not as expensive as you may think. The knack is avoid stat'ing each file, and (optionally) sorting the output of ls.
/bin/ls -1U | wc -l
avoids aliases in your shell, doesn't sort the output, and lists 1 file-per-line (not strictly necessary when piping the output into wc).
The original question can be rephrased as "does the data structure of a directory store a count of the number of entries?", to which the answer is no. There isn't a more efficient way of counting files than readdir(2)/getdents(2).
One can get the number of subdirectories of a given directory without traversing the whole list by stat'ing (stat(1) or stat(2)) the given directory and observing the number of links to that directory. A given directory with N child directories will have a link count of N+2, one link for the ".." entry of each subdirectory, plus two for the "." and ".." entries of the given directory.
However one cannot get the number of all files (whether regular files or subdirectories) without traversing the whole list -- that is correct.
The "/bin/ls -1U" command will not get all entries however. It will get only those directory entries that do not start with the dot (.) character. For example, it would not count the ".profile" file found in many login $HOME directories.
One can use either the "/bin/ls -f" command or the "/bin/ls -Ua" command to avoid the sort and get all entries.
Perhaps unfortunately for your purposes, either the "/bin/ls -f" command or the "/bin/ls -Ua" command will also count the "." and ".." entries that are in each directory. You will have to subtract 2 from the count to avoid counting these two entries, such as in the following:
expr `/bin/ls -f | wc -l` - 2 # Those are back ticks, not single quotes.
The --format=single-column (-1) option is not necessary on the "/bin/ls -Ua" command when piping the "ls" output, as in to "wc" in this case. The "ls" command will automatically write its output in a single column if the output is not a terminal.
The -U option for ls is not in POSIX, and in OS X's ls it has a different meaning from GNU ls, which is that it makes -t and -l use creation times instead of modification times. -f is in POSIX as an XSI extension. The manual of GNU ls describes -f as do not sort, enable -aU, disable -ls --color and -U as do not sort; list entries in directory order.
POSIX describes -f like this:
Force each argument to be interpreted as a directory and list the name found in each slot. This option shall turn off -l, -t, -s, and -r, and shall turn on -a; the order is the order in which entries appear in the directory.
Commands like ls|wc -l give the wrong result when filenames contain newlines.
In zsh you can do something like this:
a=(*(DN));echo ${#a}
D (glob_dots) includes files whose name starts with a period and N (null_glob) causes the command to not result in an error in an empty directory.
Or the same in bash:
shopt -s dotglob nullglob;a=(*);echo ${#a[#]}
If IFS contains ASCII digits, add double quotes around ${#a[#]}. Add shopt -u failglob to ensure that failglob is unset.
A portable option is to use find:
find . ! -name . -prune|grep -c /
grep -c / can be replaced with wc -l if filenames do not contain newlines. ! -name . -prune is a portable alternative to -mindepth 1 -maxdepth 1.
Or here's another alternative that does not usually include files whose name starts with a period:
set -- *;[ -e "$1" ]&&echo "$#"
The command above does however include files whose name starts with a period when an option like dotglob in bash or glob_dots in zsh is set. When * matches no file, the command results in an error in zsh with the default settings.
I used this command..works like a charm..only to change the maxdepth..that is sub directories
find * -maxdepth 0 -type d -exec sh -c "echo -n {} ' ' ; ls -lR {} | wc -l" \;
I think you can have more control on this using find:
find <path> -maxdepth 1 -type f -printf "." | wc -c
find -maxdepth 1 will not go deeper into the hierarchy of files.
-type f allows filtering to just files. Similarly, you can use -type d for directories.
-printf "." prints a dot for every match.
wc -c counts the characters, so it counts the dots created by the print... which means counting how many files exist in the given path.
For the number of all file in a current directory try this:
ls -lR * | wc -l
As far as I know, there is no better alternative. This information might be off-topic to this question and you may already know this that under Linux (in general under Unix) directories are just special file which contains the list of other files (I understand that the exact details will be dependent on specific file system but this is the general idea). And there is no call to find the total number of entries without traversing the whole list. Please make me correct if I'm wrong.
use ls -1 | wc -l