I want to write a bash script which will use a list of all the directories containing specific files. I can use find to echo the path of each and every matching file. I only want to list the path to the directory containing at least one matching file.
For example, given the following directory structure:
dir1/
matches1
matches2
dir2/
no-match
The command (looking for 'matches*') will only output the path to dir1.
As extra background, I'm using this to find each directory which contains a Java .class file.
find . -name '*.class' -printf '%h\n' | sort -u
From man find:
-printf format
%h Leading directories of file’s name (all but the last element). If the file name contains no slashes (since it is in the current directory) the %h specifier expands to ".".
On OS X and FreeBSD, with a find that lacks the -printf option, this will work:
find . -name *.class -print0 | xargs -0 -n1 dirname | sort --unique
The -n1 in xargs sets to 1 the maximum number of arguments taken from standard input for each invocation of dirname
GNU find
find /root_path -type f -iname "*.class" -printf "%h\n" | sort -u
Ok, i come way too late, but you also could do it without find, to answer specifically to "matching file with Bash" (or at least a POSIX shell).
ls */*.class | while read; do
echo ${REPLY%/*}
done | sort -u
The ${VARNAME%/*} will strip everything after the last / (if you wanted to strip everything after the first, it would have been ${VARNAME%%/*}).
Regards.
find / -name *.class -printf '%h\n' | sort --unique
Far too late, but this might be helpful to future readers:
I personally find it more helpful to have the list of folders printed into a file, rather than to Terminal (on a Mac).
For that, you can simply output the paths to a file, e.g. folders.txt, by using:
find . -name *.sql -print0 | xargs -0 -n1 dirname | sort --unique > folders.txt
How about this?
find dirs/ -name '*.class' -exec dirname '{}' \; | awk '!seen[$0]++'
For the awk command, see #43 on this list
Related
So I have a directory with files and sub-directories in it. I want to get all the files recursively and then list them in long format, sorted by the modified date. Here's what I came up with.
find . -type f | xargs -d "\n" | ls -lt
However this only lists the files in the current directory and not the sub-directories. I don't understand why, given that the following prints out all the files.
find . -type f | xargs -d "\n" | cat
Any help appreciated.
xargs can only start ls if it's passed ls as an argument. When you pipe from xargs into ls, only one copy of ls is started -- by the parent shell -- and it isn't given any of the filenames from find | xargs as arguments -- instead they're on its stdin, but ls never reads its stdin, so it doesn't even know that they're there.
Thus, you need to remove the | character:
# Does what you specified in the common case, but buggy; don't use this
# (filenames can contain newlines!)
# ...also, xargs -d is GNU-only
find . -type f | xargs -d '\n' ls -lt
...or, better:
# uses NUL separators, which cannot exist inside filenames
# also, while a non-POSIX extension, this is supported in both GNU and BSD xargs
find . -type f -print0 | xargs -0 ls -lt
...or, even better than that:
# no need for xargs at all here; find -exec can do the same thing
# -exec ... {} + is POSIX-mandated functionality since 2008
find . -type f -exec ls -lt {} +
Much of the content in this answer is also covered in the Actions, Complex Actions, and Actions in Bulk sections of Using Find, which is well worth reading.
I have a structure like that:
/usr/local/a/1.txt
/usr/local/a/2.txt
/usr/local/a/3.txt
/usr/local/b/4.txt
/usr/local/b/3.txt
/usr/local/c/1.txt
/usr/local/c/7.txt
/usr/local/c/6.txt
/usr/local/c/12.txt
...
I want to delete all the files *.txt in subfolders except the last three files with the greatest modification date, but here I am in current directory
ls -tr *.txt | head -n-3 |xargs rm -f
I need to combine that with the code:
find /usr/local/**/* -type f
Should I use the maxdepth option?
Thanks for helping,
aola
Added maxdepth options to find for one level, sorting files by last modification time, tail to ignore the oldest modified 3 files and xargs with -r to remove the files only if they are found.
for folder in $(find /usr/local/ -type d)
do
find $folder -maxdepth 1 -type f -name "*.txt" | xargs -r ls -1tr | tail -n+3 | xargs -r rm -f
done
Run the above command once without rm to ensure that the previous commands pick the proper files for deletion.
You've almost got the solution: use find to get the files,ls to sort them by modification date and tail to omit three most recently modified ones:
find /usr/lib -type f | xargs ls -t | tail -n +4 | xargs rm
If you would like to remove only the files at a specified depth add -mindepth 4 -maxdepth 4 to find parameters.
You can use find's -printf option, to print the modification time in front of the file name and then sort and strip the date off. This avoids using ls at all.
find /usr/local -type f -name '*.txt' -printf '%T#|%p\n' | sort -r | cut -d '|' -f 2 | head -n-3 | xargs rm -f
The other Answers using xargs ls -t can lead to incorrect results, when there are more results than xargs can put in a single ls -t command.
but for each subfolder, so when I have
/usr/local/a/1.txt
/usr/local/a/2.txt
/usr/local/a/3.txt
/usr/local/a/4.txt
/usr/local/b/4.txt
/usr/local/b/3.txt
/usr/local/c/1.txt
/usr/local/c/7.txt
/usr/local/c/6.txt
/usr/local/c/12.txt
I want to to use the code for each subfolder separately
head -n-3 |xargs rm -f
so I bet if I have it sorted by date then the files to delete:
/usr/local/a/4.txt
/usr/local/c/12.txt
I want to leave in any subfolder three newest files
Hi I want to delete a file from any of the subdirectories except one of the subdirectory.
For ex
folder a->a.txt
folder b->subdir 1 -> msgdir-> a.txt
folder c->
Now i want to delete a.txt only in folder a but not the file in msgdir .msgdir can be in any level of subdirectories as it would be changing.
Please help me to resolve this.
This will ignore specifically the msgdir at any level and remove a.txt except in msgdir.
find . ! -path '*/msgdir/*' -name a.txt -type f -delete
Tested with GNU find 4.4.2 and BSD find (Mac Yosemite).
The following approach is overkill if you have GNU find (or a newer BSD one), with the -path option. Otherwise, read on...
You haven't specified which shell you're using but if you have bash, you could go with something like this:
find -name a.txt -exec bash -c "[[ '{}' != */msgdir/* ]]" \; -print
This filters out paths containing /msgdir/, as the test will only pass if the file path doesn't contain the string. If you're happy with the results, you can change -print to -delete.
Without bash, you could use grep to determine the match:
find -name a.txt -exec sh -c "printf '%s' '{}' | grep -qv '/msgdir/'" \; -print
I wrote a script to find all folders that contain executable files. I was first seeking a oneliner command but could find one. (I especially tried to use sort -k -u).
. The script works fine but my initial question remains: Is there a oneliner command to do that?
#! /bin/bash
find $1 -type d | while read Path
do
X=$(ls -l "$Path" | grep '^-rwx' | wc -l)
if ((X>0))
then
echo $Path
fi
done
Using find:
find $1 -type f -perm /111 -exec dirname {} \; | sort -u
This finds all files with permission 111 (i.e. rwx) but then we output only the directory name. To avoid duplicates, sort -u is used.
As pointed out by Paulo Almeida in the comments, this would also work:
find $1 -type f -perm /111 -printf "%h\n" | sort -u
I am trying to remove files within a directory. Some of the files have double-quotes around their name while others do not. An example of these files would be:
"DDD344".csv
D2DW.csv
Both these files are located in sub-directories within the directory YM.
To find such files and remove them, I invoke find like so:
find YM -name "*.csv" -print | xargs rm
The above command results in a lot of No such file or directory errors.
I tried using sed in the following way:
find yum/yum_hyd -name "\"*\".csv" | sed 's/"/\"/g' | xargs rm
but to no avail. How do I remove the files?
The problem is that you're using xargs. xargs is a horribly broken program that should never be used for anything except in conjunction with the nonstandard -0 option. Even so, I can't think of any advantages to doing that in this case. You should just execute rm directly from find.
find . -type f -name '"*".csv' -exec rm -f -- {} +
Will work. If you have GNU find, you may also use -delete.
try this:
find yum/yum_hyd -name "\"*\".csv" |sed 's/"/\\"/g'|xargs rm
explanation:
you want to replace " with \". but if you write \" directly, sed considers it as plain ", you have to escape the backslash. so \\" works.
I wasn't aware of this option until recently but you can list the inode of the file in the following way:
$ ls –il
In the output you will see that the first column contains the inode value. You can then use that value to find -inum the offending files and remove them.
Output
2616366 -rw-r--r-- 1 etc etc
$ find . -inum 2616366 -exec rm -f {} \;
This will remove the file with that specific inum.
As a test you can run the following to locate your files.
ls -il \"* | awk '{print $1}' | xargs -n1 -I {} find -inum {}
Replace the final portion of this command (the "find -inum {}") with the "rm" command once you are satisfied.
This is also similar to the question on SuperUser