so I have the code right now down below, and I'm running into a few problems with it
I'm having trouble excluding the directories being outputted by
find ${1-.}
It is giving me the directories too instead of only names; I've tried different methods such as -prune etc.
I'm having trouble with deleting the empty files
The data given to me by
EMPTY_FILE=$(find ${1-.} -size 0)
Does not give me the correct path
Here is the output for that
TestFolder/TestFile
in this case I can't just do:
rm TestFolder/TestFile
As it is invalid path; since it needs ./TestFolder/TestFile
How would I add on the ./ or is there away to get the full path.
#!/bin/bash
echo "Here are all the files in the directory specified\n"
find ${1-.}
EMPTY_FILE=$(find ${1-.} -size 0)
echo "Here are the list of empty files\n"
echo "$EMPTY_FILE \n"
echo "Do you want to delete those empty files?(yes/no)"
read text
if [ "$text" == "yes" ]; then $(rm -- $EMPTY_FILE); fi
Any help is appreciated!
You want this:
#!/bin/bash
echo -e "Here are all the files in the directory specified\n"
# Use -printf "%f\n" to print the filename without leading directories
# Use -type f to restrict find to files
find "${1-.}" -type f -printf " %f\n"
echo -e "Here are the list of empty files\n"
# Again, use -printf "%f\n"
find "${1-.}" -type f -size 0 -printf " %f\n"
echo -e "Do you want to delete those empty files?(yes/no)"
read answer
# Delete files using the `-delete` option
[ "$answer" = "yes" ] && find "${1-.}" -type f -size 0 -delete
Also note that I've quotes "${1-.}" at all occurrences. Since it is user input, you can't rely on the input. Even if it is a path, it might still contain problematic characters, like spaces.
I'm having trouble excluding the directories being outputted by
find ${1-.}
It is giving me the directories too instead of only names
You are looking for the -type test. To instruct find to report only regular files, you could say
find ${1-.} -type f
That's probably what you really want, but what you actually asked (to exclude only directories) would be
find ${1-.} -not -type d
Excluding only directories will list symbolic links and special files, too.
in this case I can't just do:
rm TestFolder/TestFile
As it is invalid path; since it needs ./TestFolder/TestFile
Nonsense. ./TestFolder/TestFile means exactly the same thing as TestFolder/TestFile.
In any event, find does print paths starting at the specified starting path(s).
I have a feeling that I'm missing something from your question, but if all you need to do is exclude directories, just tell find to only look for files:
find . -type f -size 0 -delete
And then adjust that to suit your script. Hope this helps.
-size 0 -type f
rm with no option will not delete directories . Your claim that rm needs ./ is wrong anyway.
Related
I have a directory containing a large number of directories. Each directory contains some files and in some cases another directory.
parent_directory
sub_dir_1
sub_dir_1.txt
sub_dir_1_1.txt
sub_dir_2
sub_dir_2.txt
sub_dir_2_1.txt
sub_dir_3
sub_dir_3.txt
sub_dir_3_1.txt
sub_dir_4
sub_dir_4.txt
sub_dir_4_1.txt
sub_dir_5
sub_dir_5.txt
sub_dir_5_1.txt
I need to check that each sub_dir contains at least one file with the exact same name. I don' need to check any further down if there are sub directories within the sub_dirs.
I was thinking of using for d in ./*/ ; do (command here); done but I dont know how to get access to the sub_dir name inside the for loop
for d in ./*/ ;
do
(if directory does not contain 1 file that is the same name as the directory then echo directory name );
done
What is the best way to do this or is there a simpler way?
from the parent directory
find -maxdepth 1 -type d -printf "%f\n" |
xargs -I {} find {} -maxdepth 1 -type f -name {}.txt
will give you the name/name.txt pair. Compare with the all dir names to find the missing ones.
UPDATE
this might be simpler, instead of scanning you can check whether file exists or not
for f in $(find -maxdepth 1 -type d -printf "%f\n");
do if [ ! -e "$f/$f.txt" ];
then echo "$f not found";
fi; done
Maybe not understand fully, but
find . -print | grep -P '/(.*?)/\1\.txt'
this will print any file which is inside of the same-named directory, e.g:
./a/b/b.txt
./a/c/d/d.txt
etc...
Similarly
find . -print | sed -n '/\(.*\)\/\1\.txt/p'
this
find . -print | grep -P '/(.*?)/\1\.'
will list all files regardless of the extension in same-named dirs.
You can craft other regexes following the backreference logic.
I have a folder contains ~30000 files and some of them are empty. I want to find them and put 'NON' as a header of the empty files.
my script is:
find -type f -empty -exec echo 'NON' {} \;
my output is:
NON ./file1
NON ./file2
NON ./file3
NON ./file4
but I want the 'NON' to write as a header of the file1 , file2 , file3 and file4.
Thanks in advance.
Why not just echo NON first and then do your find, with no need for exec?
Also I am not sure how you are running find without specifying a directory to search, so I have added the current directory to search below.
echo 'NON'
find . -type f -empty
The other possible interpretation of your question is that you want to add the string NON to all zero-length files. You can do that like this.
find . -type f -empty -exec sh -c "echo NON > {}" \;
Here is the code that i have soo far :
echo $(pwd > adress)
var=$(head -1 adress)
rm adress
found=0 #Flag
fileshow()
{
cd $1
for i in *
do
if [ -d $i ]
then
continue
elif [ -w $i ]
then
echo $i
found=1
fi
done
cd ..
}
fileshow $1
if [ $found -eq 0 ]
then
clear
echo "$(tput setaf 1)There arent any executable files !!!$(tput sgr0)"
fi
Its working but it find files only in current directory.
I was told that i need to use some kind of recursive method to loop through all sub-directories but i dont know how to do it.
So if any one can help me i will be very grateful.
Thanks!
The effect of your script is to find the files below the current working directory that are not directories and are writeable to the current user. This can be achieved with the command:
find ./ -type f -writable
The advantage of using -type f is that it also excludes symbolic links and other special kinds of file, if that's what you want. If you want all files that are not directories (as suggested by your script), then you can use:
find ./ ! -type d -writable
If you want to sort these files (added question, assuming lexicographic ascending order), you can use sort:
find ./ -type f -writable | sort
If you want to use these sorted filenames for something else, the canonical pattern would be (to handle filenames with embedded newlines and other seldom-used characters):
while read -r -d $'\0'; do
echo "File '$REPLY' is an ordinary file and is writable"
done < <(find ./ -type f -writable -print0 | sort -z)
If you're using a very old version of find that does not support the handy -writable predicate (added to v.4.3 in 2005), then you only have file permissions to go on. You then have to be clear about what you mean by “writable” in the specific context (writable to whom?), and you can replace the -writable predicate with the -perm predicates described in #gregb's answer. If you decide that you mean “writable by anyone” you could use -perm /u=w,g=w,o=w or -perm /222, but there's actually no way of getting all the benefits of -writable just using permissions. Also note that the + form of permission tests to -perm is deprecated and should no longer be used; the / form should be used instead.
You could use find:
find /path/to/directory/ -type f -perm -o=w
Where the -o=w implies that each file has the "other write-permission" set.
or,
find /path/to/directory/ -type f -perm /u+w,g+w,o+w
Where /u+w,g+w,o+w implies that each file either has user, group, or other write-permissions set.
How can I find all zero-byte files in a directory and its subdirectories?
I have done this:
#!/bin/bash
lns=`vdir -R *.* $dir| awk '{print $8"\t"$5}'`
temp=""
for file in $lns; do
if test $file = "0"; then
printf $temp"\t"$file"\n"
fi
temp=$file
done
But, I only get results in the current directory, not subdirs,
and if any file name contains a space then I get only first word followed by tab
To print the names of all files in and below $dir of size 0:
find "$dir" -size 0
Note that not all implementations of find will produce output by default, so you may need to do:
find "$dir" -size 0 -print
Two comments on the final loop in the question:
Rather than iterating over every other word in a string and seeing if the alternate values are zero, you can partially eliminate the issue you're having with whitespace by iterating over lines. eg:
printf '1 f1\n0 f 2\n10 f3\n' | while read size path; do
test "$size" -eq 0 && echo "$path"; done
Note that this will fail in your case if any of the paths output by ls contain newlines, and this reinforces 2 points: don't parse ls, and have a sane naming policy that doesn't allow whitespace in paths.
Secondly, to output the data from the loop, there is no need to store the output in a variable just to echo it. If you simply let the loop write its output to stdout, you accomplish the same thing but avoid storing it.
As addition to the answers above:
If you would like to delete those files
find $dir -size 0 -type f -delete
No, you don't have to bother grep.
find $dir -size 0 ! -name "*.xml"
Bash 4+ tested -
This is the correct way to search for size 0:
find /path/to/dir -size 0 -type f -name "*.xml"
Search for multiple file extensions of size 0:
find /path/to/dir -size 0 -type f \( -iname \*.css -o -iname \*.js \)
Note: If you removed the \( ... \) the results would be all of the files that meet this requirement hence ignoring the size 0.
I have a list of directories
/home
/dir1
/dir2
...
/dir100
Some of them have no files in it. How can I use Unix find to do it?
I tried
find . -name "*" -type d -size 0
Doesn't seem to work.
Does your find have predicate -empty?
You should be able to use find . -type d -empty
If you're a zsh user, you can always do this. If you're not, maybe this will convince you:
echo **/*(/^F)
**/* will expand to every child node of the present working directory and the () is a glob qualifier. / restricts matches to directories, and F restricts matches to non-empty ones. Negating it with ^ gives us all empty directories. See the zshexpn man page for more details.
-empty reports empty leaf dirs.
If you want to find empty trees then have a look at:
http://code.google.com/p/fslint/source/browse/trunk/fslint/finded
Note that script can't be used without the other support scripts,
but you might want to install fslint and use it directly?
You can also use:
find . -type d -links 2
. and .. both count as a link, as do files.
The answer of Pimin Konstantin Kefalou prints folders with only 2 links and other files (d, f, ...).
The easiest way I have found is:
for directory in $(find . -type d); do
if [ -n "$(find $directory -maxdepth 1 -type f)" ]; then echo "$directory"
fi
done
If you have name with spaces use quotes in "$directory".
You can replace . by your reference folder.
I haven't been able to do it with one find instruction.