I would like to search a directory and all its subdirectories for files that are structured like this: ABC.001.XYZ, ABC.001.DEF, ABC.002.XYZ and so fourth.
It should search for all files beginning with ABC.001, count the results, and if it is less than x, delete all files beginning with that. Then move on to ABC.002 and so on.
dir = X
counter=1
while [ $counter -le 500 ]
do
if [find ${dir} -type f -name 'ABC*' | wc -l -eq 5]
then
for file in $(find ${dir} -type f -name 'ABC*')
do
/bin/rm -i ${file}
fi
((counter++))
done
My question is
I. how do I plug in the variable counter for -name 'ABC*' so it increments up. (Like a string placeholder)
II. How would I make it so if the counter is less than 10 or 100, I place 00 or 0 before the counter, so it would actually search for ABC001*, instead of ABC1*
You can use printf to print formatted numbers as in most languages:
printf "ABC%03d" "$counter"
Simple substitution can put this into the arguments to find. Also worth mentioning that find can delete files directly, and just personal preference, but a for loop is probably neater.
#!/bin/bash
dir=X
for counter in $(seq 1 500); do
if [[ $(find "$dir" -type f -name "$(printf "ABC%03d" "$counter")" | wc -l) -eq 5 ]]; then
find "$dir" -type f -name "$(printf "ABC%03d" "$counter")" -delete
fi
done
Related
i have a n files in a different folders like abc.mp3 acc.mp3 bbb.mp3 and i want to rename them 01-abc.mp3, 02-acc.mp3, 03-bbb.mp3... i tried this
#!/bin/bash
IFS='
'
COUNT=1
for file in ./uff/*;
do mv "$file" "${COUNT}-$file" let COUNT++ done
but i keep getting errors like for syntax error near 'do and sometimes for not found... Can someone provide single line solution to this using "find" from terminal. i'm looking for a solution using find only due to certain constraints... Thanks in advance
I'd probably use:
#!/bin/bash
cd ./uff || exit 1
COUNT=1
for file in *.mp3;
do
mv "$file" $(printf "%.2d-%s" ${COUNT} "$file")
((COUNT++))
done
This avoids a number of issues and also includes a 2-digit number for the first 9 files (the next 90 get 2-digit numbers anyway, and after that you get 3-digit numbers, etc).
you can try this;
#!/bin/bash
COUNT=1
for file in ./uff/*;
do
path=$(dirname $file)
filename=$(basename $file)
if [ $COUNT -lt 10 ]; then
mv "$file" "$path"/0"${COUNT}-$filename";
else
mv "$file" "$path"/"${COUNT}-$filename";
fi
COUNT=$(($COUNT+1));
done
Eg:
user#host:/tmp/test$ ls uff/
abc.mp3 acc.mp3 bbb.mp3
user#host:/tmp/test$ ./test.sh
user#host:/tmp/test$ ls uff/
01-abc.mp3 02-acc.mp3 03-bbb.mp3
Ok, here's the version without loops:
paste -d'\n' <(printf "%s\n" *) <(printf "%s\n" * | nl -w1 -s-) | xargs -d'\n' -n2 mv -v
You can also use find if you want:
paste -d'\n' <(find -mindepth 1 -maxdepth 1 -printf "%f\n") <(find -mindepth 1 -maxdepth 1 -printf "%f\n" | nl -w1 -s-) | xargs -d'\n' -n2 mv -v
Replace mv with echo mv for the "dry run":
paste -d'\n' <(printf "%s\n" *) <(printf "%s\n" * | nl -w1 -s-) | xargs -d'\n' -n2 echo mv -v
Here's a solution.
i=1
for f in $(find ./uff -mindepth 1 -maxdepth 1 -type f | sort)
do
n=$i
[ $i -lt 10 ] && n="0$i"
echo "$f" "$n-$(basename "$f")"
((i++))
done
And here it is as a one-liner (but in real life if you ever tried anything remotely like what's below in a coding or ops interview you'd not only fail to get the job, you'd probably give the interviewer PTSD. They'd wake up in cold sweats thinking about how terrible your solution was).
i=1; for f in $(find ./uff -mindepth 1 -maxdepth 1 -type f | sort); do n=$i; [ $i -lt 10 ] && n="0$i"; echo "$f" "$n-$(basename "$f")" ; ((i++)); done
Alternatively, you could just cd ./uff if you wanted the rename them in the same directory, and then use find . (along with the other find arguments) to clear everything up. I'm assuming you only want files moved, not directories. And I'm assuming you don't want to recursively rename files / directories.
I have a Base Directory that has 4 directories : Dir1 Dir2 Dir3 Dir4. Each of these directories have files in the format: "Sometext_YYYMMMDD". I'm writing a UNIX script to search through the files in all these directories that have a particular string say "20151215", and then printing it on the console.
find . -name "*20151215" -print
Example of files: File1_20151215 (this will be printed);
File2_20151214 (this will not be printed)
I want to write a script that runs through these directories and checks if Dir1 contains 4 files with string "20151215", Dir2 contains 3 files with string "20151215" and Dir3 & Dir4 contains 4 files with string "20151215". If the directories don't contain that number of files with that string, then I want to print those directories.
How do I do that? Please help!
UPDATE: I have an addition to this: There are also some files that are not in the format "Sometext_YYYMMMDD" So, for those I used something like:
find . -name "FILENAME*" -mtime -1 -exec ls -ltr '{}' \;
to extract the timestamp when that file was created. But, I want to know how do I add it to the script so that if the timestamp is 15 Dec 2015, then this file should also be counted in the search?
You got the find part, but now you need to count how many files match the pattern. Since find prints one line per match to the output, you can use "wc -l" to count how many lines there are. Assign that to a variable, that you can use in a comparison, and you're 90% of the way there. E.g.
d1=$(find ./dir1 -name '*20151215*' | wc -l)
if [ $d1 != 4 ]; then echo "dir1" ; fi
For extra credit, you can imagine turning this into a function with inputs of
Directory to search
Filename pattern to match on
How many matches to expect
Which would look like:
check_dir () {
d1=$(find $1 -name "*$2*" | wc -l)
if [ $d1 != $3 ]; then echo $1 ; fi
}
check_dir ./dir1 20151215 4
check_dir ./dir2 20151215 3
Update: with the new requirement to find files based either on the name of the file or the last modification (creation isn't possible), here's two approaches:
The first uses a fairly modern feature of find that isn't available in all versions, newermt:
check_dir () {
d1=$2
d2=$((d1+1))
n=$(find $1 \( -name "*$d1*" \) -o \( -newermt $d1 ! -newermt $d2 \) | wc -l)
if [ $n != $3 ]; then echo $1 ; fi
}
check_dir ./dir1 20151215 4
check_dir ./dir2 20151215 3
Which looks a little confusing, but break it down into small steps and it makes sense:
d1=$2 # So d1=20151215
d2=$((d1+1)) # d2=20151216 (lucky you're specifying the date format this way!)
The find command now has two predicates, to match based on the filename or the modification time:
\( -name "*$2*" \) # Matches filenames that contain 20151215
-o # Or
\( -newermt $d1 ! -newermt $d2 \)
The modification time is greater than midnight on the first day, and not greater than midnight on the next day
The second approach uses a couple of temp files, and sets the timestamps on them using the -d option of the touch command
#!/bin/bash
check_dir () {
d1=$2
d2=$((d1+1))
f1=`mktemp`
f2=`mktemp`
touch -d $d1 $f1
touch -d $d2 $f2
n=$(find $1 \( -name "*$d1*" \) -o \( -newer $f1 ! -newer $f2 \) | wc -l)
if [ $n != $3 ]; then echo $1 "=" $n ; fi
rm -f $f1 $f2
}
Again, it's lucky that the date is in YYYYMMDD since that works with the -d option of the touch command. If not, you would need to do some string manipulation to get the date into the correct format for "touch -t".
Hi i'm new in Unix and bash and I'd like to ask q. how can i do this
The specified directory is given as arguments. Locate the directory
where the sum of the number of lines of regular file is greatest.
Browse all specific directories and their subdirectories. Amounts
count only for files that are directly in the directory.
I try somethnig but it's not working properly.
while [ $# -ne 0 ];
do case "$1" in
-h) show_help ;;
-*) echo "Error: Wrong arguments" 1>&2 exit 1 ;;
*) directories=("$#") break ;;
esac
shift
done
IFS='
'
amount=0
for direct in "${directories[#]}"; do
for subdirect in `find $direct -type d `; do
temp=`find "$subdirect" -type f -exec cat {} \; | wc -l | tr -s " "`
if [ $amount -lt $temp ]; then
amount=$temp
subdirect2=$subdirect
fi
done
echo Output: "'"$subdirect2$amount"'"
done
the problem is here when i use as arguments this dirc.(just example)
/home/usr/first and there are this direct.
/home/usr/first/tmp/first.txt (50 lines)
/home/usr/first/tmp/second.txt (30 lines)
/home/usr/first/tmp1/one.txt (20 lines)
it will give me on Output /home/usr/first/tmp1 100 and this is wrong it should be /home/usr/first/tmp 80
I'd like to scan all directories and all its subdirectories in depth. Also if multiple directories meets the maximum should list all.
Given your sample files, I'm going to assume you only want to look at the immediate subdirectories, not recurse down several levels:
max=-1
# the trailing slash limits the wildcard to directories only
for dir in */; do
count=0
for file in "$dir"/*; do
[[ -f "$file" ]] && (( count += $(wc -l < "$file") ))
done
if (( count > max )); then
max=$count
maxdir="$dir"
fi
done
echo "files in $maxdir have $max lines"
files in tmp/ have 80 lines
In the spirit of Unix (caugh), here's an absolutely disgusting chain of pipes that I personally hate, but it's a lot of fun to construct :):
find . -mindepth 1 -maxdepth 1 -type d -exec sh -c 'find "$1" -maxdepth 1 -type f -print0 | wc -l --files0-from=- | tail -1 | { read a _ && echo "$a $1"; }' _ {} \; | sort -nr | head -1
Of course, don't use this unless you're mentally ill, use glenn jackman's nice answer instead.
You can have great control on find's unlimited filtering possibilities, too. Yay. But use glenn's answer!
I have been looking for a way to list file that do not exist from a list of files that are required to exist. The files can exist in more than one location. What I have now:
#!/bin/bash
fileslist="$1"
while read fn
do
if [ ! -f `find . -type f -name $fn ` ];
then
echo $fn
fi
done < $fileslist
If a file does not exist the find command will not print anything and the test does not work. Removing the not and creating an if then else condition does not resolve the problem.
How can i print the filenames that are not found from a list of file names?
New script:
#!/bin/bash
fileslist="$1"
foundfiles="~/tmp/tmp`date +%Y%m%d%H%M%S`.txt"
touch $foundfiles
while read fn
do
`find . -type f -name $fn | sed 's:./.*/::' >> $foundfiles`
done < $fileslist
cat $fileslist $foundfiles | sort | uniq -u
rm $foundfiles
#!/bin/bash
fileslist="$1"
while read fn
do
FPATH=`find . -type f -name $fn`
if [ "$FPATH." = "." ]
then
echo $fn
fi
done < $fileslist
You were close!
Here is test.bash:
#!/bin/bash
fn=test.bash
exists=`find . -type f -name $fn`
if [ -n "$exists" ]
then
echo Found it
fi
It sets $exists = to the result of the find. the if -n checks if the result is not null.
Try replacing body with [[ -z "$(find . -type f -name $fn)" ]] && echo $fn. (note that this code is bound to have problems with filenames containing spaces).
More efficient bashism:
diff <(sort $fileslist|uniq) <(find . -type f -printf %f\\n|sort|uniq)
I think you can handle diff output.
Give this a try:
find -type f -print0 | grep -Fzxvf - requiredfiles.txt
The -print0 and -z protect against filenames which contain newlines. If your utilities don't have these options and your filenames don't contain newlines, you should be OK.
The repeated find to filter one file at a time is very expensive. If your file list is directly compatible with the output from find, run a single find and remove any matches from your list:
find . -type f |
fgrep -vxf - "$1"
If not, maybe you can massage the output from find in the pipeline before the fgrep so that it matches the format in your file; or, conversely, massage the data in your file into find-compatible.
I use this script and it works for me
#!/bin/bash
fileslist="$1"
found="Found:"
notfound="Not found:"
len=`cat $1 | wc -l`
n=0;
while read fn
do
# don't worry about this, i use it to display the file list progress
n=$((n + 1))
echo -en "\rLooking $(echo "scale=0; $n * 100 / $len" | bc)% "
if [ $(find / -name $fn | wc -l) -gt 0 ]
then
found=$(printf "$found\n\t$fn")
else
notfound=$(printf "$notfound\n\t$fn")
fi
done < $fileslist
printf "\n$found\n$notfound\n"
The line counts the number of lines and if its greater than 0 the find was a success. This searches everything on the hdd. You could replace / with . for just the current directory.
$(find / -name $fn | wc -l) -gt 0
Then i simply run it with the files in the files list being separated by newline
./search.sh files.list
Currently I have multiple directories
Directory1 Directory2 Directory3 Directory4
each of these directories contain files (the files are somewhat cryptic)
what i wish to do is scan files within the folders to see if certain files are present, if they are then leave that folder alone, if the certain files are not present then just delete the entire directory. here is what i mean:
im searching for the files that have the word .pass. in the filename.
Say Directory 4 has that file that im looking for
Direcotry4:
file1.temp.pass.exmpl
file1.temp.exmpl
file1.tmp
and the rest of the Directories do not have that specific file:
file.temp
file.exmp
file.tmp.other
so i would like to delete Directory1,2 and3 But only keep Directory 4...
So far i have come up with this code
(arr is a array of all the directory names)
for x in ${arr[#]}
do
find $x -type f ! -name "*pass*" -exec rd {} $x\;
done
another way i have thought of doing this is like this:
for x in ${arr[#]}
do
cd $x find . -type f ! -name "*Pass*" | xargs -i rd {} $x/
done
SO far these don't seem to work, and im scared that i might do something wrong and have all my files deleted.....(i have backed up)
is there any way that i can do this? remember i want Directory 4 to be unchanged, everything in it i want to keep
To see if your directory contains a pass file:
if [ "" = "$(find directory -iname '*pass*' -type f | head -n 1)" ]
then
echo notfound
else
echo found
fi
To do that in a loop:
for x in "${arr[#]}"
do
if [ "" = "$(find "$x" -iname '*pass*' -type f | head -n 1)" ]
then
rm -rf "$x"
fi
done
Try this:
# arr is a array of all the directory names
for x in ${arr[#]}
do
ret=$(find "$x" -type f -name "*pass*" -exec echo "0" \;)
# expect zero length $ret value to remove directory
if [ -z "$ret" ]; then
# remove dir
rm -rf "$x"
fi
done