I'm writing a Linux script to copy files from a folder structure in to one folder. I want to use a varying folder name as the prefix of the file name.
My current script looks like this. But, I can't seem to find a way to use the folder name from the wildcard as the file name;
for f in /usr/share/storage/*/log/myfile.log*; do cp "$f" /myhome/docs/log/myfile.log; done
My existing folder structure/files as follows and I want the files copied as;
>/usr/share/storage/100/log/myfile.log --> /myhome/docs/log/100.log
>/usr/share/storage/100/log/myfile.log.1 --> /myhome/docs/log/100.log.1
>/usr/share/storage/102/log/myfile.log --> /myhome/docs/log/102.log
>/usr/share/storage/103/log/myfile.log --> /myhome/docs/log/103.log
>/usr/share/storage/103/log/myfile.log.1 --> /myhome/docs/log/103.log.1
>/usr/share/storage/103/log/myfile.log.2 --> /myhome/docs/log/103.log.2
You could use a regular expression match to extract the desired component, but it is probably easier to simply change to /usr/share/storage so that the desired component is always the first one on the path.
Once you do that, it's a simple matter of using various parameter expansion operators to extract the parts of paths and file names that you want to use.
cd /usr/share/storage
for f in */log/myfile.log*; do
pfx=${f%%/*} # 100, 102, etc
dest=$(basename "$f")
dest=$pfx.${dest#*.}
cp -- "$f" /myhome/docs/log/"$pfx.${dest#*.}"
done
One option is to wrap the for loop in another loop:
for d in /usr/share/storage/*; do
dir="$(basename "$d")"
for f in "$d"/log/myfile.log*; do
file="$(basename "$f")"
# test we found a file - glob might fail
[ -f "$f" ] && cp "$f" /home/docs/log/"${dir}.${file}"
done
done
for f in /usr/share/storage/*/log/myfile.log*; do cp "$f" "$(echo $f | sed -re 's%^/usr/share/storage/([^/]*)/log/myfile(\.log.*)$%/myhome/docs/log/\1\2%')"; done
Related
I have 20 files like:
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
...
Files have a similar format in their names. They begin with 01, and they have 01*AAA*.sh format.
I wish to copy and rename files in the same directory, changing the number 01 to 02, 03, 04, and 05:
02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh
...
03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh
...
04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh
...
05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh
...
I wish to copy 20 of 01*.sh files to 02*.sh, 03*.sh, and 04*.sh. This will make the total number of files to 100 in the folder.
I'm really not sure how can I achieve this. I was trying to use for loop in the bash script. But not even sure what should I need to select as a for loop index.
for i in {1..4}; do
cp 0${i}*.sh 0${i+1}*.sh
done
does not work.
There are going to be a lot of ways to slice-n-dice this one ...
One idea using a for loop, printf + brace expansion, and xargs:
for f in 01*.sh
do
printf "%s\n" {02..05} | xargs -r -I PFX cp ${f} PFX${f:2}
done
The same thing but saving the printf in a variable up front:
printf -v prefixes "%s\n" {02..05}
for f in 01*.sh
do
<<< "${prefixes}" xargs -r -I PFX cp ${f} PFX${f:2}
done
Another idea using a pair of for loops:
for f in 01*.sh
do
for i in {02..05}
do
cp "${f}" "${i}${f:2}"
done
done
Starting with:
$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
All of the proposed code snippets leave us with:
$ ls -1 0*.sh
01a_AAA_qwe.sh
01b_AAA_asd.sh
01c_AAA_zxc.sh
01d_AAA_rty.sh
02a_AAA_qwe.sh
02b_AAA_asd.sh
02c_AAA_zxc.sh
02d_AAA_rty.sh
03a_AAA_qwe.sh
03b_AAA_asd.sh
03c_AAA_zxc.sh
03d_AAA_rty.sh
04a_AAA_qwe.sh
04b_AAA_asd.sh
04c_AAA_zxc.sh
04d_AAA_rty.sh
05a_AAA_qwe.sh
05b_AAA_asd.sh
05c_AAA_zxc.sh
05d_AAA_rty.sh
NOTE: blank lines added for readability
You can't do multiple copies in a single cp command, except when copying a bunch of files to a single target directory. cp will not do the name mapping automatically. Wildcards are expanded by the shell, they're not seen by the commands themselves, so it's not possible for them to do pattern matching like this.
To add 1 to a variable, use $((i+1)).
You can use the shell substring expansion operator to get the part of the filename after the first two characters.
for i in {1..4}; do
for file in 0${i}*.sh; do
fileend=${file:2}
cp "$file" "0$((i+1))$fileend"
done
done
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 5 years ago.
Write a simple script that will automatically rename a number of files. As an example we want the file *001.jpg renamed to user defined string + 001.jpg (ex: MyVacation20110725_001.jpg) The usage for this script is to get the digital camera photos to have file names that make some sense.
I need to write a shell script for this. Can someone suggest how to begin?
An example to help you get off the ground.
for f in *.jpg; do mv "$f" "$(echo "$f" | sed s/IMG/VACATION/)"; done
In this example, I am assuming that all your image files contain the string IMG and you want to replace IMG with VACATION.
The shell automatically evaluates *.jpg to all the matching files.
The second argument of mv (the new name of the file) is the output of the sed command that replaces IMG with VACATION.
If your filenames include whitespace pay careful attention to the "$f" notation. You need the double-quotes to preserve the whitespace.
You can use rename utility to rename multiple files by a pattern. For example following command will prepend string MyVacation2011_ to all the files with jpg extension.
rename 's/^/MyVacation2011_/g' *.jpg
or
rename <pattern> <replacement> <file-list>
this example, I am assuming that all your image files begin with "IMG" and you want to replace "IMG" with "VACATION"
solution : first identified all jpg files and then replace keyword
find . -name '*jpg' -exec bash -c 'echo mv $0 ${0/IMG/VACATION}' {} \;
for file in *.jpg ; do mv $file ${file//IMG/myVacation} ; done
Again assuming that all your image files have the string "IMG" and you want to replace "IMG" with "myVacation".
With bash you can directly convert the string with parameter expansion.
Example: if the file is IMG_327.jpg, the mv command will be executed as if you do mv IMG_327.jpg myVacation_327.jpg. And this will be done for each file found in the directory matching *.jpg.
IMG_001.jpg -> myVacation_001.jpg
IMG_002.jpg -> myVacation_002.jpg
IMG_1023.jpg -> myVacation_1023.jpg
etcetera...
find . -type f |
sed -n "s/\(.*\)factory\.py$/& \1service\.py/p" |
xargs -p -n 2 mv
eg will rename all files in the cwd with names ending in "factory.py" to be replaced with names ending in "service.py"
explanation:
In the sed cmd, the -n flag will suppress normal behavior of echoing input to output after the s/// command is applied, and the p option on s/// will force writing to output if a substitution is made. Since a sub will only be made on match, sed will only have output for files ending in "factory.py"
In the s/// replacement string, we use "& " to interpolate the entire matching string, followed by a space character, into the replacement. Because of this, it's vital that our RE matches the entire filename. after the space char, we use "\1service.py" to interpolate the string we gulped before "factory.py", followed by "service.py", replacing it. So for more complex transformations youll have to change the args to s/// (with an re still matching the entire filename)
Example output:
foo_factory.py foo_service.py
bar_factory.py bar_service.py
We use xargs with -n 2 to consume the output of sed 2 delimited strings at a time, passing these to mv (i also put the -p option in there so you can feel safe when running this). voila.
NOTE: If you are facing more complicated file and folder scenarios, this post explains find (and some alternatives) in greater detail.
Another option is:
for i in *001.jpg
do
echo "mv $i yourstring${i#*001.jpg}"
done
remove echo after you have it right.
Parameter substitution with # will keep only the last part, so you can change its name.
Can't comment on Susam Pal's answer but if you're dealing with spaces, I'd surround with quotes:
for f in *.jpg; do mv "$f" "`echo $f | sed s/\ /\-/g`"; done;
You can try this:
for file in *.jpg;
do
mv $file $somestring_${file:((-7))}
done
You can see "parameter expansion" in man bash to understand the above better.
I am having a file named temp.txt where inside this file it contains the following content
https://abcdef/12345-xyz
https://ghifdfg/5426525-abc
I need to create a directories automatically in linux by using only th number part from each line in the file.
So the output should be something like 12345 and 5426525 directories created.
Any approach on how to do this could be helpful.
This is the code that i searched and got from internet,wherein this code, new directories will be created by the file name that starts with BR and W0 .
for file in {BR,W0}*.*; do
dir=${file%%.*}
mkdir -p "$dir"
mv "$file" "$dir"
done
Assuming each URL is of the form
http[s]://any/symbols/some_digits-some_letters
Then you indeed could use the simple prefix and suffix modifiers in shell variable expansion.
${x##*/} expands to the suffix part of x that starts after the last slash /.
${y%%-*} expands to the prefix part of y before the first -.
while read x ; do
y=${x##*/}
z=${y%%-*}
mkdir $z
done < temp.txt
I want to move sequence 20 file in different folder
so i use below code to move file but not work...
echo "Moving {$(( $j*20 -19))..$(( $j*20 ))}.png ------- > Folder $i"
mv {$(( $j*20 -19))..$(( $j*20 ))}.png $i;
So i get output in terminal
Moving {1..20}.png ------- > Folder 1
mv: cannot stat ‘{1..20}.png’: No such file or directory
But there is already 1.png to 20.png image file + Folder...
So how to move sequence file like
{1..20}.png -> Folder 1
{21..40}.png -> Folder 2
Thank you!!!
I don't think that it is possible to combine brace expansion with arithmetic expressions as you are doing. Specifically, ranges like {a..b} must contain literal values, not variables.
I would suggest that instead, you used a for loop:
for ((n=j*20-19;n<=j*20;++n)); do mv "$n.png" "$i"; done
The disadvantage of the above approach is that mv is called many times, rather than once. As suggested in the comments (thanks chepner), you could use an array to reduce the number of calls:
files=()
for ((n=j*20-19;n<=j*20;++n)); do files+=( "$n.png" ); done
mv "${files[#]}" "$i"
"${files[#]}" is the full contents of the array, so all of the files are moved in one call to mv.
You have to evaluate the resulting string again with eval
For example
eval "mv {$(( $j*20 -19))..$(( $j*20 ))}.png folder$j"
I have a folder that is full of .bak files and some other files also. I need to remove the extension of all .bak files in that folder. How do I make a command which will accept a folder name and then remove the extension of all .bak files in that folder ?
Thanks.
To remove a string from the end of a BASH variable, use the ${var%ending} syntax. It's one of a number of string manipulations available to you in BASH.
Use it like this:
# Run in the same directory as the files
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
That works nicely as a one-liner, but you could also wrap it as a script to work in an arbitrary directory:
# If we're passed a parameter, cd into that directory. Otherwise, do nothing.
if [ -n "$1" ]; then
cd "$1"
fi
for FILENAME in *.bak; do mv "$FILENAME" "${FILENAME%.bak}"; done
Note that while quoting your variables is almost always a good practice, the for FILENAME in *.bak is still dangerous if any of your filenames might contain spaces. Read David W.'s answer for a more-robust solution, and this document for alternative solutions.
There are several ways to remove file suffixes:
In BASH and Kornshell, you can use the environment variable filtering. Search for ${parameter%word} in the BASH manpage for complete information. Basically, # is a left filter and % is a right filter. You can remember this because # is to the left of %.
If you use a double filter (i.e. ## or %%, you are trying to filter on the biggest match. If you have a single filter (i.e. # or %, you are trying to filter on the smallest match.
What matches is filtered out and you get the rest of the string:
file="this/is/my/file/name.txt"
echo ${file#*/} #Matches is "this/` and will print out "is/my/file/name.txt"
echo ${file##*/} #Matches "this/is/my/file/" and will print out "name.txt"
echo ${file%/*} #Matches "/name.txt" and will print out "/this/is/my/file"
echo ${file%%/*} #Matches "/is/my/file/name.txt" and will print out "this"
Notice this is a glob match and not a regular expression match!. If you want to remove a file suffix:
file_sans_ext=${file%.*}
The .* will match on the period and all characters after it. Since it is a single %, it will match on the smallest glob on the right side of the string. If the filter can't match anything, it the same as your original string.
You can verify a file suffix with something like this:
if [ "${file}" != "${file%.bak}" ]
then
echo "$file is a type '.bak' file"
else
echo "$file is not a type '.bak' file"
fi
Or you could do this:
file_suffix=$(file##*.}
echo "My file is a file '.$file_suffix'"
Note that this will remove the period of the file extension.
Next, we will loop:
find . -name "*.bak" -print0 | while read -d $'\0' file
do
echo "mv '$file' '${file%.bak}'"
done | tee find.out
The find command finds the files you specify. The -print0 separates out the names of the files with a NUL symbol -- which is one of the few characters not allowed in a file name. The -d $\0means that your input separators are NUL symbols. See how nicely thefind -print0andread -d $'\0'` together?
You should almost never use the for file in $(*.bak) method. This will fail if the files have any white space in the name.
Notice that this command doesn't actually move any files. Instead, it produces a find.out file with a list of all the file renames. You should always do something like this when you do commands that operate on massive amounts of files just to be sure everything is fine.
Once you've determined that all the commands in find.out are correct, you can run it like a shell script:
$ bash find.out
rename .bak '' *.bak
(rename is in the util-linux package)
Caveat: there is no error checking:
#!/bin/bash
cd "$1"
for i in *.bak ; do mv -f "$i" "${i%%.bak}" ; done
You can always use the find command to get all the subdirectories
for FILENAME in `find . -name "*.bak"`; do mv --force "$FILENAME" "${FILENAME%.bak}"; done