I was trying to rename some files to another extension:
# mv *.sqlite3_done *.sqlite3
but got an error:
mv: target '*.sqlite3' is not a directory
Why?
the easy way is use find
find . -type f -name '*.sqlite3_done' -exec sh -c 'x="{}"; mv "$x" "${x%_done}"' \;
mv can only move multiple files into a single directory; it can’t move each one to a different name. You can loop in bash instead:
for x in *.sqlite3_done; do
mv -- "$x" "${x%_done}"
done
${x%_done} removes _done from the end of $x.
The wildcard expansion results in multiple names being passed to the command. The shell thinks you are trying to move multiple files to the *.sqlite3 directory.
You need to use a loop:
for nam in *sqlite3_done
do
newname=${nam%_done}
mv $nam $newname
done
The %_done says to remove the last occurrence of _done from the string.
If you may have spaces in your filenames you will want to quote the filenames.
Related
I would like to batch copy specific files that ends with fastq.gz from each folder (with unique names) to a new directory, but it keeps giving me an error saying that the files cannot be found. Is it because I am using a wildcard wrong?
for f in ./*/split-adapter-quality-trimmed/*.fastq.gz; do
cp *fastq.gz ../../new;
done
Executing for f in ./*/split-adapter-quality-trimmed/*.fastq.gz will already contain the filenames ending with *.fastq.gz in variable f. So use it directly in cp (cp $f destination) inside the loop. If you put an echo $f inside the loop, you can see all the files and verify it before cp.
for f in ./*/split-adapter-quality-trimmed/*.fastq.gz; do
cp $f ../../new;
done
Except if you absolutely want to use a for-loop, you could perform that with one find command:
find ./*/split-adapter-quality-trimmed -name "*fastq.gz" -exec cp {} ../../new \;
It will browse the directories matching ./*/split-adapter-quality-trimmed, looking for each file terminating with fastq.gz, and then execute the needed cp command (in the current directory of the shell, the command line ends with a semi-colon):
cp <found-path> ../../new
(The wildcarded term *fastq.gz is surrounded by quotes to prevent Bash to interpret it, just in case. So is it with the semi-colon.)
I'm in a directory with 3 subdirectories: sub1, sub2, and sub3. Each subdirectory has files in it. I would like to rename each file by prepending sample_ to it.
Here's what I have:
for d in */; do
for f in "$d"; do
mv "$f" "sample_$f"
done
done
This prepends to the folder name, which isn't what I want. What am I doing incorrectly?
Thanks!
You can easily accomplish this with find and brace expansion (part of shell expansion):
find . -type f -execdir mv {,sample_}{} \;
This should recursively find only files (-type f) within each subdirectory then move them (renaming them) using the -execdir option (see below), prepending sample_ to each filename. The remaining mv {,_sample}{} is the Cartesian product way of doing mv {} sample_{}.
-execdir command {} + Like -exec, but the specified command is run from the subdirectory
containing the matched file, which is not normally the directory in
which you started find. This a much more secure method for invoking
commands, as it avoids race conditions during resolution of the paths
to the matched files. As with the -exec option, the '+' form of
-execdir will build a command line to process more than one matched file, but any given invocation of command will only list files that
exist in the same subdirectory. If you use this option, you must
ensure that your $PATH environment variable does not reference the
current directory; otherwise, an attacker can run any commands they
like by leaving an appropriately-named file in a directory in which
you will run -execdir.
↳ GNU : Brace / Shell Expansions
you need to use dirname and basename to split your file name.
for d in */; do
for f in $d/*; do
mv "$f" "$d/sample_$(basename $f)"
done
done
From the current directory I have multiple sub directories:
subdir1/
001myfile001A.txt
002myfile002A.txt
subdir2/
001myfile001B.txt
002myfile002B.txt
where I want to strip every character from the filenames before myfile so I end up with
subdir1/
myfile001A.txt
myfile002A.txt
subdir2/
myfile001B.txt
myfile002B.txt
I have some code to do this...
#!/bin/bash
for d in `find . -type d -maxdepth 1`; do
cd "$d"
for f in `find . "*.txt"`; do
mv "$f" "$(echo "$f" | sed -r 's/^.*myfile/myfile/')"
done
done
however the newly renamed files end up in the parent directory
i.e.
myfile001A.txt
myfile002A.txt
myfile001B.txt
myfile002B.txt
subdir1/
subdir2/
In which the sub-directories are now empty.
How do I alter my script to rename the files and keep them in their respective sub-directories? As you can see the first loop changes directory to the sub directory so not sure why the files end up getting sent up a directory...
Your script has multiple problems. In the first place, your outer find command doesn't do quite what you expect: it outputs not only each of the subdirectories, but also the search root, ., which is itself a directory. You could have discovered this by running the command manually, among other ways. You don't really need to use find for this, but supposing that you do use it, this would be better:
for d in $(find * -maxdepth 0 -type d); do
Moreover, . is the first result of your original find command, and your problems continue there. Your initial cd is without meaningful effect, because you're just changing to the same directory you're already in. The find command in the inner loop is rooted there, and descends into both subdirectories. The path information for each file you choose to rename is therefore stripped by sed, which is why the results end up in the initial working directory (./subdir1/001myfile001A.txt --> myfile001A.txt). By the time you process the subdirectories, there are no files left in them to rename.
But that's not all: the find command in your inner loop is incorrect. Because you do not specify an option before it, find interprets "*.txt" as designating a second search root, in addition to .. You presumably wanted to use -name "*.txt" to filter the find results; without it, find outputs the name of every file in the tree. Presumably you're suppressing or ignoring the error messages that result.
But supposing that your subdirectories have no subdirectories of their own, as shown, and that you aren't concerned with dotfiles, even this corrected version ...
for f in `find . -name "*.txt"`;
... is an awfully heavyweight way of saying this ...
for f in *.txt;
... or even this ...
for f in *?myfile*.txt;
... the latter of which will avoid attempts to rename any files whose names do not, in fact, change.
Furthermore, launching a sed process for each file name is pretty wasteful and expensive when you could just use bash's built-in substitution feature:
mv "$f" "${f/#*myfile/myfile}"
And you will find also that your working directory gets messed up. The working directory is a characteristic of the overall shell environment, so it does not automatically reset on each loop iteration. You'll need to handle that manually in some way. pushd / popd would do that, as would running the outer loop's body in a subshell.
Overall, this will do the trick:
#!/bin/bash
for d in $(find * -maxdepth 0 -type d); do
pushd "$d"
for f in *.txt; do
mv "$f" "${f/#*myfile/myfile}"
done
popd
done
You can do it without find and sed:
$ for f in */*.txt; do echo mv "$f" "${f/\/*myfile/\/myfile}"; done
mv subdir1/001myfile001A.txt subdir1/myfile001A.txt
mv subdir1/002myfile002A.txt subdir1/myfile002A.txt
mv subdir2/001myfile001B.txt subdir2/myfile001B.txt
mv subdir2/002myfile002B.txt subdir2/myfile002B.txt
If you remove the echo, it'll actually rename the files.
This uses shell parameter expansion to replace a slash and anything up to myfile with just a slash and myfile.
Notice that this breaks if there is more than one level of subdirectories. In that case, you could use extended pattern matching (enabled with shopt -s extglob) and the globstar shell option (shopt -s globstar):
$ for f in **/*.txt; do echo mv "$f" "${f/\/*([!\/])myfile/\/myfile}"; done
mv subdir1/001myfile001A.txt subdir1/myfile001A.txt
mv subdir1/002myfile002A.txt subdir1/myfile002A.txt
mv subdir1/subdir3/001myfile001A.txt subdir1/subdir3/myfile001A.txt
mv subdir1/subdir3/002myfile002A.txt subdir1/subdir3/myfile002A.txt
mv subdir2/001myfile001B.txt subdir2/myfile001B.txt
mv subdir2/002myfile002B.txt subdir2/myfile002B.txt
This uses the *([!\/]) pattern ("zero or more characters that are not a forward slash"). The slash has to be escaped in the bracket expression because we're still inside of the pattern part of the ${parameter/pattern/string} expansion.
Maybe you want to use the following command instead:
rename 's#(.*/).*(myfile.*)#$1$2#' subdir*/*
You can use rename -n ... to check the outcome without actually renaming anything.
Regarding your actual question:
The find command from the outer loop returns 3 (!) directories:
.
./subdir1
./subdir2
The unwanted . is the reason why all files end up in the parent directory (that is .). You can exclude . by using the option -mindepth 1.
Unfortunately, this was onyl the reason for the files landing in the wrong place, but not the only problem. Since you already accepted one of the answers, there is no need to list them all.
a slight modification should fix your problem:
#!/bin/bash
for f in `find . -maxdepth 2 -name "*.txt"`; do
mv "$f" "$(echo "$f" | sed -r 's,[^/]+(myfile),\1,')"
done
note: this sed uses , instead of / as the delimiter.
however, there are much faster ways.
here is with the rename utility, available or easily installed wherever there is bash and perl:
find . -maxdepth 2 -name "*.txt" | rename 's,[^/]+(myfile),/$1,'
here are tests on 1000 files:
for `find`; do mv 9.176s
rename 0.099s
that's 100x as fast.
John Bollinger's accepted answer is twice as fast as the OPs, but 50x as slow as this rename solution:
for|for|mv "$f" "${f//}" 4.316s
also, it won't work if there is a directory with too many items for a shell glob. likewise any answers that use for f in *.txt or for f in */*.txt or find * or rename ... subdir*/*. answers that begin with find ., on the other hand, will also work on directories with any number of items.
I'm looping through certain files (all files starting with MOVIE) in a folder with this bash script code:
for i in MY-FOLDER/MOVIE*
do
which works fine when there are files in the folder. But when there aren't any, it somehow goes on with one file which it thinks is named MY-FOLDER/MOVIE*.
How can I avoid it to enter the things after
do
if there aren't any files in the folder?
With the nullglob option.
$ shopt -s nullglob
$ for i in zzz* ; do echo "$i" ; done
$
for i in $(find MY-FOLDER/MOVIE -type f); do
echo $i
done
The find utility is one of the Swiss Army knives of linux. It starts at the directory you give it and finds all files in all subdirectories, according to the options you give it.
-type f will find only regular files (not directories).
As I wrote it, the command will find files in subdirectories as well; you can prevent that by adding -maxdepth 1
Edit, 8 years later (thanks for the comment, #tadman!)
You can avoid the loop altogether with
find . -type f -exec echo "{}" \;
This tells find to echo the name of each file by substituting its name for {}. The escaped semicolon is necessary to terminate the command that's passed to -exec.
for file in MY-FOLDER/MOVIE*
do
# Skip if not a file
test -f "$file" || continue
# Now you know it's a file.
...
done
I am writing the following script to copy *.nzb files to a folder to queue them for Download.
I wrote the following script
#!/bin/bash
#This script copies NZB files from Downloads folder to HellaNZB queue folder.
${DOWN}="/home/user/Downloads/"
${QUEUE}="/home/user/.hellanzb/nzb/daemon.queue/"
for a in $(find ${DOWN} -name *.nzb)
do
cp ${a} ${QUEUE}
rm *.nzb
done
it gives me the following error saying:
HellaNZB.sh: line 5: =/home/user/Downloads/: No such file or directory
HellaNZB.sh: line 6: =/home/user/.hellanzb/nzb/daemon.queue/: No such file or directory
Thing is that those directories exsist, I do have right to access them.
Any help would be nice.
Please and thank you.
Variable names on the left side of an assignment should be bare.
foo="something"
echo "$foo"
Here are some more improvements to your script:
#!/bin/bash
#This script copies NZB files from Downloads folder to HellaNZB queue folder.
down="/home/myusuf3/Downloads/"
queue="/home/myusuf3/.hellanzb/nzb/daemon.queue/"
find "${down}" -name "*.nzb" | while read -r file
do
mv "${file}" "${queue}"
done
Using while instead of for and quoting variables that contain filenames protects against filenames that contain spaces from being interpreted as more than one filename. Removing the rm keeps it from repeatedly producing errors and failing to copy any but the first file. The file glob for -name needs to be quoted. Habitually using lowercase variable names reduces the chances of name collisions with shell variables.
If all your files are in one directory (and not in multiple subdirectories) your whole script could be reduced to the following, by the way:
mv /home/myusuf3/Downloads/*.nzb /home/myusuf3/.hellanzb/nzb/daemon.queue/
If you do have files in multiple subdirectories:
find /home/myusuf3/Downloads/ -name "*.nzb" -exec mv {} /home/myusuf3/.hellanzb/nzb/daemon.queue/ +
As you can see, there's no need for a loop.
The correct syntax is:
DOWN="/home/myusuf3/Downloads/"
QUEUE="/home/myusuf3/.hellanzb/nzb/daemon.queue/"
for a in $(find ${DOWN} -name *.nzb)
# escape the * or it will be expanded in the current directory
# let's just hope no file has blanks in its name
do
cp ${a} ${QUEUE} # ok, although I'd normally add a -p
rm *.nzb # again, this is expanded in the current directory
# when you fix that, it will remove ${a}s before they are copied
done
Why don't you just use rm $(a}?
Why use a combination of cp and rm anyway, instead of mv?
Do you realize all files will end up in the same directory, and files with the same name from different directories will overwrite each other?
What if the cp fails? You'll lose your file.