Task: concatinate array of string with delimiter, dilimeter is "/".
Metatask: i've a folder with many files. Need to copy them into another folder.
So i need to get "name of file" and "path to folder".
What's wrong: delimiter "/" works incorrectly. It doesn't concatinate with my strings. If i try to use "\/" - string disappeare at all.
What's going on?
loc_path='./test/*'
delim='\/'
for itt in $loc_path; do
IFS=$delim
read -ra res <<< "$itt"
str=''
for ((i = 1; i \<= ${#res[#]}; i++)); do
#str=($str${res[$i]}$delim)
str="$str${res[$i]}$delim"
done
echo $str
done
Please, give to two answers:
how to solve task-problem
better way to solve metatask
There is an issue in delim='\/'. Firstly, you need not to protect slash. Secondly all characters are already protected between simple quotes.
There is a syntax issue with your concatenation. You must not use parenthesis here! They can be used to open a sub shell. We need not that.
To solve your 'meta-task', you should avoid to use IFS, or read. They are complex to use (for example by modifying IFS globally as you do, you change how echo display the res array. It can mislead you while you troubleshoot...) I suggest you use more simple tool like: basename, etc.
Here few scripts to solve your meta (scholar?) task:
# one line :-)
cp src/* dst/
# to illustrate basename etc
for file in "$SRC/"*; do
dest="$DST/$(basename $file)"
cp "$file" "$dest"
done
# with a change of directory
cd "$SRC"
for file in *; do cp "$file" "$DST/$file"; done
cd -
# Change of directory and a sub shell
(cd "$SRC" ; for file in *; do cp "$file" "$DST/$file"; done)
Task solution:
arr=( string1 string2 string3 ) # array of strings
str=$( IFS='/'; printf '%s' "${arr[*]}" ) # concatenated with / as delimiter
$str will be the single string string1/string2/string3.
Meta task solution:
Few files:
cp path/to/source/folder/* path/to/dest/folder
Note that * matches any type of file and that it does not match hidden names. For hidden names, use shopt -s dotglob in bash. This will fail if there are thousands of files (argument list too long).
Few or many files files, only non-directories:
for pathaname in path/to/source/folder/*; do
[ ! -type d "$pathame" ] && cp "$pathname" path/to/dest/folder
done
or, with find,
find path/to/source/folder -maxdepth 1 ! -type d -exec cp {} path/to/dest/folder \;
The difference between these two is that the shell loop will refuse to copy symbolic links that resolve to directories, while the find command will copy them.
Related
If any of this isn't particularly clear, please let me know and I'll do my best to clarify.
I basically need to sort a set of files with various extensions and similar patterns to the filename, into directories and subdirectories that match the pattern and type of extension.
To elaborate a bit:
All files, regardless of extension, begin with the pattern "zz####" where #### is a number from 1 to 900; "zz1.zip through zz950.zip, zz1.mov through zz950.mov, zz1.mp4 through zz950.mp4"
Some files contain additional characters; "zz360_hello_world.zip"
Some files contain spaces; "zz370_hello world.zip"
I need these files to be sorted and moved into directories and subdirectories following a particular format: "/home/hello/zz1/zip, /home/hello/zz1/vid"
If the directories and/or subdirectories don't exist, I need them created.
Example:
zz400_testing.zip ----> /home/hello/zz400/zip
zz400 testing video.mov ----> /home/hello/zz400/vid
zz500.zip ----> /home/hello/zz500/zip
zz500_testing another video.mp4 ----> /home/hello/zz500/vid
I found a few answers around here for simpler use-cases, but wasn't able to get anything working for my particular needs.
Any help at all would be much appreciated.
Thank you!
EDIT: Adding the code I've been messing with
for f in *.zip; do
set=`echo "$f"|sed 's/[0-9].*//'`
dir="/home/demo/$set/photos"
mkdir -p "$dir"
mv "$f" "$dir"
done
I think I'm just having trouble wrapping my head around how to match with regex. I've got this far with it:
[demo#alpha grep]$ echo zz433.zip|sed 's/[0-9].*//'
zz
The script will run the mkdir, and even move the zip files into their proper place. I just can't get it to create the proper top-level directory (zz433).
The sed command here doesn't do what you're trying to achieve:
set=`echo "$f"|sed 's/[0-9].*//'`
The meaning of the regular expression [0-9].* is "a digit followed by anything".
The s/// command of sed performs a replacement.
The result is effectively removing everything from the input starting from the first digit.
So for "zz360_hello_world.zip" it removes everything starting from "3",
leaving only "zz".
Note also that to match the files, the pattern *.zip doesn't match your description. You're looking for files starting with "zz" and a number from 1 up to 900. If you don't mind including numbers > 900 then you can write the loop expression like this:
for f in zz[0-9][^0-9]* zz[0-9][0-9][^0-9]* zz[0-9][0-9][0-9][^0-9]*; do
Or the same thing more compactly:
for f in zz{[0-9],[0-9][0-9],[0-9][0-9][0-9]}[^0-9]*; do
These are glob patterns.
zz[0-9][^0-9]* means "start with 'zz', followed by a digit, followed by a non-digit, followed by anything".
In the above example I use three patterns to cover the cases of "zz" followed by 1, 2 or 3 digits, followed by a non-digit.
The second example is a more compact form of the first,
the idea is that a{b,c}d expands to abd and acd.
Next, to get the appropriate prefix, you could use pattern matching with a case statement and extract substrings.
The syntax of these patterns is the same glob syntax as in the previous example in the for statement.
case "$f" in
zz[0-9][0-9][0-9]*) prefix=${f:0:5} ;;
zz[0-9][0-9]*) prefix=${f:0:4} ;;
zz[0-9]*) prefix=${f:0:3} ;;
esac
It seems you also want to create grouping by file type. You could get the file extension by chopping off the beginning of the name until the dot with ext=${f##*.}, and then use a case statement as in the earlier example to map extensions to the desired directory names.
Putting the above together:
for f in zz{[0-9],[0-9][0-9],[0-9][0-9][0-9]}[^0-9]*; do
case "$f" in
zz[0-9][0-9][0-9]*) prefix=${f:0:5} ;;
zz[0-9][0-9]*) prefix=${f:0:4} ;;
zz[0-9]*) prefix=${f:0:3} ;;
esac
ext=${f##*.}
case "$ext" in
mov|mp4) group=vid ;;
*) group=$ext ;;
esac
dir="/home/demo/$prefix/$group"
mkdir -p "$dir"
mv "$f" "$dir"
done
I've answered part of my own question!
for f in *.zip; do
set=`echo "$f"|grep -o -P 'zz[0-9]+.{0,0}'`
dir="/home/demo/$set/photos"
mkdir -p "$dir"
mv "$f" "$dir"
done
Basically, the following script will grab files like:
zz232.zip
zz233test.zip
zz234 test.zip
Then it will create the top-level directory (zz####), the photos sub-directory, and move the file into place:
/home/demo/zz232/photos/zz232.zip
/home/demo/zz233/photos/zz233test.zip
/home/demo/zz234/photos/zz234 test.zip
Moving on to expanding the script for additional functionality.
Thanks all!
How about:
#!/bin/bash
IFS=$'\n'
for file in *; do
if [[ $file =~ ^(zz[0-9]+).*\.(zip|mov|mp4)$ ]]; then
ext=${BASH_REMATCH[2]}
if [ $ext = "mov" -o $ext = "mp4" ]; then
ext="vid"
fi
dir="/home/hello/${BASH_REMATCH[1]}/$ext"
mkdir -p $dir
mv "$file" $dir
fi
done
Hope this helps.
I am looking for a way to make a simple loop in bash over everything my directory contains, i.e. files, directories and links including hidden ones.
I will prefer if it could be specifically in bash but it has to be the most general. Of course, file names (and directory names) can have white space, break line, symbols. Everything but "/" and ASCII NULL (0×0), even at the first character. Also, the result should exclude the '.' and '..' directories.
Here is a generator of files on which the loop has to deal with :
#!/bin/bash
mkdir -p test
cd test
touch A 1 ! "hello world" \$\"sym.dat .hidden " start with space" $'\n start with a newline'
mkdir -p ". hidden with space" $'My Personal\nDirectory'
So my loop should look like (but has to deal with the tricky stuff above):
for i in * ;
echo ">$i<"
done
My closest try was the use of ls and bash array, but it is not working with, is:
IFS=$(echo -en "\n\b")
l=( $(ls -A .) )
for i in ${l[#]} ; do
echo ">$i<"
done
unset IFS
Or using bash arrays but the ".." directory is not exclude:
IFS=$(echo -en "\n\b")
l=( [[:print:]]* .[[:print:]]* )
for i in ${l[#]} ; do
echo ">$i<"
done
unset IFS
* doesn't match files beginning with ., so you just need to be explicit:
for i in * .[^.]*; do
echo ">$i<"
done
.[^.]* will match all files and directories starting with ., followed by a non-. character, followed by zero or more characters. In other words, it's like the simpler .*, but excludes . and ... If you need to match something like ..foo, then you might add ..?* to the list of patterns.
As chepner noted in the comments below, this solution assumes you're running GNU bash along with GNU find GNU sort...
GNU find can be prevented from recursing into subdirectories with the -maxdepth option. Then use -print0 to end every filename with a 0x00 byte instead of the newline you'd usually get from -print.
The sort -z sorts the filenames between the 0x00 bytes.
Then, you can use sed to get rid of the dot and dot-dot directory entries (although GNU find seems to exclude the .. already).
I also used sed to get read of the ./ in front of every filename. basename could do that too, but older systems didn't have basename, and you might not trust it to handle the funky characters right.
(These sed commands each required two cases: one for a pattern at the start of the string, and one for the pattern between 0x00 bytes. These were so ugly I split them out into separate functions.)
The read command doesn't have a -z or -0 option like some commands, but you can fake it with -d "" and blanking the IFS environment variable.
The additional -r option prevents a backslash-newline combo from being interpreted as a line continuation. (A file called backslash\\nnewline would otherwise be mangled to backslashnewline.) It might be worth seeing if other backslash-combos get interpreted as escape sequences.
remove_dot_and_dotdot_dirs()
{
sed \
-e 's/^[.]\{1,2\}\x00//' \
-e 's/\x00[.]\{1,2\}\x00/\x00/g'
}
remove_leading_dotslash()
{
sed \
-e 's/^[.]\///' \
-e 's/\x00[.]\//\x00/g'
}
IFS=""
find . -maxdepth 1 -print0 |
sort -z |
remove_dot_and_dotdot_dirs |
remove_leading_dotslash |
while read -r -d "" filename
do
echo "Doing something with file '${filename}'..."
done
It may not be the most favorable way but I tried bellow thing
while read line ; do echo $line; done <<< $(ls -a | grep -v -w ".")
check the below trail which I did
Try the find command, something like:
find .
That will list all the files in all recursive directories.
To output only files excluding the leading . or .. try:
find . -type f -printf %P\\n
I want to copy multiple files in one go using cp. The problem is that the filenames contain spaces here and there. So I tried cp $(ls -1|tr ' ' '') dest but apparently I cannot truncate to nothing. Then I tried removing null space to spaces. That didnt work either. I tried running it over a for loop. That too didnt work. Can some one please help me????
You might want to try using find in combination with xargs and cp. See How can I use xargs to copy files that have spaces and quotes in their names? for more info.
Ahn, ok.
I was writing this:
for i in *
do
if echo $i | grep " "
then
NEWNAME=`echo $i | tr ' ' '_'`
mv "$i" $NEWNAME
fi
done
But in this case you rename the file (and of course you have to adapt the code to your needs)
I have this function in my .bashrc
The problem is that you can't tr the source file names for cp - you have to give it the names of the files as they exist on disk or it won't find them. So it sounds like what you effectively want is to cp "file with spaces" destdir/filewithspaces for each individual file:
ls -1 | while read filename; do
cp "$filename" "$dest/${filename// /}"
done
The ${filename// /} is an instance of the bash-ism ${variable//search/replacement} (see shell parameter expansion in the Bash manual) to give the value of a variable with all instances of the search string replaced with the replacement string - so in this case replace all spaces with nothing.
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
I have a directory that contains numerically named subdirectories ( eg. 1, 2, 3, 32000, 43546 ). I need to delete all directories over a certain number. For example, I need to delete all subdirectories that have a name that is numerically larger than 14234. Can this be done with a single command line action?
rm -r /directory/subdirectories_over_14234 ( how can I do this? )
In bash, I'd write
for dir in *; do [[ -d $dir ]] && (( dir > 14234 )) && echo rm -r $dir; done
Remove the echo at your discretion.
Well you can do a bash for loop instruction so as to iterate over the directory filename and use the test command then after extracting the target number of the file name.
Should be something like this :
for $file in /your/path
do
#extract number here with any text processing command (ed ?)
if test [$name -leq your_value]
then
rm -R $file
fi
done
You don't mention which shell you're using. I'm using Zsh and it has a very cool feature: it can select files based on numbers just like you want! So you can do
$ rm -r /directory/<14234->(/)
to select all the subdirectories of /directory with a numeric value over 14234.
In general, you use
<a-b>
to select paths with a numeric values between a and b. You append a (/) to only match directories. Use (.) to only match files. The glob patterns in Zsh are very powerful and can mostly (if not always) replace the good old find command.