I'am trying to write simple script that will get files name from one folder and search them in another folder and remove if found them in that folder.
Got two folder like
/home/install/lib
/home/install/bin
/home/install/include
and
/usr/local/lib
/usr/local/bin
/usr/local/include
I want to remove all file's from /usr/local/lib{bin,include} that contains in /home/install/lib{bin,include}. For example having
/home/install/lib/test1
/usr/local/lib/test1
scritp will remove /usr/local/lib/test1. I tried to do it from each separate directory
/home/install/lib:ls -f -exec rm /usr/local/lib/{} \;
but nothing. Can you help me to manage with this simple script?
Create script rmcomm
#!/bin/bash
a="/home/install/$1"
b="/usr/local/$1"
comm -12 <(ls "$a") <(ls "$b") | while read file; do
rm "$b/$file"
done
Then call this script for every pair:
for dir in lib bin include; do rmcomm "$dir"; done
Here's something simple. Remove the echo from the line containing rm to run it after you've ensured it's doing what you want:
#!/bin/bash
dirs[0]=lib
dirs[1]=bin
dirs[2]=include
pushd /home/install
for dir in "${dirs[#]}"
do
for file in $(find $dir -type f)
do
# Remove 'echo' below once you're satisfied the correct files
# are being removed
echo rm /usr/local/$file
done
done
popd
Related
I have a folder structure like this:
A big parent folder named Photos. This folder contains 900+ subfolders named a_000, a_001, a_002 etc.
Each of those subfolders contain more subfolders, named dir_001, dir_002 etc. And each of those subfolders contain lots of pictures (with unique names).
I want to move all these pictures contained in the subdirectories of a_xxx inside a_xxx. (where xxx could be 001, 002 etc)
After looking in similar questions around, this is the closest solution I came up with:
for file in *; do
if [ -d $file ]; then
cd $file; mv * ./; cd ..;
fi
done
Another solution I got is doing a bash script:
#!/bin/bash
dir1="/path/to/photos/"
subs= `ls $dir1`
for i in $subs; do
mv $dir1/$i/*/* $dir1/$i/
done
Still, I'm missing something, can you help?
(Then it would be nice to discard the empty dir_yyy, but not much of a problem at the moment)
You could try the following bash script :
#!/bin/bash
#needed in case we have empty folders
shopt -s nullglob
#we must write the full path here (no ~ character)
target="/path/to/photos"
#we use a glob to list the folders. parsing the output of ls is baaaaaaaddd !!!!
#for every folder in our photo folder ...
for dir in "$target"/*/
do
#we list the subdirectories ...
for sub in "$dir"/*/
do
#and we move the content of the subdirectories to the parent
mv "$sub"/* "$dir"
#if you want to remove subdirectories once the copy is done, uncoment the next line
#rm -r "$sub"
done
done
Here is why you don't parse ls in bash
Make sure the directory where the files exist is correct (and complete) in the following script and try it:
#!/bin/bash
BigParentDir=Photos
for subdir in "$BigParentDir"/*/; do # Select the a_001, a_002 subdirs
for ssdir in "$subdir"/*/; do # Select dir_001, … sub-subdirs
for f in "$ssdir"/*; do # Select the files to move
if [[ -f $f ]]; do # if indeed are files
echo \
mv "$ssdir"/* "$subdir"/ # Move the files.
fi
done
done
done
No file will be moved, just printed. If you are sure the script does what you want, comment the echo line and run it "for real".
You can try this
#!/bin/bash
dir1="/path/to/photos/"
subs= `ls $dir1`
cp /dev/null /tmp/newscript.sh
for i in $subs; do
find $dir1/$i -type f -exec echo mv \'\{\}\' $dir1/$i \; >> /tmp/newscript.sh
done
then open /tmp/newscript.sh with an editor or less and see if looks like what you are trying to do.
if it does then execute it with sh -x /tmp/newscript.sh
I'm trying to create a script that retrieves files (including subfolders) from CVS and stores them into a temporary directory /tmp/projectdir/ (OK), then removes copies of those files from my project directory /home/projectdir/ (not OK) without touching any other files in the project directory or the folder structure itself.
I've been attempting two methods, but I'm running into problems with both. Here's my script so far:
#!/usr/bin/bash
cd /tmp/
echo "removing /tmp/projectdir/"
rm -rf /tmp/projectdir
# CVS login goes here, code redacted
# export files to /tmp/projectdir/dir_1/file_1 etc
cvs export -kv -r $1 projectdir
# method 1
for file in /tmp/projectdir/*
do
# check for zero-length string
if [-n "$file"]; then
echo "removing $file"
rm /home/projectdir/"$file"
fi
done
# method 2
find /tmp/projectdir/ -exec rm -i /home/projectdir/{} \;
Neither method works as intended, because I need some way of stripping /tmp/projectdir/ from the filename (to be replaced with /home/projectdir/) and to prevent them from executing rm /home/projectdir/dir_1 (i.e. the directory and not a specific file), but I'm not sure how to achieve this.
(In case anybody is wondering, the zero-length string bit was an attempt to avoid rm'ing the directory, before I realised /tmp/projectdir/ would also be a part of the string)
You can use:
cd /tmp/projectdir/
find . -type f -exec rm -i /home/projectdir/{} \;
Often after unzipping a file I end up with a directory containing nothing but another directory (e.g., mkdir foo; cd foo; tar xzf ~/bar.tgz may produce nothing but a bar directory in foo). I wanted to write a script to collapse that down to a single directory, but if there are dot files in the nested directory it complicates things a bit.
Here's a naive implementation:
mv -i $1/* $1/.* .
rmdir $1
The only problem here is that it'll also try to move . and .. and ask overwrite ./.? (y/n [n]). I can get around this by checking each file in turn:
IFS=$'\n'
for file in $1/* $1/.*; do
if [ "$file" != "$1/." ] && [ "$file" != "$1/.." ]; then
mv -i $file .
fi
done
rmdir $1
But this seems like an inelegant workaround. I tried a cleaner method using find:
for file in $(find $1); do
mv -i $file .
done
rmdir $1
But find $1 will also give $1 as a result, which gives an error of mv: bar and ./bar are identical.
While the second method seems to work, is there a better way to achieve this?
Turn on the dotglob shell option, which allows the your pattern to match files beginning with ..
shopt -s dotglob
mv -i "$1"/* .
rmdir "$1"
First, consider that many tar implementations provide a --strip-components option that allows you to strip off that first path. Not sure if there is a first path?
tar -tf yourball.tar | awk -F/ '!s[$1]++{print$1}'
will show you all the first-level contents. If there is only that one directory, then
tar --strip-components=1 -tf yourball.tar
will extract the contents of that directory in tar into the current directory.
So that's how you can avoid the problem altogether. But it's also a solution to your immediate problem. Having extracted the files already, so you have
foo/bar/stuff
foo/bar/.otherstuff
you can do
tar -cf- foo | tar --strip-components=2 -C final_destination -xf-
The --strip-components feature is not part of the POSIX specification for tar, but it is on both the common GNU and OSX/BSD implementations.
I have a folder that after an rsync will have a zip in it. I want to unzip it to its own folder(if the zip is L155.zip, to unzip its content to L155 folder). The problem is that I dont know it's name beforehand(although i know it will be "letter-number-number-number"), so I have to unzip an uknown file to its unknown folder and this to be done automatically.
The command “unzip *”(or unzip *.zip) works in terminal, but not in a script.
These are the commands that have worked through terminal one by one, but dont work in a script.
#!/bin/bash
unzip * #also tried .zip and /path/to/file/* when script is on different folder
i=$(ls | head -1)
y=${i:0:4}
mkdir $y
unzip * -d $y
First I unzip the file, then I read the name of the first extracted file through ls and save it in a variable.I take the first 4 chars and make a directory with it and then again unzip the files to that specific folder.
The whole procedure after first unzip is done, is because the files inside .zip, all start with a name that the zip already has, so if L155.ZIP is the zip, the files inside with be L155***.txt.
The zip file is at /path/to/file/NAME.zip.
When I run the script I get errors like the following:
unzip: cannot find or open /path/to/file/*.ZIP
unzip: cannot find or open /path/to/file//*.ZIP.zip
unzip: cannot find or open /path/to/file//*.ZIP.ZIP. No zipfiles found.
mkdir: cannot create directory 'data': File exists data
unzip: cannot find or open data, data.zip or data.ZIP.
Original answer
Supposing that foo.zip contains a folder foo, you could simply run
#!/bin/bash
unzip \*.zip \*
And then run it as bash auto-unzip.sh.
If you want to have these files extracted into a different folder, then I would modify the above as
#!/bin/bash
cp *.zip /home/user
cd /home/user
unzip \*.zip \*
rm *.zip
This, of course, you would run from the folder where all the zip files are stored.
Another answer
Another "simple" fix is to get dtrx (also available in the Ubuntu repos, possibly for other distros). This will extract each of your *.zip files into its own folder. So if you want the data in a different folder, I'd follow the second example and change it thusly:
#!/bin/bash
cp *.zip /home/user
cd /home/user
dtrx *.zip
rm *.zip
I would try the following.
for i in *.[Zz][Ii][Pp]; do
DIRECTORY=$(basename "$i" .zip)
DIRECTORY=$(basename "$DIRECTORY" .ZIP)
unzip "$i" -d "$DIRECTORY"
done
As noted, the basename program removes the indicated suffix .zip from the filename provided.
I have edited it to be case-insensitive. Both .zip and .ZIP will be recognized.
for zfile in $(find . -maxdepth 1 -type f -name "*.zip")
do
fn=$(echo ${zfile:2:4}) # this will give you the filename without .zip extension
mkdir -p "$fn"
unzip "$zfile" -d "$fn"
done
If the folder has only file file with the extension .zip, you can extract the name without an extension with the basename tool:
BASE=$(basename *.zip .zip)
This will produce an error message if there is more than one file matching *.zip.
Just to be clear about the issue here, the assumption is that the zip file does not contain a folder structure. If it did, there would be no problem; you could simply extract it into the subfolders with unzip. The following is only needed if your zipfile contains loose files, and you want to extract them into a subfolder.
With that caveat, the following should work:
#!/bin/bash
DIR=${1:-.}
BASE=$(basename "$DIR/"*.zip .zip 2>/dev/null) ||
{ echo More than one zipfile >> /dev/stderr; exit 1; }
if [[ $BASE = "*" ]]; then
echo No zipfile found >> /dev/stderr
exit 1
fi
mkdir -p "$DIR/$BASE" ||
{ echo Could not create $DIR/$BASE >> /dev/stderr; exit 1; }
unzip "$DIR/$BASE.zip" -d "$DIR/$BASE"
Put it in a file (anywhere), call it something like unzipper.sh, and chmod a+x it. Then you can call it like this:
/path/to/unzipper.sh /path/to/data_directory
simple one liner I use all the time
$ for file in `ls *.zip`; do unzip $file -d `echo $file | cut -d . -f 1`; done
My directory structure is as follows
Directory1\file1.jpg
\file2.jpg
\file3.jpg
Directory2\anotherfile1.jpg
\anotherfile2.jpg
\anotherfile3.jpg
Directory3\yetanotherfile1.jpg
\yetanotherfile2.jpg
\yetanotherfile3.jpg
I'm trying to use the command line in a bash shell on ubuntu to take the first file from each directory and rename it to the directory name and move it up one level so it sits alongside the directory.
In the above example:
file1.jpg would be renamed to Directory1.jpg and placed alongside the folder Directory1
anotherfile1.jpg would be renamed to Directory2.jpg and placed alongside the folder Directory2
yetanotherfile1.jpg would be renamed to Directory3.jpg and placed alongside the folder Directory3
I've tried using:
find . -name "*.jpg"
but it does not list the files in sequential order (I need the first file).
This line:
find . -name "*.jpg" -type f -exec ls "{}" +;
lists the files in the correct order but how do I pick just the first file in each directory and move it up one level?
Any help would be appreciated!
Edit: When I refer to the first file what I mean is each jpg is numbered from 0 to however many files in that folder - for example: file1, file2...... file34, file35 etc... Another thing to mention is the format of the files is random, so the numbering might start at 0 or 1a or 1b etc...
You can go inside each dir and run:
$ mv `ls | head -n 1` ..
If first means whatever the shell glob finds first (lexical, but probably affected by LC_COLLATE), then this should work:
for dir in */; do
for file in "$dir"*.jpg; do
echo mv "$file" "${file%/*}.jpg" # If it does what you want, remove the echo
break 1
done
done
Proof of concept:
$ mkdir dir{1,2,3} && touch dir{1,2,3}/file{1,2,3}.jpg
$ for dir in */; do for file in "$dir"*.jpg; do echo mv "$file" "${file%/*}.jpg"; break 1; done; done
mv dir1/file1.jpg dir1.jpg
mv dir2/file1.jpg dir2.jpg
mv dir3/file1.jpg dir3.jpg
Look for all first level directories, identify first file in this directory and then move it one level up
find . -type d \! -name . -prune | while read d; do
f=$(ls $d | head -1)
mv $d/$f .
done
Building on the top answer, here is a general use bash function that simply returns the first path that resolves to a file within the given directory:
getFirstFile() {
for dir in "$1"; do
for file in "$dir"*; do
if [ -f "$file" ]; then
echo "$file"
break 1
fi
done
done
}
Usage:
# don't forget the trailing slash
getFirstFile ~/documents/
NOTE: it will silently return nothing if you pass it an invalid path.