I would like to create a .tar file from all .gz files in $a/$b and delete those files afterwards.
I have come up with the following code but it's not working:
cd "$a"
#tar cf $a/$b'.tar' "$b_sql.gz" "$b_moh.tar.gz"
tar cf $a/$b'.tar' $b'_*.gz'
gzip $a/$b'.tar'
rm -f $b'_*.gz'
I try this way and answer :
cd "$a"
for f in $b'_*.gz'
do
tar cf $a/$b'.tar' $f
done
The problem is with your use of * within a string. It's treating * as a character rather than doing the glob in the shell.
You can test this yourself by running ls *.gz and then ls '*.gz' within your shell.
Related
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$1}'
will show you all the first-level contents. If there is only that one directory, then
tar --strip-components=1 -tf yourball.tar
will extract the contents of that directory in tar into the current directory.
So that's how you can avoid the problem altogether. But it's also a solution to your immediate problem. Having extracted the files already, so you have
foo/bar/stuff
foo/bar/.otherstuff
you can do
tar -cf- foo | tar --strip-components=2 -C final_destination -xf-
The --strip-components feature is not part of the POSIX specification for tar, but it is on both the common GNU and OSX/BSD implementations.
How can I untar all tar files in one command using Putty.
I Tried the following but its not un-tarring (all files start with alcatelS*)
tar -xfv alcatelS*.tar
It is not working i don't get no errors and it is not un-tarring.
Thank you,
-xfv is wrong since v is being referred as the file instead. Also, tar can't accept multiple files to extract at once. Perhaps -M can be used but it's a little stubborn when I tried it. Also, it would be difficult to pass multiple arguments that were extracted from pathname expansion i.e. you have to do tar -xvM -f file1.tar -f file2.tar.
Do this instead:
for F in alcatelS*.tar; do
tar -xvf "$F"
done
Or one-line: (EDIT: Sorry that -is- a "one"-liner but I find that not technically a real one-liner, just a condensed one so I should haven't referred to that as a one-liner. Avoid the wrong convention.)
for F in alcatelS*.tar; do tar -xvf "$F"; done
You can use following command for extract all tar.gz files in directory in unix
find . -name 'alcatelS*.tar.gz' -exec tar -xvf {} \;
Following is my favorite way to untar multiple tar files:
ls *tar.gz | xargs -n1 tar xvf
Can be done in one line:
cat *.tar | tar -xvf - -i
I have a folder that after an rsync will have a zip in it. I want to unzip it to its own folder(if the zip is L155.zip, to unzip its content to L155 folder). The problem is that I dont know it's name beforehand(although i know it will be "letter-number-number-number"), so I have to unzip an uknown file to its unknown folder and this to be done automatically.
The command “unzip *”(or unzip *.zip) works in terminal, but not in a script.
These are the commands that have worked through terminal one by one, but dont work in a script.
#!/bin/bash
unzip * #also tried .zip and /path/to/file/* when script is on different folder
i=$(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
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
I have a few subdirectories in a given folder, where a file d2.sh~ exists. I want to delete this file via following shell script, which, rather than writing in a .sh file I wrote on terminal, on one line. [Edit: been formatted properly here for clarity]
for i in `ls *`; do
if [ -d $i ]; then
cd $i
rm d2.sh~
cd ..
fi
done
This did not give me any errors but it failed to delete d2.sh~ from the subdirectories. So I want to know what mistake I have made above?
find /some/path -type f -name "d2.sh~" -delete
Your first mistake is trying to parse ls. See this link as to why.
Just use for i in *; do ....
If you need recursion then you need to look to find or if you have Bash 4.X you can do:
shopt -s globstar; for i in **/d2.sh~; do rm "$i"; done