Bash script of unzipping unknown name files - bash

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

Related

How do i assign filename to a variable after unzipping unknown zip file in bash shell?

Here I have a zip file but I have no idea what the name it is. I want to assign filename to a variable after unzipping the zip file in bash shell but I got some message like
Archive: name.zip inflating: ...inflating: ..
Here is my script and anyone knows whats the wrong with it?
filename=$(unzip -o *.zip)
echo $filename
If I understand you correctly: you have one ZIP file but you don't know its name. That ZIP file contains one directory at top level whose name you want to put into the variable $filename.
Try:
filename=$(unzip -Z -1 *.zip | head -n1 | sed 's/\/$//')
echo $filename
But if you are sure that the name of the top directory inside your ZIP archive is the same as the name of the archive itself without the suffix .zip, which is the default behaviour of zip, you can just try:
filename=$(basename -s .zip *.zip)
echo $filename
You just need to loop over the .zip files:
for f in *.zip; do echo "unzipping $f"; unzip -o "$f"; done;
(unzip will not return the name of the zip archive. Doing this will give you the name in $f.)

Unix - batch file to unzip folder with specific name

I need a batch script for unix but I don't know it very well.
I have folder A and his subfolder
A\a1\b\c\file.zip
A\a2\b\c\otherFile.zip
A\a3\b\c\thirdFile.zip
Each zip file contains a xml file and a text file
The script have to do 2 things:
unzip all the zip files that are in all folder named 'c' of all sub
folder of 'A' ; the unzipped files should stay in the same folder in
which was the zip
all the unzipped files that have xml extension have to been renamed
someone can help me?
Thank you very much
You can do like this.
#find the folder 'c' and unzip all zip files
for folder in `find ./A -name c -type d`; do unzip $folder/data.zip -d $foler; done
#find all .xml files and change the extension to .edefg
for file in `find ./A -name *.xml -type f`; do mv "$file" "${file%.xml}.edefg"; done
You must go to all dir's.
Would be something like
find A -type d -name c | while read dir; do
cd ${dir} || continue
unzip -u *.zip
call_rename_xml_function
cd - # not needed in bash where this code is performed in a subshell
done
EDIT: Added -u flag for when the script is called twice.

Remove files from one folder that contained in another folder

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

Unzip ZIP file and extract unknown folder name's content

My users will be zipping up files which will look like this:
TEMPLATE1.ZIP
|--------- UnknownName
|------- index.html
|------- images
|------- image1.jpg
I want to extract this zip file as follows:
/mysite/user_uploaded_templates/myrandomname/index.html
/mysite/user_uploaded_templates/myrandomname/images/image1.jpg
My trouble is with UnknownName - I do not know what it is beforehand and extracting everything to the "base" level breaks all the relative paths in index.html
How can I extract from this ZIP file the contents of UnknownName?
Is there anything better than:
1. Extract everything
2. Detect which "new subdidrectory" got created
3. mv newsubdir/* .
4. rmdir newsubdir/
If there is more than one subdirectory at UnknownName level, I can reject that user's zip file.
I think your approach is a good one. Step 2 could be improved my extracting to a newly created directory (later deleted) so that "detection" is trivial.
# Bash (minimally tested)
tempdest=$(mktemp -d)
unzip -d "$tempdest" TEMPLATE1.ZIP
dir=("$tempdest"/*)
if (( ${#dir[#]} == 1 )) && [[ -d $dir ]]
# in Bash, etc., scalar $var is the same as ${var[0]}
mv "$dir"/* /mysite/user_uploaded_templates/myrandomname
else
echo "rejected"
fi
rm -rf "$tempdest"
The other option I can see other than the one you suggested is to use the unzip -j flag which will dump all paths and put all files into the current directory. If you know for certain that each of your TEMPLATE1.ZIP files includes an index.html and *.jpg files then you can just do something like:
destdir=/mysite/user_uploaded_templates/myrandomname
unzip -j -d "$destdir"
mkdir "${destdir}/images"
mv "${destdir}/*.jpg" "${destdir}/images"
It's not exactly the cleanest solution but at least you don't have to do any parsing like you do in your example. I can't seem to find any option similar to patch -p# that lets you specify the path level.
Each zip and unzip command differs, but there's usually a way to list the file contents. From there, you can parse the output to determine the unknown directory name.
On Windows, the 1996 Wales/Gaily/van der Linden/Rommel version it is unzip -l.
Of course, you could just simply allow the unzip to unzip the files to whatever directory it wants, then use mv to rename the directory to what you want it as.
$tempDir = temp.$$
mv $zipFile temp.$$
cd $tempDir
unzip $zipFile
$unknownDir = * #Should be the only directory here
mv $unknownDir $whereItShouldBe
cd ..
rm -rf $tempDir
It's always a good idea to create a temporary directory for these types of operations in case you end up running two instances of this command.

bash find subfolder and backup

How could I read the content of a parent folder and if sub-folders are found, then make a tar.gz files of those subfolders found. However, I know that subfolders will have the following filename format: name-1.2.3 What I want is to create a tar.gz file that looks like: name-1.2.3-20100928.tar.gz Any help will be appreciated.
#!/bin/sh
DATE=`/bin/date +%y%m%e`
cd /path/to/your/folder
for folder in *; do
if [ -d $folder ]; then
tar -cvzf $folder-$DATE.tar.gz $folder
fi
done
Better immunize your scripts against spaces in names:
#!/bin/bash
# Yes nested quoting is allowed with $(...) syntax, no matter whether syntax coloration does
DATE="$(/bin/date "+%y%m%e")"
cd /path/to/your/folder
for folder in * ; do
if test -d "$folder" ; then
tar cvzf "$folder-$DATE.tar.gz" "$folder"
fi
done

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