I want to get the content of the file1.txt of my archive.zip without extrancting the file.
For all these commands I obtain caution: filename not matched: file1.txt
unzip -p ~/archive.zip ~/archive.zip/file1.txt | less
unzip -p ~/archive.zip ~/archive/file1.txt | less
unzip -p ~/archive.zip file1.txt | less
the archive.zip is at the home directory, and the respective names are correct.
Hardcoding the path, produces the same undesired outcome:
unzip -p /home/pi/archive.zip ~/archive.zip/file1.txt | less
unzip -p /home/pi/archive.zip ~/archive/file1.txt | less
unzip -p /home/pi/archive.zip file1.txt | less
I am trying to do this in a raspberry-pi.
The expected output is the content of the file1.txt.
It's possible that the zip extracts into a directory and the file is present there, eg file.zip creates a someproject-name top level directory and the contents are under that. So you can do something like this:
unzip -p /home/pi/archive.zip '*/file1.txt' So it would look at the top level directory aswell due to the glob.
Related
cat paste_output.txt | while read -r file_name path_name file;
do mkdir -p -- "$path_name";
wget "$file_name";
mv "$file" "$path_name";
done;
Hi! I have this piece of code that reads field by field from the file specified. What I am trying to do here is I am creating a directory that is specified in second field and then I am downloading file specified in first field and then after having that file downloaded I am that file in the directory specified in second field.
Output: I am getting the desired directory structure and files downloaded however files are downloading in the directory I am executing the commands from.
How to move files in the desired directories?
You can use the -P flag of wget to put the file in the target directory.
If the directory doesn't exist, it will create it,
so this also let's you save on the mkdir.
while read -r file_name path_name file; do
wget -P "$path_name" "$file_name"
done < paste_output.txt
I made some other improvements to the script:
The cat is useless, input redirection is better
The semicolons at end of lines are unnecessary
It's good to indent the body of loops, for readability
I found the slimier post from STO but those does not filter files with extension. So writing again.
I an writing a shell script to keep last (most latest) 3 .txt files in directory and wants to remove all other .txt files.
For Example... In Directory "Home" I have following files.
test.txt
my.txt
image.jpg
test.avi
sample.txt
country.txt
study.txt
When I run linux script, output should be like as below....
Keep File (keep only last 3 .txt files only)
test.txt
my.txt
image.jpg
test.avi
sample.txt
Delete File
country.txt
study.txt
Thanks
List entries by ctime (newest first), skip the first three items, delete the rest:
ls -c *.txt | tail -n +4 | xargs rm
I have some zip files that are really large and I want to print them without extracting first. I am using zcat and zless to do that and then I redirect the output to a different application. When my zip file contains more than one text file I receive the following error:
zcat tweets.zip >a
gzip: tweets.zip has more than one entry--rest ignored
How can I do what I want with zip files that contain more than one text file?
You can do this to output a file without extracting:
$ unzip -p <zip_file> <file_to_print>
For example:
$ unzip -p MyEar.ear META-INF/MANIFEST.MF
As cur4so mentioned you can also list all files using:
$ unzip -l <zip_file>
Use the -p option of unzip to pipe the output. Multiple files are concatenated. The -c option does the same thing, but includes the file name in front of each file.
If you just want to see a list of files in your zip archive use:
unzip -l tweets.zip
if you want to extract just some file:
unzip tweets.zip file-of-interest-as-it-is-pointed-in-the-archive
if you want something else, could you clarify your question?
I'm trying to backup just one file that is generated by other application in dynamic named folders.
for example:
parent_folder/
back_01 -> file_blabla.zip (timestam 2013.05.12)
back_02 -> file_blabla01.zip (timestam 2013.05.14)
back_03 -> file_blabla02.zip (timestam 2013.05.22)
and I need to get the latest generated zip, just that one it doesnt matter the name of the file as long as is the latest, is a zip and is inside "parent_folder" get that one.
as well when I do the rsync the folder structure + file name is generated and I want to omit that I want to backup that file in a folder and with a name so I know where is the latest and it will be always named the same.
now im doing this with a perl that get the latest generated folder with
"ls -tAF | grep '/$' | head -1"
and perform the rsync but it does brings the last zip but with the folder structure that I dont want because it doesnt override my latest zip file.
rsync -rvtW --prune-empty-dirs --delay-updates --no-implied-dirs --modify-window=1 --include='*.zip' --exclude='*.*' --progress /source/ /myBackup/
as well it would be great if I could do the rsync without needing to use perl or any other script.
thanks
The file names will differ each time ?
This would be hard for any type of syncing to work.
What you could do is :
create a new folder outside of where it is found, then :
Before you start remove the last sym linked file in that folder
When the file is found i.e. ls -tAF | grep '/$' | head -1 ....
symlink it this folder
then rsync,ssh,unison file across to new node.
If the symlink name is file-latest.zip then it will always be this
one file sent across.
But why do all that when you can just scp and you can take a look at here:
https://github.com/vahidhedayati/definedscp
for a more long winded approach, and not for this situation but it uses the real file date/time stamp then converts to seconds... It might be useful if you wish to do the stat in a different way
Using stat to work out file, work out latest file then simply scp it across, here is something to get you started:
One liner:
scp $(find /path/to/parent_folder -name \*.zip -exec stat -t {} \;|awk '{print $1" "$13}'|sort -k2nr|head -n1|awk '{print $1}') remote_server:/path/to/name.zip
More long winded way, maybe of use to understand what above is doing:
#!/bin/bash
FOUND_ARRAY=()
cd parent_folder;
for file in $(find . -name \*.zip); do
ptime=$(stat -t $file|awk '{print $13}');
FOUND_ARRAY+=($file" "$ptime)
done
IFS=$'\n'
FOUND_FILE=$(echo "${FOUND_ARRAY[*]}" | sort -k2nr | head -n1|awk '{print $1}');
scp $FOUND_FILE remote_host:/backup/new_name.zip
I have a bash shell script that unzips a zip file, and manipulates the resulting files. Because of the process, I expect all the content I am interested to be within a single folder like so:
file.zip
/file
/contentFolder1
/contentFolder2
stuff1.txt
stuff2.txt
...
I've noticed users on Windows typically don't create a sub folder but instead submit an exploding zip file that looks like:
file.zip
/contentFolder1
/contentFolder2
stuff1.txt
stuff2.txt
...
How can I detect these exploding zips, so that I may handle them accordingly? Is it possible without unzipping the file first?
If you want to check, unzip -l will print the contents of the zip file without extracting them. You'll have to massage the output a bit, though, since it's printing all sorts of additional crud.
Unzip to a directory first, and then remove the extra layer if the zip is not a bomb.
tempdir=`mktemp -d`
unzip -d $tempdir file.zip
if [ $(ls $tempdir | wc -l) = 1 ]; then
mv $tempdir/* .
rmdir $tempdir
else
mv $tempdir file
fi
I wouldn't try to detect it. I'd just force unzip to do what I want. With InfoZip:
$ unzip -j -d unzip-output-dir FileFromUntrustedSource.zip
-j makes it ignore any directory structure within the file, and -d tells it to put files in a particular directory, creating it if necessary.
If there are two files with the same name but in different subdirectories, the above command will make unzip ask if you want to overwrite the first with the second. You can add -o to force it to overwrite without asking, or -f to only overwrite if the second file is newer.