Example:
Here's list of files in "/tmp/test_dir"
file1
zip -r Test_Files.zip *
When I unzip Test_Files.zip I'm getting the below output
Current working directory "/tmp/test_dir"
/tmp/test_dir/file1
What I'm expecting when I unzip Test_Files.zip;
/tmp/test_dir/Test_Files/file1
Can anyone help how do i get expected result as mentioned above?
Use unzip. You can use -o to overwrite the existing files and -q to make it quiet. In doubt? Just use terminal and type in unzip (or try /usr/bin/unzip) to see helpful information.
Related
I want to find a specific line of text inside of a file and print it on my screen using Linux commands. I know I could do:
find -name '[filename]' | xargs grep -i ‘[text I'm looking for inside file]'
However, my file is inside a .zip file. I know unzip -l [.zip file name] will list of all files inside of .zip file, but it won't let me access in order to "grep" the information I need.
Is there a solution to this?
The find_zip tool from the open-source Zip-Ada project.
get / download the sources
build with the command gnatmake -P zipada (you get GNAT through apt or yum)
you have now a binary find_zip.
Usage: find_zip archive[.zip] ["]text["]
I want to download all the files in a specific directory of my site.
Let's say I have 3 files in my remote SFTP directory
www.site.com/files/phone/2017-09-19-20-39-15
a.txt
b.txt
c.txt
My goal is to create a local folder on my desktop with ONLY those downloaded files. No parents files or parents directory needed. I am trying to get the clean report.
I've tried
wget -m --no-parent -l1 -nH -P ~/Desktop/phone/ www.site.com/files/phone/2017-09-19-20-39-15 --reject=index.html* -e robots=off
I got
I want to get
How do I tweak my wget command to get something like that?
Should I use anything else other than wget ?
Ihue,
Taking a shell programatic perspective I would recommend you try the following command line script, note I also added the citation so you can see the original threads.
wget -r -P ~/Desktop/phone/ -A txt www.site.com/files/phone/2017-09-19-20-39-15 --reject=index.html* -e robots=off
-r enables recursive retrieval. See Recursive Download for more information.
-P sets the directory prefix where all files and directories are saved to.
-A sets a whitelist for retrieving only certain file types. Strings and patterns are accepted, and both can be used in a comma separated list. See Types of Files for more information.
Ref: #don-joey
https://askubuntu.com/questions/373047/i-used-wget-to-download-html-files-where-are-the-images-in-the-file-stored
Have got a problem executing the command as below:
tar -xvf arch.tar.gz -s '/^bundle//'
Could be the
-s '/^bundle//'
is a problem as I've got errors like:
$ tar -xvf arch.tar.gz -s '/^bundle//'
tar: /^bundle: Not found in archive
tar: Exiting with failure status due to previous errors
The tried to run the command under Cygwin/Win10.
It's part of the longer script but I'm not sure what was the idea of original author. Archive does include the 'bundle' folder inside... and it's the only first level file there.
Thank you in advance :)
-s does not mean to do a substitution, which seems to be how you're trying to use it. You probably want --xform='s/^bundle//'
-s has the following entry in the help listing:
-s, --preserve-order, --same-order
member arguments are listed in the same order as
the files in the archive
With your code it's actually trying to find a file with the name /^bundle// which does not exist, even if bundle does. Also, the --xform option I gave will rewrite the names of files to strip the string bundle from the front. If you are just trying to not extract the file bundle you would want the flag --exclude='bundle'
In this case, if bundle is a top level directory in the archive, and it's the only one, you could also use the flag --strip-components=1, though this would get rid of all the top level directories, so might not be exactly what you want depending on your archive
Thanks all,
Problem solved other, then 'tar', way but for those who may be interested here is the answer I have found on the web:
If you are developing on Linux, or using GNU tar, this command should work:
tar -xvf arch.tar.gz --transform 's|^bundle/||'
For Mac or BSD-based operating systems:
tar -xvf arch.tar.gz -s '/^bundle//'
Yes, the idea was to remove the /bundle/ folder from files paths.
I used command zip in linux (RedHat), this is my command:
zip -r /home/username/folder/compress/zip.zip /home/username/folder/compressed/*
Then, i open file zip.zip, i see architecture as path folder compress.
I want to in folder zip only consist list file *.txt
Because i used this command in script crontab hence i can't use command cd to path folder before run command zip
Please help me
I skimmed the zip man page and this is what I have found. There is not an option archive files relative to a different directory. The closest I have found is zip -j which removes the entire path and stores the files directly in the zip rather than sub directories. I do not know what happens in the case of file name conflicts such as if /home/username/folder/compressed/a.txt and /home/username/folder/compressed/subdir/a.txt both exist. If this is not a problem for you, you can use this option, but I am concerned because you did specify the -r option indicating that you expect zip to traverse sub folders.
I also thought of the possibility that your script could somehow call zip with a different working directory, but I took a look at this unix stack exchange page and it looks like their options use cd.
I have to admit I do not understand why you cannot use cd and I am very curious about it. You said something about using crontab, but I have never heard of anything wrong with changing directories in a crontab script.
I used option -j in command zip
zip -jr /home/username/folder/compress/zip.zip /home/username/folder/compressed/*
and i was yet settled this problem, thanks
I'm writing a PHP script that downloads a series of generated files (using wget) into a directory, and then zips then up, using the zip command.
The downloads work perfectly, and the zipping mostly works. I run the command:
zip -r /var/www/oraviewer/rgn_download/download/fcst_20100318_0319.zip /var/www/oraviewer/rgn_download/download/fcst_20100318_0319
which yields a zip file with all the downloaded files, but it contains the full /var/www/oraviewer/rgn_download/download/ directories, before reaching the fcst_20100318_0319/ directory.
I'm probably just missing a flag, or something small, from the zip command, but how do I get it to use fcst_20100318_0319/ as the root directory?
I don't think zip has a flag to do that. I think the only way is something like:
cd /var/www/oraviewer/rgn_download/download/ && \
zip -r fcst_20100318_0319.zip fcst_20100318_0319
(The backslash is just for clarity, you can remove it and put everything on one line.)
Since PHP is executing the command in a subshell, it won't change your current directory.
I have also get it worked by using this command
exec('cd '.$_SERVER['DOCUMENT_ROOT'].' && zip -r com.zip "./"');
cd /home/public_html/site/upload/ && zip -r sub_upload.zip sub_upload/
Use the -j or --junk-paths option in your zip command.
From the zip man page:
-j
--junk-paths
Store just the name of a saved file (junk the path), and do not store
directory names. By default, zip will store the full path (relative
to the current directory).