I have 500 files and I want to zip into one file output file.
Input files:
abc.pdf
afa.pdf
dgh.pdf
sdg.pdf
....so on
Desired output file should be:
abc.zip
Or
sdg.zip
Or any starting name of input file.
just use the standard zip method for linux:
zip out.zip *.pdf
Assuming all your files are in the same directory, you could do this (run from the same directory):
zip abc.zip *.pdf
Related
What is the Unix bash command to get the list of files (like ls) from archive file of type .bz2 (without unzipping the archive)?
First bzip2, gzip, etc compress only one file. So probably you have compressed tar file. To list the files you need command like:
tar tjvf file.bz2
This command uncompress the archive and test the content of tar.
Note that bzip2 compresses each file, and a simple .bz2 file always contains a single file of the same name with the ".bz2" part stripped off. When using bzip2 to compress a file, there is no option to specify a different name, the original name is used and .bz2 appended. So there are no files, only 1 file. If that file is a tar archive, it can contain many files, and the whole contents of the .tar.bz2 file can be listed with "tar tf file.tar.bz2" without unpacking the archive.
I'm in a directory and I have a zip containing files and directories.
I need to unzip that file, into current directory, but preserving the file structure.
unzip myfile.zip will create a myfile directory in current directory which is not what I want.
unzip -j myfile.zip will kill all the file strucure, which is not what I want.
unzip myfile.zip extracts files in the working directory by keeping path names from the zip file.
So if you get a subdirectory myfile it means it is part of the relative path of compressed files. Check it by listing the zip content
unzip -l myfile.zip
So you can unzip the file from the directory above, or, from the target directory unzip with -d option, where -d is the directory above
cd myfile
unzip myfile.zip -d ..
Dont select the folder while zipping.
For example
myfile/abc.txt and myfile/efg.txt
so while zipping select the files (abc.txt,efg.txt) and zip dont select the myfile folder to zip.
So that when you unzip the file, the parent dir for each file or folder will be the directory in which you unzip.
The myfile directory was zipped into the zip file when it was created and looking at the unzip options there isn't a way to do this without adding additional steps.
If this entire process is under your control you should look at either creating the zip without using including the parent directory or you could use an alternative like tar (to create and extract) which allows you to extract content from the repo with greater precision.
I have several files in a folder with extension .img and I have only one file with extension data.txt
What I need is to copy data.txt and rename it as the names of the .img files.
For instance for the first file in my folder:
`Meaurmen_2154_data.img` >>> copy data.txt >>> rename it Meaurmen_2154_data.txt
Now I have :
Meaurmen_2154_data.img
Meaurmen_2154_data.txt ## the content is the same as data.txt
and do the same for all other files. The content of he text files will be the same for all files just we change the name according to the .img files in my folder.
Run this script
#!/bin/bash
imageFiles=( *.img );
for i in ${imageFiles[*]}
do
withoutExtension=${i%.img};
cp data.txt "$withoutExtension.txt";
done
inside the relevant directory and it will do it for you.
Try
for i in *.img; do cp data.txt $i.txt; done
rename 's/.img.txt/.txt/' *.img.txt
In some distro's rename is different, requiring
rename .img.txt .txt *.img.txt
As always, you might find yourself in need of installing additional packages.
If I have a list of files I want to zip, how can I pass the list to zip?
cookbook/application/views/index.php
cookbook/application/controller/index.php
cookbook/js/index.js
....
cookbook/css/index.css
To do the above list one by one at the command-line would be like zip -r my.zip cookbook/css/index.css, where my.zip is in the same root directory as cookbook
Try
zip -r# my.zip < listfile
The -# flag tells zip to read file names from stdin.
If all files are in the same folder, you don't need to type each file that you want to include in the archive. Just invoke the command and specify their common folder like this:
zip -r cookbook.zip cookbook
All files inside the cookbook directory will be included in the zip archive.
Let's say we have a zip file contains a directory named aq and in the current working directory we have files:
./
|- aq/a.txt
|- b.txt
When i use this command:
zip test.zip aq/* the a.txt file will be zipped into the aq directory that's inside the zip file
The question is how then can I add b.txt into the aq directory that's inside the test.zip file without putting the b.txt in the aq directory first which is in the current working directory like what I did with the a.txt?
Make a temp directory called, e.g. /tmp/$$/aq/, symlink into the temp directory and then do:
(cd /tmp/$$ && zip -r $ZIPDEST aq/)
i.e. zip using the temp dir. zip by default follows symbolic links, so it puts the file into the zip without making a copy.
This is pretty much how I construct complicated hierarchical zip files without copying everything to make the archive.
Tar has better options for renaming items as you're putting them into the archive, but you asked about zip.