Using Ruby (1.9.3) I need to replace a single file in a zip archive.
The situation is as follows. I have ~1000 zip archives that need to be updated, specifically one file in each of them needs to be replaced. The archives are all of the same structure. Is there a quick and dirty way for Ruby, or a library/gem for Ruby, to simply say "replace the file in this zip archive with this file on the filesystem"?
I'll work on a solution of my own in the meantime.
You can use the zip command, called from the ruby, which probably will be the best solution. From the zip manpage zip manpage
-d
--delete
Remove (delete) entries from a zip archive. For example:
zip -d foo foo/tom/junk foo/harry/\* \*.o
will remove the entry foo/tom/junk, all of the files that start with foo/harry/, and all of the files that end with .o (in any path). Note that shell path‐
name expansion has been inhibited with backslashes, so that zip can see the asterisks, enabling zip to match on the contents of the zip archive instead of the
contents of the current directory. (The backslashes are not used on MSDOS-based platforms.) Can also use quotes to escape the asterisks as in
zip -d foo foo/tom/junk "foo/harry/*" "*.o"
Not escaping the asterisks on a system where the shell expands wildcards could result in the asterisks being converted to a list of files in the current
directory and that list used to delete entries from the archive.
Under MSDOS, -d is case sensitive when it matches names in the zip archive. This requires that file names be entered in upper case if they were zipped by
PKZIP on an MSDOS system. (We considered making this case insensitive on systems where paths were case insensitive, but it is possible the archive came from
a system where case does matter and the archive could include both Bar and bar as separate files in the archive.) But see the new option -ic to ignore case
in the archive.
If you want a pure ruby solution take a look at ZipFileSystem
Zip::ZipFile looks promising. It appears to have a way to delete and add files to a zip archive.
Related
I have a lot of markdown files in various directories each with the same format (# title, then ## sub-title).
can I make the --toc respect the folder layout, in that the folder itself is the name of chapter, and each markdown file is content of this chapter.
so far pandoc totally ignores my folder names, it works the same as putting all the markdown files within the same folder.
My approach to this is to create index files in each folder with first level heading and downgrade headings in other files by one level.
I use Git and by default I'm using default structure, having first level headings in files, but when I want to generate ebook using pandoc I'm modifying files via automated Linux shell script. After that, I revert changed files via Git.
Here's the script:
find ./docs/*/ -name "*.md" ! -name "*index.md" -exec perl -pi -e "s/^(#)+\s/#$&/g" {} \;
./docs/*/ means I'm looking only for files inside subfolders of docs directory like docs/foo/file1.md, docs/bar/file2.md.
I'm also interested only in *.md files, excluding *index.md files.
In index.md files (that I name usually 00-index.md to make them appear as first), I put a first level heading # and because those files are excluded from find portion of the script, their headings aren't downgraded.
Next, there's a perl's search and replace command with regular expression s/^(#)+\s/#$&/g that looks for all lines starting from one or more # and adds another # to them.
In the end, I'm running pandoc with --toc-depth=2 so the table of content contains only first and second level headings.
pandoc ./docs/**/*.md --verbose --fail-if-warnings --toc-depth=2 --table-of-contents -o ./ebook.epub
To revert all changes made to files, I restore changes in the Git repo.
git restore .
If I run...
$ myTest="bar"
$ zip -r foo-${myTest} path/*
...then I get a zip file named foo-bar.zip. (note the .zip extension!) However, if I run...
$ myTest="1.0.1"
$ zip -r foo-${myTest} path/*
...then I get a zip file named foo-1.0.1. (no .zip extension!)
I can obviously add .zip to my script, but I would like to understand what is going on here. Why doesn't zip add the extension when the filename is built from a variable with numbers in it?
It dawned on me as I wrote that last question that this isn't about numbers. Quoting from man zip:
If the name of the zip archive does not contain an extension, the extension .zip is added. If the name already contains an extension other than .zip, the existing extension is kept unchanged. However, split archives (archives split over multiple files) require the .zip extension on the last split.
The problem is that I have .'s in the variable, which zip interprets as filename extensions. Luckily, my script constructs the variable with .'s so I can confidently add .zip to the end. Otherwise, I would need to test for .'s to name the file correctly.
Kind of easy question, but I can't find the answer. I want to extract the contents of multiple zipped folders into a single directory. I am using the bash console, which is the only tool available on the particular website I am using.
For example, I have two folders: a.zip (which contains a1.txt and a2.txt) and b.zip (which contains b1.txt and b2.txt). I want to get extract all four text files into a single directory.
I have tried
unzip \*.zip -d \newdirectory
But it creates two directories (a and b) with two text files in each.
I also tried concatenating the two zipped folders into one big folder and extracting it, but it still creates two directories, even when I specify a new directory.
I can't figure what I am doing wrong. Any help?
Thanks in advance!
Use the -j parameter to ignore any directory structure.
unzip -j -d /path/to/your/directory '*.zip*'
I have a folder of data folders with the following structure:
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data2.gz
sampleName2-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
I want to modify all the data.gz within each sample folder by appending the sample name but not the random numbers to get:
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName1_data1.gz
sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName1_data2.gz
sampleName2-randomNumbers/subfolder1/subfolder2/subfolder3/sampleName2_data1.gz
It seems like this should be a simple mv for loop but I haven't been able to figure out how to pull part of a folder name using basename.
for i in */Data/Intensities/BaseCalls/*.gz; do mv $i "fastq""/"${i%%-*}"."`basename $i`; done
I couldn't figure out how to make the files stay in their original folder but for my purposes it works to have all the files go to a new folder ("fastq")
I suppose the "sampleName" part doesn't include dashes. In that case, use the standard pattern removal expansion: %%. That is, suppose your full path (relative to directory root) is stored in $path, just do ${path%%-*} to extract the "sampleName" part. Search for %% in the Bash Reference Manual for more details. As a simple example:
> path=sampleName1-randomNumbers/subfolder1/subfolder2/subfolder3/data1.gz
> echo ${path%%-*}
sampleName1
Otherwise, you could also use more advanced substring extraction based on regex. See BashFAQ/100 or Manipulating Strings from the TLDP Advanced Bash Scripting Guide.
Update. Here's the full command to perform the job described, and it is entirely native to the shell:
for file in */Data/Intensities/BaseCalls/*.gz; do
mv "$file" "${file%/*}/${file%%-*}_${file##*/}"
done
I have to use bash scripting to copy files from one folder to another. If the destination folder has a file with the same name but older timestamp, it should not copy. Only newer files should be copied. I could have used cp -u, but I was asked not to use it. Essentially I have to use the test command testing for "ot". Please let me know how could this be done. I believe two for loops one to read the files in the source and one for the destination directories can be used and the the time stamp compared. The problem is that both for loops produce the absolute path names along with the file name. So not sure how to compare them
Thanks
You can profit from the parameter substitution:
for file in "$folder1"/* ; do
filename=${file##*/} # Remove everything to the last slash.
Or, you can change the directory:
cd "$folder1"
for file in * ; do
## you have to use full or relative path to $folder2 here