How to zip files and folders which are in a particular directory using Node JS ? I have abc directory under which there are some .xml, .rels etc. files, and a few folder workbook, styles, etc. Now I want to zip files under abc folder with the name abc.zip
I don't want the zip to happen with the abc directory i.e. abc.zip -> abc -> and those files; I want abc.zip-> those files.
Currently using JSZIP which is creating the files on generate. but i want to create the files and folders first and then ZIP them
Expected->
on unzipping the abc.zip file ->
workbook (Folder)
styles (Folder)
content.xml (File)
should be extracted instead of "abc" folder.
You can use zip-lib.
var zl = require("zip-lib");
zl.archiveFolder("path/to/abc", "path/to/abc.zip").then(function () {
console.log("done");
}, function (err) {
console.log(err);
});
path/to/abc.zip archive file directory will be as follows:
path/to/abc.zip
.
├── file.xml
├── workbook
├── styles
└── file2.rels
Related
I'm using the 7zip command line interface to extract archives, like so:
7za.exe x -y {path_to_zipfile} -o{path_to_target_folder}
If my zipfile is named my_archive.7z, then I get the following filestructure in the target folder:
🗁 target_folder
└─ 🗁 my_archive
├─ 🗋 foo.png
├─ 🗁 bar
│ ├─ 🗋 baz.txt
│ └─ 🗋 qux.txt
...
However, I don't want the subfolder 🗁 my_archive. I'm looking for flags to apply on the 7zip command such that everything extracts directly in the target folder, without creating the 🗁 my_archive subfolder.
NOTES
I can't replace x with e because the filestructure shouldn't be lost (the e flag pushes all files to the toplevel).
I'm working on a Windows 10 computer, but the solution must also work on Linux.
I'm using the following version: 7-Zip (a) 19.00 (x64)
Some background info: I'm calling 7zip from a Python program, like so:
# Variables:
# 'sevenzip_abspath': absolute path to 7za executable
# 'zipfile_abspath': absolute path to zipped file (`.7z` format)
# 'targetdir_abspath': absolute path to target directory
commandlist = [
sevenzip_abspath,
'x',
'-y',
zipfile_abspath,
f'-o{targetdir_abspath}',
]
output = subprocess.Popen(
commandlist,
stdout=subprocess.PIPE,
shell=False,
).communicate()[0]
if output is not None:
print(output.decode('utf-8'))
I know I could do all kinds of things in Python after the unzipping has finished (move/rename directories, etc etc), but that's for plan B. First I want to check if there is an elegant solution.
I'd like to stick to 7zip for reasons that would lead us too far here.
You can rename the top level folder to match the target folder before extracting the archive.
7za rn {path_to_zipfile} my_archive target_folder
This will permanently change the archive. If you don't want that, take a copy first.
I have folder in like CW1234.zip and it has various folders and subfolders like below. So, CW1234.zip has CW_All folder which in turn has CW123 and CW234 folders and so on
CW1234.zip
CW_All
CW123
xyz.pdf
CW234
abc.doc
and to extract I use this code:
from zipfile import ZipFile
with ZipFile(r'CW41234.zip', 'r') as zipObj:
# Extract all the contents of zip file in current directory
zipObj.extract()
The only problem is the unzipped folder I get is from CW_All and all the subfolders and file.
What I want is to get it from CW1234 as one folder and then the structure follows?
Current Output
CW_All
CW123
xyz.pdf
CW234
abc.doc
Expected Output
CW1234
CW_All
CW123
xyz.pdf
CW234
abc.doc
Couldn't find anything in the documentation also!!
Using ZipFile.extractall() we can simply provide a new path to extract the contents of the archive to, which we can base on the filename of the archive.
I have a .zip file with the following structure:
archive1024.zip:.
│
└───Folder_with_script
stuff.py
Here is the script to extract all of the files inside of the archive into a sub-folder:
from zipfile import ZipFile
file = "archive1024.zip"
with ZipFile(file, "r") as zFile:
zFile.extractall(path=file.split(".")[0])
I now have a folder-structure like this:
J:.
│ archive1024.zip
│ unzip.py
│
└───archive1024
└───Folder_with_script
stuff.py
I would like to recursively search through "project" directories for "Feedback Report" folder and if that folder has no more sub directories I would like to process the files in a particular manner.
After we have reached the target directory, I want to find the latest feedback report.xlsx in that directory(which will contain many previous versions of it)
the data is really huge and inconsistent in its directory structure. I believe the following algorithm should bring me close to my desired behavior but still not sure. I have tried multiple scrappy code scripts to convert into json path hierarchy and then parse from it but the inconsistency makes the code really huge and not readable
The path of the file is important.
My algorithm that I would like to implement is:
dictionary_of_files_paths = {}
def recursive_traverse(path):
//not sure if this is a right base case
if(path.isdir):
if re.match(dir_name, *eedback*port*) and dir has no sub directory:
process(path,files)
return
for contents in os.listdir(path):
recursive_traverse(os.path.join(path, contents))
return
def process(path,files):
files.filter(filter files only with xlsx)
files.filter(filter files only that have *eedback*port* in it)
files.filter(os.path.getmtime > 2016)
files.sort(key=lambda x:os.path.getmtime(x))
reversed(files)
dictionary_of_files_paths[path] = files[0]
recursive_traverse("T:\\Something\\Something\\Projects")
I need guidance before I actually implement and need to validate if this is correct.
There is another snippet that I got for path hierarchy from stackoverflow which is
try:
for contents in os.listdir(path):
recursive_traverse(os.path.join(path, contents))
except OSError as e:
if e.errno != errno.ENOTDIR:
raise
//file
Use pathlib and glob.
Test directory structure:
.
├── Untitled.ipynb
├── bar
│ └── foo
│ └── file2.txt
└── foo
├── bar
│ └── file3.txt
├── foo
│ └── file1.txt
└── test4.txt
Code:
from pathlib import Path
here = Path('.')
for subpath in here.glob('**/foo/'):
if any(child.is_dir() for child in subpath.iterdir()):
continue # Skip the current path if it has child directories
for file in subpath.iterdir():
print(file.name)
# process your files here according to whatever logic you need
Output:
file1.txt
file2.txt
I have a folder structure like this:
- project:
-- folder 01:
--- file1.cpp
--- file2.cpp
-- folder 02:
--- file1.cpp
--- file2.cpp
I want to zip the content of the project folder in a way I get(when i unzip) this structure:
- folder 01:
-- file1.cpp
-- file2.cpp
- folder 02:
-- file1.cpp
-- file2.cpp
My Problem is now that I always get a parent folder with the same name as my zip file which contains folder 01 and 02. Is there a way I can zip without getting this parent folder ?
zip -r foo ./
assuming **
./
** i.e., present working directory is project in your case.
-r for recursively zipping
foo is the name of your zip file, i.e., foo.zip is the final zipped product you want.
I'm using node-archiver to zip some files and in the zip file I find the archived files under the directory where was zipped from.
So in the zip file, 'myfile.txt' is under this sort of directory:
'home/ubuntu/some_folder/another_folder/'
I want to have the files in the root directory of the zip file.
This is the code I suspect that is in charge with this:
archive.bulk(
{src: filePaths}
);
Its documentation is here although I don't see how to do it.
Any ideas?
Just add expand:true, flatten:true and indicate what name you want to use for destination folder (dest: 'destinationfoldername/')
archive.bulk(
{src: filePaths, dest:'destinationfoldername/', expand:true, flatten:true}
);
It workerd for me