zipdir('/path/to/be/zipped', { saveTo: '~/myzip.zip' }, function (err, buffer) {
// `buffer` is the buffer of the zipped file
// And the buffer was saved to `~/myzip.zip`
});
I have try to compress folder with files but it has compressed file but when extract this file give me /* files only with out parent folder zipped/*, how can i do this process !?
Related
I call an API which returns a buffer data of a .zip file. I want to read files' buffer data which resides inside the .zip file using its buffer data without saving the .zip file. Is it possible?
Try the zlib library (its a core node.js library - docs: https://nodejs.org/api/zlib.html#zlib), with this example I took from the documentation
const {unzip } = require('node:zlib');
const buffer = Buffer.from('eJzT0yMAAGTvBe8=', 'base64');
unzip(buffer, (err, buffer) => {
if (err) {
console.error('An error occurred:', err);
process.exitCode = 1;
}
console.log(buffer.toString());
});
I have two folders, input and output folder with many text files in the below format. How do I read all the files from the input folder,run the regex and write all the updated files to another output folder?I am using nodejs.
Input: $.Carpool[0].NoofSeats], [$.Carpool[1].NoofSeats]
So after replace with regex the updated text file should be:
Regex: str = str.replace(/\.[A-Z]/g, (m0) => m0.toLowerCase());
Output: [$.carpool[0].noOfSeats], [$.carpool[1].noOfSeats]
So far I got to reading files from the directory:
const fs= require("fs");
let directory = "Input" // Desktop/Input
let files = fs.readdirSync(directory)
console.log(files);
You want to loop through the files, assuming if the contents are a text file in UTF-8 format here is an example.
You use fs.readFile to read a specific file after listing directory.
Then use fs.writeFile to write a new file with contents.
I use /directory/${f} for the new file directory path and ${f} for filename that was opened.
const fs = require("fs");
// Directory
let directory = "/";
// Files
let files = fs.readdirSync(directory);
// Loop through the files
files.forEach(f => {
// Read the contents in UTF-8 format
fs.readFile(f, 'utf8', function(err, contents) {
if (err) { console.log(err); }
// Output Contents
console.log(contents);
// Perform regex here
contents = contents.replace(/\.[A-Z]/g, (m0) => m0.toLowerCase());
// Write new file to path /new, with contents
fs.writeFile(`/directory/${f}`, contents, function(err) {
if (err) {
// Error writing
return console.log(err);
}
console.log("The file was saved!");
});
});
});
I have a zip file that has a folder like
1234/pic1.png
1234/pic2.png
1234/data.xlsx
I am trying to extract the spreadsheet (failing that, all files) using node-stream-zip.
const StreamZip = require('node-stream-zip');
const zip = new StreamZip({
file: path.join(downloadsDir, fileToFind),
storeEntries: true
});
zip.on('ready', () => {
if(!fs.existsSync('extracted')) {
fs.mkdirSync('extracted');
}
zip.extract('1234/', './extracted', err => {
console.log(err);
});
zip.close();
});
This produces
EBADF: bad file descriptor, read
In the extracted folder is one of the png files. But when following the guide to extract just the xlsx file it appears that the xlsx file is the one causing this error.
zip.extract('1234/data.xlsx', './extracted.xlsx', err => {
console.log(err);
});
Is the problem with the xlsx file? I can open it manually. Is it permissions-related? Node? This particular package?
Your problem is related to zip.close(). You're closing it on the same tick as you're invoking zip.extract().
I want to use the following library to compress images
https://github.com/imagemin/imagemin
The problem is when the user uploads using a form, how do I plug in the file details from the form to the image min plugin? Like for example, if the file form field is call example-image, how do I plug that file form field to image min plugin so that it can compress the images?
I tried:
req is from the express/nodejs req
var filebus = new Busboy({ headers: req.headers }),
promises = [];
filebus.on('file', function(fieldname, file, filename, encoding, contentType) {
var fileBuffer = new Buffer(0),
s3ImgPath;
if (file) {
file.on('data', function (d) {
fileBuffer = Buffer.concat([fileBuffer, d]);
}).on('end', function () {
Imagemin.buffer(fileBuffer, {
plugins: [
imageminMozjpeg(),
imageminPngquant({quality: '80'})
]
}).then(function (data) {
console.log(data[0]);
if (s3ImgPath) {
promises.push(this.pushImageToS3(s3ImgPath, data[0].data, contentType));
}
});
}.bind(this));
}
});
But the problem is I rather have a buffer of the file that I can upload to S3. I don't want to come the files to a build/images folder. I want to get a buffer for the file, compress it, and upload that buffer to s3. How can I use image min to get a buffer of the file upload via html form and upload that to s3?
The documentation for the output parameter shows that it is optional (though admittedly, the function declaration did not, which might be confusing).
output
Type: string
Set the destination folder to where your files will be written. If no
destination is specified no files will be written.
Therefore, you can opt out of writing the files to storage and just use the output in memory:
imagemin([file.path], {
plugins: [
imageminMozjpeg(),
imageminPngquant({quality: '65-80'})
]
}).then(files => {
// upload file to S3
});
I zipped a folder in nodejs with code :
fstream = require('fstream'),
tar = require('tar'),
zlib = require('zlib');
fstream.Reader(toZipDetails) /* Read the source directory */
.pipe(tar.Pack()) /* Convert the directory to a .tar file */
.pipe(zlib.Gzip()) /* Compress the .tar file */
.pipe(fstream.Writer(zipOutDetails)); /* Give the output file name */
Then i unzipped it with :
fs.createReadStream(inFileName)
.pipe(zlib.Gunzip())
.pipe(tar.Extract({ path: "C:\\temp\\extract" }))
.on("end", function () {
alert("done");
});
The folder name is toZip with file a.txt.
I wanted a folder toZip with a.txt in extract folder, but
i got a a.txt file in extract folder.
How can i get the toZip folder ?
You can try :
var unzip = require('unzip')
var fs = require('fs');
fs.createReadStream('<<.zip folder path>>').pipe(unzip.Extract({ path: <<save unzipped data path>> }));