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How to extract one file from zipped directory?
Zlib doesn't have any file browsing features, neither does extract-zip
So I don't know what to use.
You can use EvanOxfeld/node-unzip parse zip file contents:
var fs = require('fs')
var unzip = require('unzip')
var path = require('path')
var mkdir = require('mkdirp')
fs.createReadStream('./archive.zip')
.pipe(unzip.Parse())
.on('entry', function (entry) {
var fileName = entry.path
var type = entry.type
if (type==='File' && fileName === 'dir/fileInsideDir.txt') {
var fullPath = __dirname + '/output/' + path.dirname( fileName )
fileName = path.basename( fileName )
mkdir.sync(fullPath)
entry.pipe(fs.createWriteStream( fullPath + '/' + fileName ))
} else {
entry.autodrain()
}
})
[ Example archive ]
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I'm fetching a JS file over the network.
ie:
// file is a JS file
// for example: console.log("I am file")
const file = await get('/api/file/1')
I'm not sure how to feed this type of data to fs. I think it needs to be a Buffer but I haven't gotten it to work:
const dataAsBuffer = Buffer.from(file, 'utf-8')
Then I try:
fs.writeFileSync(filePath), dataAsBuffer, (err) => { ...
But get: TypeError [ERR_INVALID_ARG_TYPE]: The "data" argument must be of type string or an instance of Buffer, TypedArray, or DataView. Received undefined
How can I write a JS file I fetched, to local directory?
EDIT:
the output of get:
// initialize code called once per entity
Test.prototype.initialize = function() {
};
// update code called every frame
Test.prototype.update = function(dt) {
};
console.log('hell oworld');
Fixed: No callback in writeFileSync
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Closed 1 year ago.
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I want to execute a command for each directory in my packages directory. In the command I want to use part of the directory name.
packages
folder-one
folder-two
folder-three
So for each folder execute command 'one' etc
Does anyone have some pointers for this?
Using fs.readdir to list directory contents and child_process.exec to run the command:
const { readdir } = require('fs/promises');
const { exec } = require('child_process');
readdir(__dirname + '/packages').then(packages => {
for (let packageName of packages) {
packageName = packageName.replace(/^folder-/, ''); // remove the 'folder-' part
exec(`your_command_here '${__dirname}/packages/${packageName}'`, (err, stdout, stderr) => {
if (err) {
console.log(err);
}
// if needed you can read the process' stdout and stderr
});
}
});
This question already has answers here:
How do I get the path to the current script with Node.js?
(15 answers)
Closed 3 years ago.
If you have a module that's being imported in some code, let's say...
var my_cool_module = require('my_directory/my_cool_module');
my_cool_module.print_directory_name();
And I'm in that module's context, say this is the file my_cool_module.js...
function get_dir_name() {
// get the directory name...
return directory_name;
}
exports.module.print_directory_name = function() {
console.log("This module is in directory " + get_dir_name());
};
How can I get the directory that the module is in (i.e. "my_directory")
I found out from the node.js documentation here that you can use __dirname
function get_directory_name() {
return __dirname;
}
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Closed 8 years ago.
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I am using the web-server.js node script to run a webserver on my local machine.
The behavior of web-server.js is to dump a complete file listing if a directory is requested.
instead I would like to emulate the apache DirectoryIndex where by if there is an index.html in the directory it will be served instead.
var fs = require('fs');
http://nodejs.org/api/fs.html You can check for files with this.
I could not find the answer so here is my code that does it.
first I had to change the handleRequest function
if (stat.isDirectory()) {
var indexFile = self.getDirectoryIndex_(path);
if (indexFile)
return self.sendFile_(req, res, indexFile);
return self.sendDirectory_(req, res, path);
}
and here is my implementation to getDirectoryIndex_
StaticServlet.prototype.getDirectoryIndex_ = function(path) {
var result = null;
var files = fs.readdirSync(path);
files.forEach(function(fileName, index) {
if (fileName.match(/^index\./gi)) {
result = path + fileName;
return false; //break foreach loop
}
});
return result;
};
This question already has answers here:
node.js require all files in a folder?
(15 answers)
Closed 9 years ago.
Help me please.
I want require all files from directory ?
How best to do it?
I read about require_paths but it does not perform those functions.
You can walk through a directory and require each one:
var mods = {};
var files = fs.readdirSync('your_directory').filter(function(x) { return x.substr(-3) == ".js"; });
for(var i = 0; i != files.length;++i) {
mods[files[i]] = require(path.join('your_directory',files[i]));
}