Is there a way to determine that a file in Windows is hidden in Electron. I need to know the hidden attribute set in Properties of the file and not file name having a dot at front. Currently using fs to get file system information I have no means to obtain this information.
Some sample use case would be (using Typescript):
import fs = require('fs');
getVisibleFilesSync(): string[] {
const folderItems = fs.readdirSync(folderPath);
return folderItems.filter(x => !isHidden(x));
}
isHidden(path: string): boolean {
// how to determine if path is hidden under Windows?
}
Try using the isHiddenSync function in the hidefile module.
Related
I'm using WebMScore to render audio of music scores (it's a fork of MuseScore that runs in the browser or node).
I can successfully load my own, local .sf2 or .sf3 files, however
Trying to load an .sfz soundfont throws error 15424120. (And error.message is simply 'undefined'.)
Unlike .sf2 and .sf3, which contain the sounds and instructions in a single file, the .sfz format is just a text instruction file that refers to a separate folder of samples.
The reason I need the .sfz is that I need to be able to edit the .sfz file textually and programatically without an intervening Soundfont generator.
Is there a way to use .sfz's? Do I need to specify Zerberus (the Musescore .sfz player)? Do I need a different file structure? Please see below.
My environment is node js, with the following test case and file structure:
File Structure
Project Folder
app.js
testScore.mscz
mySFZ.sfz
samples
one.wav
two.wav
etc.wav
Test Case (Works with .sf3 , errors with .sfz)
const WebMscore = require('webmscore');
const fs = require('fs');
// free example scores available at https://musescore.com/openscore/scores
const name = 'testScore.mscz';
const exportedPrefix = 'exported';
const filedata = fs.readFileSync(`./${name}`);
WebMscore.ready.then(async () => {
const score = await WebMscore.load('mscz', filedata, [], false);
await score.setSoundFont(fs.readFileSync('./mySFZ.sfz'));
try { fs.writeFileSync(`./${exportedPrefix}.mp3`, await score.saveAudio('mp3')); }
catch (err) { console.log(err) }
score.destroy();
});
I have the following logic that imports a typescript module and makes some changes to it. Below is the simplified version of code.
const fs = require('fs')
// No deep nestings, and no arrays (only simple key: `string-values`)
const POSTCSS_SELECTORS = {
propA: `value`,
probB: `value`
}
fs.write('fileName.js', POSTCSS_SELECTORS)
After obtaining POSTCSS_SELECTORS I would like to save the output as commonjs module as described below.
Expected output:
fileName.js:
module.exports = {
propA: `value`,
probB: `value`
}
I will appreciate a lot if you could suggest a valid workaround for this case :)
You're 95% there, you just need to write out the rest of the text that you want in your file:
const fs = ...
const POSTCSS_SELECTORS = ...
// prepare that object for writing to file:
const json = JSON.stringify(POSTCSS_SELECTORS, false, 2);
// and then "template" it into the final text you want written.
fs.writeFile(`filename.js`, `module.exports = ${jsonForm}`);
But, if you have a file with a hardcoded variable called POSTCSS_SELECTORS, why not make a file called postcss_selectors.js instead, and put the hardcoded object there, instead? No need to "make a file based on this object" if what you want is that object, as a module.
Just make that module.
I use browserify in order to be able to use require. To use fs functions with browserify i need to transform it with brfs but as far as I understood this results in only being able to input static strings as parameters inside my fs function. I want to be able to use variables for this.
I want to search for xml files in a specific directory and read them. Either by searching via text field or showing all of their data at once. In order to do this I need fs and browserify in order to require it.
const FS = require('fs')
function lookForRoom() {
let files = getFileNames()
findSearchedRoom(files)
}
function getFileNames() {
return FS.readdirSync('../data/')
}
function findSearchedRoom(files) {
const SEARCH_FIELD_ID = 'room'
let searchText = document.getElementById(SEARCH_FIELD_ID).value
files.forEach((file) => {
const SEARCHTEXT_FOUND = file.includes(searchText.toLowerCase())
if (SEARCHTEXT_FOUND) loadXML(file)
})
}
function loadXML(file) {
const XML2JS = require('xml2js')
let parser = new XML2JS.Parser()
let data = FS.readFile('../data/' + file)
console.dir(data);
}
module.exports = { lookForRoom: lookForRoom }
I want to be able to read contents out of a directory containing xml files.
Current status is that I can only do so when I provide a constant string to the fs function
The brfs README contains this gotcha:
Since brfs evaluates your source code statically, you can't use dynamic expressions that need to be evaluated at run time.
So, basically, you can't use brfs in the way you were hoping.
I want to be able to read contents out of a directory containing xml files
If by "a directory" you mean "any random directory, the name of which is determined by some form input", then that's not going to work. Browsers don't have direct access to directory contents, either locally or on a server.
You're not saying where that directory exists. If it's local (on the machine the browser is running on): I don't think there are standardized API's to do that, at all.
If it's on the server, then you need to implement an HTTP server that will accept a directory-/filename from some clientside code, and retrieve the file contents that way.
I am trying to normalize the paths in NodeJS, so that irrespective of user input (*nix/windows), the path should be accessible by node path.
Ex: c:/test/test.xml
c:\\test\test.xml
c:\test/test.xml
/usr/var/test.xml
/usr/var\test.xml
These should be normalized so that path lib can access them. I tried using path.normalize on the input /usr/var\test.xml, it didn't work. the output path string is same as input instead of /usr/var/test.xml
I have written myself simple function to achieve this
export function pathToNix(pathStr: string) {
var p = path.normalize(pathStr);
var path_regex = /\/\//;
p = p.replace(/\\/g, "/");
while (p.match(path_regex)) {
p = p.replace(path_regex, "/");
}
return p;
}
To clarify few things - Windows supports *nix type file structure (c:/test/a.txt). Thus converted all paths to *nix format.
How can I get the file name from an absolute path in Nodejs?
e.g. "foo.txt" from "/var/www/foo.txt"
I know it works with a string operation, like fullpath.replace(/.+\//, ''),
but I want to know is there an explicit way, like file.getName() in Java?
Use the basename method of the path module:
path.basename('/foo/bar/baz/asdf/quux.html')
// returns
'quux.html'
Here is the documentation the above example is taken from.
To get the file name portion of the file name, the basename method is used:
var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var file = path.basename(fileName);
console.log(file); // 'python.exe'
If you want the file name without the extension, you can pass the extension variable (containing the extension name) to the basename method telling Node to return only the name without the extension:
var path = require("path");
var fileName = "C:\\Python27\\ArcGIS10.2\\python.exe";
var extension = path.extname(fileName);
var file = path.basename(fileName,extension);
console.log(file); // 'python'
var path = require("path");
var filepath = "C:\\Python27\\ArcGIS10.2\\python.exe";
var name = path.parse(filepath).name;
console.log(name); //python
var base = path.parse(filepath).base;
console.log(base); //python.exe
var ext = path.parse(filepath).ext;
console.log(ext); //.exe
For those interested in removing extension from filename, you can use
https://nodejs.org/api/path.html#path_path_basename_path_ext
path.basename('/foo/bar/baz/asdf/quux.html', '.html');
If you already know that the path separator is / (i.e. you are writing for a specific platform/environment), as implied by the example in your question, you could keep it simple and split the string by separator:
'/foo/bar/baz/asdf/quux.html'.split('/').pop()
That would be faster (and cleaner imo) than replacing by regular expression.
Again: Only do this if you're writing for a specific environment, otherwise use the path module, as paths are surprisingly complex. Windows, for instance, supports / in many cases but not for e.g. the \\?\? style prefixes used for shared network folders and the like. On Windows the above method is doomed to fail, sooner or later.
path is a nodeJS module meaning you don't have to install any package for using its properties.
import path from 'path'
const dir_name = path.basename('/Users/Project_naptha/demo_path.js')
console.log(dir_name)
// returns
demo_path.js
In NodeJS, __filename.split(/\|//).pop() returns just the file name from the absolute file path on any OS platform.
Why need to care about remembering/importing an API while this regex approach also letting us recollect our regex skills.
So Nodejs comes with the default global variable called '__fileName' that holds the current file being executed
My advice is to pass the __fileName to a service from any file , so that the retrieval of the fileName is made dynamic
Below, I make use of the fileName string and then split it based on the path.sep. Note path.sep avoids issues with posix file seperators and windows file seperators (issues with '/' and '\'). It is much cleaner. Getting the substring and getting only the last seperated name and subtracting it with the actulal length by 3 speaks for itself.
You can write a service like this (Note this is in typescript , but you can very well write it in js )
export class AppLoggingConstants {
constructor(){
}
// Here make sure the fileName param is actually '__fileName'
getDefaultMedata(fileName: string, methodName: string) {
const appName = APP_NAME;
const actualFileName = fileName.substring(fileName.lastIndexOf(path.sep)+1, fileName.length - 3);
//const actualFileName = fileName;
return appName+ ' -- '+actualFileName;
}
}
export const AppLoggingConstantsInstance = new AppLoggingConstants();