I have a file named test.apk this file in located in C:\\workspace\app\build\output\apk\test.apk
in node js i tried following code
var workDir = 'C:\\workspace\app'
var path = require('path');
var localTagPath = path.dirname(workDir);
console.log(path.resolve(localTagPath, 'test.apk'))
what i need is to get the test.apk without specifyinh the path in code
(var workDir = 'C:\\workspace\app\build\output\apk\test.apk')
instead of this ihave to specify like this
where local path is
var localpath = workDir +(\..+\..+\..\)+test.apk
path.resolve(localTagPath, 'test.apk')
You cannot use local file paths as browsers do not have access to the computer's filesystem.
Related
I want to browse directories for json files and merge all json files into one json file per directory.
My javascript skill is at beginner level.
The javascript code needs to run by calling it in the Terminal, not in the browser.
you can make use of glob npm package to retrieve all json file paths in the directories.
glob : https://www.npmjs.com/package/glob
you can simply go with :
const glob = require("glob");
const fs = require('fs-extra');
async function processJsonData(){
let parentDirectoryPath = '<specify path here>';
let data = {};
let jsonFilePaths = glob.sync(`${parentDirectoryPath}/*/*.json`);
//this will return paths for all the json files
for (const jsonFilePath of jsonFilePaths) {
let content = await fs.readJsonSync(jsonFilePath);
Object.assign(data,content);
}
await fs.writeJson(<filePath>, data, {spaces: 2});
//here specify your file path
}
ps. I am a new contributor, might have misread the question.
This question already has answers here:
Proper way to reference files relative to application root in Node.JS
(5 answers)
Closed 1 year ago.
My question is, is it possible to do so that when I have a code and a text file, but when I want to open or locate the text file, I dont need to specify it everytime where it is or like when I send it to someone else, they dont need to change the let file = C:\Users\etc...
My code:
const { readFile, readFileSync } = require('fs');
let file = 'C:\\Users\\eeroj\\Desktop\\word-counter\\TextFile2.txt';
function countRepeatedWords(sentence) {
const words = sentence.split(" ").filter(word => !!word);
const wordMap = {};
for (let word of words) {
const key = word.trim().toLowerCase();
const currentWordCount = wordMap[key];
wordMap[key] = (currentWordCount ?? 0) + 1;
}
const sortedEntries = Object.entries(wordMap).sort(([a,], [b,]) => a.localeCompare(b));
const sortedWordMap = Object.fromEntries(sortedEntries);
//returning the wordMap and the sorting of the wordMap
return sortedWordMap
return wordMap;
}
words = readFileSync(file).toString();
console.log(countRepeatedWords(words));
Yes you can, through the path (built-in Nodejs) module and the __dirname constant. The __dirname environment variable will automatically get the current folder's path for you, and then you can use path.join to join the current directory with the filename you want to access. Something like this
const path = require("path");
// Assuming that the current script file are inside the 'C:\\Users\\eeroj\\Desktop\\word-counter' folder
const file = path.join(__dirname, "TextFile2.txt");
// file = 'C:\\Users\\eeroj\\Desktop\\word-counter\\TextFile2.txt'
If you put "TextFile2.txt" in a relative position not in the current folder, you can also use path.join with the folder traversal syntax such as .. to go back a folder
const path = require("path");
// Assuming that the current script file are inside the 'C:\\Users\\eeroj\\Desktop\\word-counter' folder
const file = path.join(__dirname,"../../", "TextFile2.txt");
// file = "C:\\Users\\eeroj\\TextFile2.txt"
As a result, this way should work regardless of anywhere you put the project, even on a different machine, as long as the project folder structure stays relatively the same and the "TextFile2.txt" file is in the right place relative to the script file.
If your txt file is in the same directory as the script you can simply use the relative path with a '.' as
let file = "./TextFile2.txt"
Just note that the textfile will have to always have the same name.
When using .js files that are run with Windows Script Host, is it possible to get the path of the .js file that is running (which is not the same as the current working directory)? In other words, I'm wanting something similar to PowerShell's $PSScriptRoot variable. Is there any way to get this?
The reason I want this is because the working directory when these scripts are executed is not always the same as the location of the .js file itself, but I want my code to reference things relative to the location of the .js file.
You can get this with Node Js
const path = require('path'); // you can install this with npm
// __dirname = C:\dev\project
// __filename = C:\dev\project\index.js
const dir= path.basename(__dirname);
const archive= path.basename(__filename);
const fullpath = path.join(dir, archive);
console.log('Dir:', dir); // Dir: test
console.log('Archive:', archive); // Archive: index.js
console.log('Dir + Archive:', fullpath); // Dir + Archive: project\index.js
I have a file under "modules/test/main/main.js" and i want to get the "test" folder :
I tried this code but all i get is the main folder
path.basename(path.dirname('modules/test/main/main.js'))
You can split on / and than drop off last item as it is file name.
and than take the middle element
let path = `modules/test/main/main.js`
let middle = path.split(/\//g)
middle.pop()
console.log(middle[Math.floor(middle.length-1)/2])
You can use path.join() to go up two directories, and then get the basename of the directory:
const p = 'modules/test/main/main.js';
const dir = path.basename(path.join(p, '../..')); // 'test'
I am attempting to zip the contents of two directories and download the resulting .zip file. One directory contains .txt files and the other .jpg. I am using archiver to zip the files and running the express framework on node js. I am inclined to think that the problem exists in the download step, as the resulting zipped file that is created in the project root expands as expected, however when the file is downloaded, I get an "Error 2 - No such file or directory."
app.get('/download',function(req, res){
zipFile = new Date() + "-Backup.zip";
var output = fs.createWriteStream(__dirname +"/backups/"+ zipFile);
var archive = archiver('zip');
output.on('close', function() {
console.log(archive.pointer() + ' total bytes');
console.log('archiver has been finalized and the output file descriptor has closed.');
});
archive.on('error', function(err) {
throw err;
});
archive.pipe(output);
var files1 = fs.readdirSync(__dirname+'/posts');
var files2 = fs.readdirSync(__dirname+'/uploads');
for(var i = 0; i< files1.length; i++){
archive.append(fs.createReadStream(__dirname+"/posts/"+files1[i]), { name: files1[i] });
}
for(var i = 0; i< files2.length; i++){
archive.append(fs.createReadStream(__dirname+"/uploads/"+files2[i]), { name: files2[i] });
}
archive.finalize();
res.download(__dirname + "/backups/" + zipFile, zipFile);
});
zipFile is a global variable.
The on 'close' logs fire properly and no errors occur, but the file will not open after being downloaded. Is there an issue with response headers or something else I am unaware of?
Thanks for the help.
I solved my own problem using node-zip as the archive utility.
var zip = require('node-zip')(); // require the node-zip utility
var fs = require('fs'); // I use fs to read the directories for their contents
var zipName = "someArbitraryName.zip"; // This just creates a variable to store the name of the zip file that you want to create
var someDir = fs.readdirSync(__dirname+"/nameOfDirectoryToZip"); // read the directory that you would like to zip
var newZipFolder = zip.folder('nameOfDirectoryToZip'); // declare a folder with the same name as the directory you would like to zip (we'll later put the read contents into this folder)
//append each file in the directory to the declared folder
for(var i = 0; i < someDir.length,i++){
newZipFolder.file(someDir[i], fs.readFileSync(__dirname+"/nameOfDirectoryToZip/"+someDir[i]),{base64:true});
}
var data = zip.generate({base64:false,compression:'DEFLATE'}); //generate the zip file data
//write the data to file
fs.writeFile(__dirname +"/"+ zipName, data, 'binary', function(err){
if(err){
console.log(err);
}
// do something with the new zipped file
}
Essentially, what is happening can broken down into 3 steps:
Use fs to read the contents of a directory that you would like to zip.
Use zip.folder to declare the folder, then use zip.file to append files to that directory. I just used a for loop to iteratively add each file in the directory that was read in step 1.
Use zip.generate to create the .zip file data, and write it to file using fs.
The resulting file can be downloaded or whatever you would like to do with it. I have seen no issues using this method.
If you want to zip more than one directory, just repeat steps 1 and 2 before you zip.generate, creating a new zip.folder for each directory.
Just use
archive.on("finish",function(){
res.download(__dirname + "/backups/" + zipFile);
})