Get file folder with Nodejs

Question

Hello, I am trying to get the file folder but without success. I’ve tried with path, with __dirname, process.cwd(), but all return the complete path, for example:

C: Node folder where is the.js file

What I want to know is the file folder:

the file.js

How can I do that?

Answer 1

Use path dirname.

    // filePath deve ser C:\node\pasta\onde\esta\o
    var filePath = require('path').dirname('C:\node\pasta\onde\esta\o\arquivo.js');
    var currentDirectory = filePath.split(path.sep).pop();

Answer 2

Use the function path.basename:

const path = require('path');

// __dirname = C:\desenvolvimento\amostra
// __filename = C:\desenvolvimento\amostra\amostra.js
const diretorio = path.basename(__dirname);
const arquivo = path.basename(__filename);
const composicao = path.join(diretorio, arquivo);

console.log('Diretório:', diretorio); // Diretório: amostra
console.log('Arquivo:', arquivo); // Arquivo: amostra.js
console.log('Diretório + Arquivo:', composicao); // Diretório + Arquivo: amostra\amostra.js

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path.basename

The path.basename() methods Returns the last Portion of a path, similar to the Unix basename command. Trailing directory separators are Ignored, see path.sep.

In free translation:

The method path.basename() returns the last portion of a path, similar to the Unix command basename. Tabs on the right are ignored, see path.sep.

Answer 3

Hello,

I don’t know if there is a direct way to get only the directory folder, but I suggest you take the result you already have and do a simple manipulation.

const caminho = 'C:\\node\\pasta\\onde\\esta\\o\\arquivo.js';

const caminhoPorNivelDeDiretorio = caminho.split('\\');

const diretorio = caminhoPorNivelDeDiretorio.slice(-2).join('\\');