我怎么写这个回到父2层去找文件?

fs.readFile(__dirname + 'foo.bar');

当前回答

你可以使用

path.join(__dirname, '../..');

其他回答

你可以用不同的方法定位父文件夹下的文件,

const path = require('path');
const fs = require('fs');

// reads foo.bar file which is located in immediate parent folder.
fs.readFile(path.join(__dirname, '..', 'foo.bar'); 

// Method 1: reads foo.bar file which is located in 2 level back of the current folder.
path.join(__dirname, '..','..');


// Method 2: reads foo.bar file which is located in 2 level back of the current folder.
fs.readFile(path.normalize(__dirname + "/../../foo.bar"));

// Method 3: reads foo.bar file which is located in 2 level back of the current folder.
fs.readFile(__dirname + '/../../foo.bar');

// Method 4: reads foo.bar file which is located in 2 level back of the current folder.
fs.readFile(path.resolve(__dirname, '..', '..','foo.bar'));

你可以使用

path.join(__dirname, '../..');

最简单的方法是使用path.resolve:

path.resolve(__dirname, '..', '..');

我在运行电子应用程序我可以通过path。resolve()获取父文件夹

父1级:路径。解析(__dirname, '..') + '/'

父2级:路径。解决(__dirname”. .', '..') + '/'

这很好

path.join(__dirname + '/../client/index.html')
const path = require('path')
const fs = require('fs')
    fs.readFile(path.join(__dirname + '/../client/index.html'))