我怎么写这个回到父2层去找文件?
fs.readFile(__dirname + 'foo.bar');
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'))