我怎么写这个回到父2层去找文件?

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

当前回答

看起来你需要路径模块。(路径。特别是正常化)

var path = require("path"),
    fs = require("fs");

fs.readFile(path.normalize(__dirname + "/../../foo.bar"));

其他回答

如果另一个模块调用你的,你仍然想知道正在运行的主文件的位置,你可以使用@Jason的代码修改:

var path = require('path'),
    __parentDir = path.dirname(process.mainModule.filename);

fs.readFile(__parentDir + '/foo.bar');

通过这种方式,您将获得实际运行的脚本的位置。

你可以使用

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

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'));

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'))

试试这个:

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

请注意相对路径开头的正斜杠。