我试图创建一个完整的路径,如果它不存在。

代码如下所示:

var fs = require('fs');
if (!fs.existsSync(newDest)) fs.mkdirSync(newDest); 

只要只有一个子目录(像'dir1'这样的newDest),这段代码就能很好地工作,但是当有一个目录路径('dir1/dir2')时,它就会失败 错误:ENOENT,没有这样的文件或目录

我希望能够用尽可能少的代码行创建完整的路径。

我读到fs上有一个递归选项,并尝试了这样做

var fs = require('fs');
if (!fs.existsSync(newDest)) fs.mkdirSync(newDest,'0777', true);

我觉得递归地创建一个不存在的目录应该这么简单。我是否遗漏了一些东西,或者我是否需要解析路径并检查每个目录,如果它不存在,则创建它?

我对Node很陌生。也许我使用的是旧版本的FS?


当前回答

Windows的示例(没有额外的依赖和错误处理)

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

let dir = "C:\\temp\\dir1\\dir2\\dir3";

function createDirRecursively(dir) {
    if (!fs.existsSync(dir)) {        
        createDirRecursively(path.join(dir, ".."));
        fs.mkdirSync(dir);
    }
}

createDirRecursively(dir); //creates dir1\dir2\dir3 in C:\temp

其他回答

const fs = require('fs');

try {
    fs.mkdirSync(path, { recursive: true });
} catch (error) {
    // this make script keep running, even when folder already exist
    console.log(error);
}

递归创建目录的异步方法:

import fs from 'fs'

const mkdirRecursive = function(path, callback) {
  let controlledPaths = []
  let paths = path.split(
    '/' // Put each path in an array
  ).filter(
    p => p != '.' // Skip root path indicator (.)
  ).reduce((memo, item) => {
    // Previous item prepended to each item so we preserve realpaths
    const prevItem = memo.length > 0 ? memo.join('/').replace(/\.\//g, '')+'/' : ''
    controlledPaths.push('./'+prevItem+item)
    return [...memo, './'+prevItem+item]
  }, []).map(dir => {
    fs.mkdir(dir, err => {
      if (err && err.code != 'EEXIST') throw err
      // Delete created directory (or skipped) from controlledPath
      controlledPaths.splice(controlledPaths.indexOf(dir), 1)
      if (controlledPaths.length === 0) {
        return callback()
      }
    })
  })
}

// Usage
mkdirRecursive('./photos/recent', () => {
  console.log('Directories created succesfully!')
})

我对fs的递归选项有问题。mkdir,所以我做了一个函数,做以下工作:

Creates a list of all directories, starting with the final target dir and working up to the root parent. Creates a new list of needed directories for the mkdir function to work Makes each directory needed, including the final function createDirectoryIfNotExistsRecursive(dirname) { return new Promise((resolve, reject) => { const fs = require('fs'); var slash = '/'; // backward slashes for windows if(require('os').platform() === 'win32') { slash = '\\'; } // initialize directories with final directory var directories_backwards = [dirname]; var minimize_dir = dirname; while (minimize_dir = minimize_dir.substring(0, minimize_dir.lastIndexOf(slash))) { directories_backwards.push(minimize_dir); } var directories_needed = []; //stop on first directory found for(const d in directories_backwards) { if(!(fs.existsSync(directories_backwards[d]))) { directories_needed.push(directories_backwards[d]); } else { break; } } //no directories missing if(!directories_needed.length) { return resolve(); } // make all directories in ascending order var directories_forwards = directories_needed.reverse(); for(const d in directories_forwards) { fs.mkdirSync(directories_forwards[d]); } return resolve(); }); }

使用reduce,我们可以验证每个路径是否存在,并在必要时创建它,而且我认为这样更容易遵循。编辑,谢谢@Arvin,我们应该使用路径。Sep来获得适当的平台特定路径段分隔符。

const path = require('path');

// Path separators could change depending on the platform
const pathToCreate = 'path/to/dir'; 
pathToCreate
 .split(path.sep)
 .reduce((prevPath, folder) => {
   const currentPath = path.join(prevPath, folder, path.sep);
   if (!fs.existsSync(currentPath)){
     fs.mkdirSync(currentPath);
   }
   return currentPath;
 }, '');

我用这种方法解决了这个问题——类似于其他递归的答案,但对我来说,这更容易理解和阅读。

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

function mkdirRecurse(inputPath) {
  if (fs.existsSync(inputPath)) {
    return;
  }
  const basePath = path.dirname(inputPath);
  if (fs.existsSync(basePath)) {
    fs.mkdirSync(inputPath);
  }
  mkdirRecurse(basePath);
}