您可以使用下一个函数

const recursiveUpload =(路径:字符串)=> { Const paths = path.split("/")

const fullPath = paths.reduce((accumulator, current) => {
  fs.mkdirSync(accumulator)
  return `${accumulator}/${current}`
  })

  fs.mkdirSync(fullPath)

  return fullPath
}

它的作用是:

创建路径变量，它将每个路径单独存储为数组的一个元素。 在数组中每个元素的末尾添加"/"。 形成循环: 从索引从0到当前迭代的数组元素的连接中创建一个目录。基本上，它是递归的。

希望有帮助!

顺便说一下，在Node v10.12.0中，可以通过将递归路径创建作为附加参数来使用。

fs。mkdir(“/ tmp / /苹果”,{递归:真},(err) = > { 如果(错误)抛出错误; })；

https://nodejs.org/api/fs.html#fs_fs_mkdirsync_path_options

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;
 }, '');

我知道这是一个老问题，但是nodejs v10.12.0现在通过将递归选项设置为true在本地支持这个选项。fs.mkdir

// Creates /tmp/a/apple, regardless of whether `/tmp` and /tmp/a exist.
fs.mkdir('/tmp/a/apple', { recursive: true }, (err) => {
  if (err) throw err;
});

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(); }); }

像这样干净:)

function makedir(fullpath) {
  let destination_split = fullpath.replace('/', '\\').split('\\')
  let path_builder = destination_split[0]
  $.each(destination_split, function (i, path_segment) {
    if (i < 1) return true
    path_builder += '\\' + path_segment
    if (!fs.existsSync(path_builder)) {
      fs.mkdirSync(path_builder)
    }
  })
}

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);
}