我一直试图请求以 fs 模块递归读取目录。我一路上遇到了问题,只给了我一个文件名。以下是我需要的方式:
任何人都请帮忙。 感谢。
答案 0 :(得分:1)
这是一个递归解决方案。您可以对其进行测试,将其保存在文件中,然后运行node yourfile.js /the/path/to/traverse
。
const fs = require('fs');
const path = require('path');
const util = require('util');
const traverse = function(dir, result = []) {
// list files in directory and loop through
fs.readdirSync(dir).forEach((file) => {
// builds full path of file
const fPath = path.resolve(dir, file);
// prepare stats obj
const fileStats = { file, path: fPath };
// is the file a directory ?
// if yes, traverse it also, if no just add it to the result
if (fs.statSync(fPath).isDirectory()) {
fileStats.type = 'dir';
fileStats.files = [];
result.push(fileStats);
return traverse(fPath, fileStats.files)
}
fileStats.type = 'file';
result.push(fileStats);
});
return result;
};
console.log(util.inspect(traverse(process.argv[2]), false, null));
输出如下:
[ { file: 'index.js',
path: '/stackoverflow/test-class/index.js',
type: 'file' },
{ file: 'message.js',
path: '/stackoverflow/test-class/message.js',
type: 'file' },
{ file: 'somefolder',
path: '/stackoverflow/test-class/somefolder',
type: 'dir',
files:
[ { file: 'somefile.js',
path: '/stackoverflow/test-class/somefolder/somefile.js',
type: 'file' } ] },
{ file: 'test',
path: '/stackoverflow/test-class/test',
type: 'file' },
{ file: 'test.c',
path: '/stackoverflow/test-class/test.c',
type: 'file' } ]