关于使用fs.readdir进行异步目录搜索有什么想法吗?我意识到我们可以引入递归,并调用read目录函数来读取下一个目录,但我有点担心它不是异步的…
什么好主意吗?我已经看了node-walk,它很棒,但它不能像readdir那样只给我数组中的文件。虽然
寻找这样的输出…
['file1.txt', 'file2.txt', 'dir/file3.txt']
关于使用fs.readdir进行异步目录搜索有什么想法吗?我意识到我们可以引入递归,并调用read目录函数来读取下一个目录,但我有点担心它不是异步的…
什么好主意吗?我已经看了node-walk,它很棒,但它不能像readdir那样只给我数组中的文件。虽然
寻找这样的输出…
['file1.txt', 'file2.txt', 'dir/file3.txt']
当前回答
a .看一下文件模块。它有一个叫walk的函数:
文件。步行(开始,回调) 导航文件树,为每个目录调用回调,传入 (null, dirPath, dirs, files)。
这可能是为你准备的!是的,它是异步的。但是,如果需要的话,我认为您必须自己聚合完整的路径。
B.另一种选择,甚至是我的最爱之一:使用unix find来查找。为什么要再做一件已经编程好的事情呢?也许不是你真正需要的,但仍然值得一试:
var execFile = require('child_process').execFile;
execFile('find', [ 'somepath/' ], function(err, stdout, stderr) {
var file_list = stdout.split('\n');
/* now you've got a list with full path file names */
});
Find有一个很好的内置缓存机制,使得后续搜索非常快,只要只有少数文件夹被更改。
其他回答
我必须将基于promise的砂光器库添加到列表中。
var sander = require('sander');
sander.lsr(directory).then( filenames => { console.log(filenames) } );
为了好玩,这里有一个基于流的版本,它与highland.js streams库一起工作。作者之一是维克多·伍。
###
directory >---m------> dirFilesStream >---------o----> out
| |
| |
+--------< returnPipe <-----------+
legend: (m)erge (o)bserve
+ directory has the initial file
+ dirListStream does a directory listing
+ out prints out the full path of the file
+ returnPipe runs stat and filters on directories
###
_ = require('highland')
fs = require('fs')
fsPath = require('path')
directory = _(['someDirectory'])
mergePoint = _()
dirFilesStream = mergePoint.merge().flatMap((parentPath) ->
_.wrapCallback(fs.readdir)(parentPath).sequence().map (path) ->
fsPath.join parentPath, path
)
out = dirFilesStream
# Create the return pipe
returnPipe = dirFilesStream.observe().flatFilter((path) ->
_.wrapCallback(fs.stat)(path).map (v) ->
v.isDirectory()
)
# Connect up the merge point now that we have all of our streams.
mergePoint.write directory
mergePoint.write returnPipe
mergePoint.end()
# Release backpressure. This will print files as they are discovered
out.each H.log
# Another way would be to queue them all up and then print them all out at once.
# out.toArray((files)-> console.log(files))
另一个答案,但这次使用的是TypeScript:
/** * Recursively walk a directory asynchronously and obtain all file names (with full path). * * @param dir Folder name you want to recursively process * @param done Callback function, returns all files with full path. * @param filter Optional filter to specify which files to include, * e.g. for json files: (f: string) => /.json$/.test(f) */ const walk = ( dir: string, done: (err: Error | null, results ? : string[]) => void, filter ? : (f: string) => boolean ) => { let results: string[] = []; fs.readdir(dir, (err: Error, list: string[]) => { if (err) { return done(err); } let pending = list.length; if (!pending) { return done(null, results); } list.forEach((file: string) => { file = path.resolve(dir, file); fs.stat(file, (err2, stat) => { if (stat && stat.isDirectory()) { walk(file, (err3, res) => { if (res) { results = results.concat(res); } if (!--pending) { done(null, results); } }, filter); } else { if (typeof filter === 'undefined' || (filter && filter(file))) { results.push(file); } if (!--pending) { done(null, results); } } }); }); }); };
谁想要一个公认答案的同步替代方案(我知道我做过):
var fs = require('fs');
var path = require('path');
var walk = function(dir) {
let results = [], err = null, list;
try {
list = fs.readdirSync(dir)
} catch(e) {
err = e.toString();
}
if (err) return err;
var i = 0;
return (function next() {
var file = list[i++];
if(!file) return results;
file = path.resolve(dir, file);
let stat = fs.statSync(file);
if (stat && stat.isDirectory()) {
let res = walk(file);
results = results.concat(res);
return next();
} else {
results.push(file);
return next();
}
})();
};
console.log(
walk("./")
)
这是一个简单的同步递归解决方案
const fs = require('fs')
const getFiles = path => {
const files = []
for (const file of fs.readdirSync(path)) {
const fullPath = path + '/' + file
if(fs.lstatSync(fullPath).isDirectory())
getFiles(fullPath).forEach(x => files.push(file + '/' + x))
else files.push(file)
}
return files
}
用法:
const files = getFiles(process.cwd())
console.log(files)
您可以异步地编写它,但是没有必要。只需确保输入目录存在并且可以访问。