在nodejs中的文件夹下按扩展名*.html查找文件

Question

我想使用 nodejs 查找 src 文件夹及其所有子文件夹中的所有 *.html 文件。最好的方法是什么？

var folder = '/project1/src';
var extension = 'html';
var cb = function(err, results) {
   // results is an array of the files with path relative to the folder
   console.log(results);

}
// This function is what I am looking for. It has to recursively traverse all sub folders. 
findFiles(folder, extension, cb);

我认为很多开发人员都应该有很棒且经过测试的解决方案，使用它比自己编写一个更好。

Answer 1

node.js，递归简单函数：

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

function fromDir(startPath,filter){

    //console.log('Starting from dir '+startPath+'/');

    if (!fs.existsSync(startPath)){
        console.log("no dir ",startPath);
        return;
    }

    var files=fs.readdirSync(startPath);
    for(var i=0;i<files.length;i++){
        var filename=path.join(startPath,files[i]);
        var stat = fs.lstatSync(filename);
        if (stat.isDirectory()){
            fromDir(filename,filter); //recurse
        }
        else if (filename.indexOf(filter)>=0) {
            console.log('-- found: ',filename);
        };
    };
};

fromDir('../LiteScript','.html');

如果你想变得花哨，请添加 RegExp，并添加一个回调以使其通用。

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

function fromDir(startPath,filter,callback){

    //console.log('Starting from dir '+startPath+'/');

    if (!fs.existsSync(startPath)){
        console.log("no dir ",startPath);
        return;
    }

    var files=fs.readdirSync(startPath);
    for(var i=0;i<files.length;i++){
        var filename=path.join(startPath,files[i]);
        var stat = fs.lstatSync(filename);
        if (stat.isDirectory()){
            fromDir(filename,filter,callback); //recurse
        }
        else if (filter.test(filename)) callback(filename);
    };
};

fromDir('../LiteScript',/\.html$/,function(filename){
    console.log('-- found: ',filename);
});

Answer 2

我喜欢使用glob 包：

const glob = require('glob');

glob(__dirname + '/**/*.html', {}, (err, files)=>{
  console.log(files)
})

Answer 3

什么，等一下？！ ...好吧，也许这对其他人也更有意义。

[nodejs 7请注意]

fs = import('fs');
let dirCont = fs.readdirSync( dir );
let files = dirCont.filter( function( elm ) {return elm.match(/.*\.(html?)/ig);});

用正则表达式做任何事情，让它成为你在函数中设置的参数，并使用默认值等。

Answer 4

根据 Lucio 的代码，我做了一个模块。它将返回一个带有特定扩展名的所有文件。把它贴在这里以防万一有人需要。

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


/**
 * Find all files recursively in specific folder with specific extension, e.g:
 * findFilesInDir('./project/src', '.html') ==> ['./project/src/a.html','./project/src/build/index.html']
 * @param  {String} startPath    Path relative to this file or other file which requires this files
 * @param  {String} filter       Extension name, e.g: '.html'
 * @return {Array}               Result files with path string in an array
 */
function findFilesInDir(startPath,filter){

    var results = [];

    if (!fs.existsSync(startPath)){
        console.log("no dir ",startPath);
        return;
    }

    var files=fs.readdirSync(startPath);
    for(var i=0;i<files.length;i++){
        var filename=path.join(startPath,files[i]);
        var stat = fs.lstatSync(filename);
        if (stat.isDirectory()){
            results = results.concat(findFilesInDir(filename,filter)); //recurse
        }
        else if (filename.indexOf(filter)>=0) {
            console.log('-- found: ',filename);
            results.push(filename);
        }
    }
    return results;
}

module.exports = findFilesInDir;

Answer 5

您可以使用 Filehound 来执行此操作。

例如：查找/tmp中的所有.html文件：

const Filehound = require('filehound');

Filehound.create()
  .ext('html')
  .paths("/tmp")
  .find((err, htmlFiles) => {
    if (err) return console.error("handle err", err);

    console.log(htmlFiles);
});

如需更多信息（和示例），请查看文档： https://github.com/nspragg/filehound

免责声明：我是作者。

Answer 6

我已经查看了上述答案，并将这个适合我的版本混合在一起：

function getFilesFromPath(path, extension) {
    let files = fs.readdirSync( path );
    return files.filter( file => file.match(new RegExp(`.*\.(${extension})`, 'ig')));
}

console.log(getFilesFromPath("./testdata", ".txt"));

此测试将从路径 ./testdata 的文件夹中找到的文件返回文件名数组。在节点版本 8.11.3 上工作。

Answer 7

以下代码在 ./ 内进行递归搜索（适当更改）并返回以 .html 结尾的绝对文件名数组

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

var searchRecursive = function(dir, pattern) {
  // This is where we store pattern matches of all files inside the directory
  var results = [];

  // Read contents of directory
  fs.readdirSync(dir).forEach(function (dirInner) {
    // Obtain absolute path
    dirInner = path.resolve(dir, dirInner);

    // Get stats to determine if path is a directory or a file
    var stat = fs.statSync(dirInner);

    // If path is a directory, scan it and combine results
    if (stat.isDirectory()) {
      results = results.concat(searchRecursive(dirInner, pattern));
    }

    // If path is a file and ends with pattern then push it onto results
    if (stat.isFile() && dirInner.endsWith(pattern)) {
      results.push(dirInner);
    }
  });

  return results;
};

var files = searchRecursive('./', '.html'); // replace dir and pattern
                                                // as you seem fit

console.log(files);

Answer 8

您可以为此使用操作系统帮助。这是一个跨平台的解决方案：

1。波纹管函数使用ls 和dir 并且不递归搜索但它具有相对路径

var exec = require('child_process').exec;
function findFiles(folder,extension,cb){
    var command = "";
    if(/^win/.test(process.platform)){
        command = "dir /B "+folder+"\\*."+extension;
    }else{
        command = "ls -1 "+folder+"/*."+extension;
    }
    exec(command,function(err,stdout,stderr){
        if(err)
            return cb(err,null);
        //get rid of \r from windows
        stdout = stdout.replace(/\r/g,"");
        var files = stdout.split("\n");
        //remove last entry because it is empty
        files.splice(-1,1);
        cb(err,files);
    });
}

findFiles("folderName","html",function(err,files){
    console.log("files:",files);
})

2。波纹管函数使用find和dir，递归搜索但在windows上它有绝对路径

var exec = require('child_process').exec;
function findFiles(folder,extension,cb){
    var command = "";
    if(/^win/.test(process.platform)){
        command = "dir /B /s "+folder+"\\*."+extension;
    }else{
        command = 'find '+folder+' -name "*.'+extension+'"'
    }
    exec(command,function(err,stdout,stderr){
        if(err)
            return cb(err,null);
        //get rid of \r from windows
        stdout = stdout.replace(/\r/g,"");
        var files = stdout.split("\n");
        //remove last entry because it is empty
        files.splice(-1,1);
        cb(err,files);
    });
}

findFiles("folder","html",function(err,files){
    console.log("files:",files);
})

Answer 9

由于声誉问题无法添加评论，但请注意以下几点：

使用 fs.readdir 或 node-glob 在包含 500,000 个文件的文件夹中查找通配符文件集大约需要 2 秒。 将 exec 与 DIR 一起使用需要 ~0.05s（非递归）或 ~0.45s（递归）。 （我在单个目录中寻找与我的模式匹配的 ~14 个文件）。

到目前为止，我还没有找到任何使用低级操作系统通配符搜索效率的 nodejs 实现。但就效率而言，上述基于 DIR/ls 的代码在 windows 中工作得非常好。但是，对于大型目录，linux 会找到 will likely be very slow。

Answer 10

看看file-regex

let findFiles = require('file-regex')
let pattern = '\.js'

findFiles(__dirname, pattern, (err, files) => {  
   console.log(files);
})

上面的 sn-p 将打印当前目录中的所有js 文件。

Answer 11

安装

你可以通过walk-sync安装这个包

yarn add walk-sync

用法

const walkSync = require("walk-sync");
const paths = walkSync("./project1/src", {globs: ["**/*.html"]});
console.log(paths);   //all html file path array

Answer 12

我的两便士，用 map 代替 for 循环

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

var findFiles = function(folder, pattern = /.*/, callback) {
  var flist = [];

  fs.readdirSync(folder).map(function(e){ 
    var fname = path.join(folder, e);
    var fstat = fs.lstatSync(fname);
    if (fstat.isDirectory()) {
      // don't want to produce a new array with concat
      Array.prototype.push.apply(flist, findFiles(fname, pattern, callback)); 
    } else {
      if (pattern.test(fname)) {
        flist.push(fname);
        if (callback) {
          callback(fname);
        }
      }
    }
  });
  return flist;
};

// HTML files   
var html_files = findFiles(myPath, /\.html$/, function(o) { console.log('look what we have found : ' + o} );

// All files
var all_files = findFiles(myPath);

Answer 13

我刚刚注意到，您正在使用同步 fs 方法，这可能会阻止您的应用程序，这是一种使用 async 和 q 的基于 Promise 的异步方式，您可以执行使用 START=/myfolder FILTER=".jpg" 节点 myfile.js，假设您将以下代码放在名为 myfile.js 的文件中：

Q = require("q")
async = require("async")
path = require("path")
fs = require("fs")

function findFiles(startPath, filter, files){
    var deferred;
    deferred = Q.defer(); //main deferred

    //read directory
    Q.nfcall(fs.readdir, startPath).then(function(list) {
        var ideferred = Q.defer(); //inner deferred for resolve of async each
        //async crawling through dir
        async.each(list, function(item, done) {

            //stat current item in dirlist
            return Q.nfcall(fs.stat, path.join(startPath, item))
                .then(function(stat) {
                    //check if item is a directory
                    if (stat.isDirectory()) {
                        //recursive!! find files in subdirectory
                        return findFiles(path.join(startPath, item), filter, files)
                            .catch(function(error){
                                console.log("could not read path: " + error.toString());
                            })
                            .finally(function() {
                                //resolve async job after promise of subprocess of finding files has been resolved
                                return done();
                             });
                    //check if item is a file, that matches the filter and add it to files array
                    } else if (item.indexOf(filter) >= 0) {
                        files.push(path.join(startPath, item));
                        return done();
                    //file is no directory and does not match the filefilter -> don't do anything
                    } else {
                        return done();
                    }
                })
                .catch(function(error){
                    ideferred.reject("Could not stat: " + error.toString());
                });
        }, function() {
            return ideferred.resolve(); //async each has finished, so resolve inner deferred
        });
        return ideferred.promise;
    }).then(function() {
        //here you could do anything with the files of this recursion step (otherwise you would only need ONE deferred)
        return deferred.resolve(files); //resolve main deferred
    }).catch(function(error) {
        deferred.reject("Could not read dir: " + error.toString());
        return
    });
    return deferred.promise;
}


findFiles(process.env.START, process.env.FILTER, [])
    .then(function(files){
        console.log(files);
    })
    .catch(function(error){
        console.log("Problem finding files: " + error);
})

Answer 14

您可以编辑此代码以适合您的意图。我为 nodejs IO 操作使用了同步版本，以便在 node 继续执行下一行代码之前返回结果：

const fs = require('fs');
const path = require('path');
    
// Path to the directory(folder) to look into
const dirPath = path.resolve(`${__dirname}../../../../../tests_output`);
        
// Read all files with .html extension in the specified folder above
const filesList = fs.readdirSync(dirPath, (err, files) => files.filter((e) => path.extname(e).toLowerCase() === '.html'));
        
// Read the content of the first file with .txt extension in the folder
const data = fs.readFileSync(path.resolve(`${__dirname}../../../../../tests_output/${filesList[0]}`), 'utf8');

res.writeHead(200, { 'Content-Type': 'text/html' });
res.write(data);
return res.end();

Answer 15

对于无数可能的解决方案，我们还可以添加非常适合构建脚本目的的 fs-jetpack 库。

const jetpack = require("fs-jetpack");

// the sync way
const files = jetpack.find("my_project", { matching: "*.html" });
console.log(files);

// or the async way
jetpack.findAsync("my_project", { matching: "*.html" }).then(files => {
  console.log(files);
});

Answer 16

旧帖子，但 ES6 现在使用 includes 方法开箱即用地处理这个问题。

let files = ['file.json', 'other.js'];

let jsonFiles = files.filter(file => file.includes('.json'));

console.log("Files: ", jsonFiles) ==> //file.json