在 nodejs 中的文件夹下按扩展名查找文件 *.html

find files by extension, *.html under a folder in nodejs

提问by Nicolas S.Xu

I'd like to find all *.html files in src folder and all its sub folders using nodejs. What is the best way to do it?

我想使用 nodejs 在 src 文件夹及其所有子文件夹中找到所有 *.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);

I think a lot developers should have great and tested solution and it is better to use it than writing one myself.

我认为很多开发人员都应该拥有出色且经过测试的解决方案，并且使用它比自己编写一个更好。

回答by Lucio M. Tato

node.js, recursive simple function:

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');

add RegExp if you want to get fancy, and a callback to make it generic.

如果你想要花哨，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);
});

回答by David Cheung

i like using the globpackage:

我喜欢使用glob包：

const glob = require('glob');

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

回答by Master James

What, hang on?! ... Okay ya, maybe this makes more sense to someones else too.

什么，坚持？！...好吧，也许这对其他人也更有意义。

[nodejs 7mind you]

[ nodejs 7 请注意]

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

Do whatever with regex make it an argument you set in the function with a default etc.

使用正则表达式执行任何操作，使其成为您在函数中使用默认值等设置的参数。

回答by Nicolas S.Xu

Based on Lucio's code, I made a module. It will return an away with all the files with specific extensions under the one. Just post it here in case anybody needs it.

基于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。把它贴在这里以防有人需要它。

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;

回答by nickool

You can use Filehoundto do this.

您可以使用Filehound来执行此操作。

For example: find all .html files in /tmp:

例如：在 /tmp 中查找所有 .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);
});

For further information (and examples), check out the docs: https://github.com/nspragg/filehound

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

Disclaimer: I'm the author.

免责声明：我是作者。

回答by Netsi1964

I have looked at the above answers and have mixed together this version which works for me:

我已经查看了上述答案，并将这个对我有用的版本混合在一起：

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

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

This test will return an array of filenames from the files found in the folder at the path ./testdata. Working on node version 8.11.3.

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

回答by Nikhil

The following code does a recursive search inside ./ (change it appropriately) and returns an array of absolute file names ending with .html

以下代码在 ./ 中进行递归搜索（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);

回答by Emil Condrea

You can use OS help for this. Here is a cross-platform solution:

为此，您可以使用操作系统帮助。这是一个跨平台的解决方案：

1. The bellow function uses lsand dirand does not search recursively but it has relative paths

1.波纹管函数使用lsanddir和，不递归搜索但有相对路径

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. The bellow function uses findand dir, searches recursively but on windows it has absolute paths

2. 波纹管函数使用findand 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);
})

回答by Akash Babu

Take a look into file-regex

看看文件正则表达式

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

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

This above snippet would print all the jsfiles in the current directory.

上面的代码段将打印js当前目录中的所有文件。

回答by jset74

my two pence, using map in place of for-loop

我的两便士，使用地图代替 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);