Question

我有这个程序打印给定目录中的文件列表，按文件的扩展名进行筛选。我需要将它更改为一个程序，该程序搜索包含指定字符串（在文件名中）的所有文件，这些文件也在当前目录下由文件扩展名过滤。基本上我需要第一个参数作为文件名而不是目录路径，目录路径应该是当前目录而不是参数。

var fs = require('fs');
var path = require('path');
var dirPath = process.argv[2];  //directory path
var fileType = '.'+process.argv[3]; //file extension
var files = [];
fs.readdir(dirPath, function(err,list){
    if(err) throw err;
    for(var i=0; i<list.length; i++)
    {
        if(path.extname(list[i])===fileType)
        {
            console.log(list[i]); //print the file
            files.push(list[i]); //store the file name into the array files
        }
    }
});

Answer 1

我建议使用glob包。请参阅：https://github.com/isaacs/node-glob

示例：

var glob = require("glob")

// options is optional
glob("**/*.js", options, function (er, files) {
  // files is an array of filenames.
  // If the `nonull` option is set, and nothing
  // was found, then files is ["**/*.js"]
  // er is an error object or null.
})

Answer 2

将 __ dirname 用于当前目录路径

var fs = require('fs');
var path = require('path');
var dirPath = __dirname;//process.argv[2];  //directory path
var fileType = '.'+process.argv[2]; //file extension
var files = [];
fs.readdir(dirPath, function(err,list){
    if(err) throw err;
    for(var i=0; i<list.length; i++)
    {
        if(path.extname(list[i])===fileType)
        {
            console.log(list[i]); //print the file
            files.push(list[i]); //store the file name into the array files
        }
    }
});

Answer 3

try this code   
var fs = require('fs');
    var path = require('path');
    var filename = '.'+process.argv[1]; //file extension
    var fileType = '.'+process.argv[2]; //file extension
    var files = [];
    fs.readdir(process.cwd(), function(err,list){
        if(err) throw err;
        for(var i=0; i<list.length; i++)
        {
            /*user your conditions AND/OR */
            if(path.extname(list[i])===fileType && list[i].indexOf(filename) != -1)
            {
                console.log(list[i]); //print the file
                files.push(list[i]); //store the file name into the array files
            }
        }
    });

node.js按文件名搜索

3 个答案: