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How to Count File Extensions and Group it using LINQ?

+1 vote
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How to Count File Extensions and Group it using LINQ?
posted Jun 5, 2017 by Rohini Agarwal

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1 Answer

+1 vote

This C# Program Counts File Extensions and Group it using LINQ. Here a service reads files generated in a folder every hour and returns a string array containing the file names and showes the count of files grouped by the file extension.

Here is source code of the C# Program to Count File Extensions and Group it using LINQ. The C# program is successfully compiled and executed with Microsoft Visual Studio. The program output is also shown below.

/*
 * C# Program to Count File Extensions and Group it using LINQ
 */
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
namespace ConsoleApplication9
{
    class Program
    {
        public static void Main()
        {
            string[] arr = { "aaa.txt", "bbb.TXT", "xyz.abc.pdf", "aaaa.PDF", "abc.xml", "ccc.txt", "zzz.txt" };
            var egrp = arr.Select(file => Path.GetExtension(file).TrimStart('.').ToLower())
                     .GroupBy(x => x,(ext, extCnt) =>new
                                                     {
                                                        Extension = ext,
                                                        Count = extCnt.Count()
                                                      });

            foreach (var v in egrp)
                Console.WriteLine("{0} File(s) with {1} Extension ",v.Count, v.Extension);
            Console.ReadLine();
        }
    }
}

Here is the output of the C# Program:

4 File(s) with txt Extension
2 File(s) with pdf Extension
1 File(s) with xml Extension

source :http://www.sanfoundry.com/csharp-program-count-file-extensions-linq/
Many answers are just a google away, Have a good day :)

answer Jun 6, 2017 by Andrews Wilfred Selvin
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+2 votes

Develop a C# Windows Form Application that allows users to do all of the following.
Read a List of patient's information from a text file (*.txt), the input text file is selected from the Open File Dialog.
Your program should accept any text file with the following format:
a. The file contains columns with Basic information about patients.
b. Columns may be separated by spaces and/or tabs.
c. The first line in the file is the column header.
d. After the header, each line represents a Patient (name, address, phone#, Bdate, gander ,wheight, height and blood type).
e. Successfully imported patients are displayed on a data grid view in a second form and added to a patient list.

  1. The user will be able to display on The Grid view :
    a) Female patients.
    b) Patients with age<45 year. c) Save Over weighted patients on a text file .  Note: To find over weighted patients you have to calculate the BMI value. BMI=Weight/ (Height*Height). If BMI <18.5 >>>>> under weighted.
    18.5<=BMI<=25 >>>>>>>Normal.
    BMI>25 >>>>>>>>>>>> Over Weighted.
+1 vote

If write this code in form 1 :

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.IO;
using System.Xml.Serialization;

namespace PostLab5_**********
{
    public partial class Form1 : Form
    {
        List<StudentInfo> Stu_Info = new List<StudentInfo>();
        Queue<StudentInfo> pass = new Queue<StudentInfo>();
        Queue<StudentInfo> fail = new Queue<StudentInfo>();
        Stack<StudentInfo> Excelent = new Stack<StudentInfo>();
        Stack<StudentInfo> VeryGood = new Stack<StudentInfo>();
        Stack<StudentInfo> Good = new Stack<StudentInfo>();
        public Form1()
        {
            InitializeComponent();
        }

        private void importXMLToolStripMenuItem_Click(object sender, EventArgs e)
        {

            OpenFileDialog Dlg = new OpenFileDialog();
            Dlg.Filter = "XML File|*.xml";
            if (Dlg.ShowDialog() == DialogResult.OK)
            {
                StreamReader Infile = new StreamReader(Dlg.FileName);
                XmlSerializer Des = new XmlSerializer(typeof(List<StudentInfo>));
                Stu_Info = (List<StudentInfo>)Des.Deserialize(Infile);
                Infile.Close();
                dataGridView1.DataSource = Stu_Info;
            }
        }

        private void passedToolStripMenuItem_Click(object sender, EventArgs e)
        {
            var R = from Item in Stu_Info
                    where Item.total >= 50
                    select Item;
            dataGridView1.DataSource = R.ToArray();
            foreach (StudentInfo item in R)
            {
                pass.Enqueue(item);
            }
        }

        private void failedToolStripMenuItem_Click(object sender, EventArgs e)
        {
            var R = from Item in Stu_Info
                    where Item.total < 50
                    select Item;
            dataGridView1.DataSource = R.ToArray();
            foreach (StudentInfo item in R)
            {
                fail.Enqueue(item);
            }
        }

        private void excellentToolStripMenuItem_Click(object sender, EventArgs e)
        {
            var R = from Item in Stu_Info
                    where Item.total >= 84
                    select Item;
            dataGridView1.DataSource = R.ToArray();
            foreach (StudentInfo Item in R)
            {
                Excelent.Push(Item);
            }
        }

        private void veryGoodToolStripMenuItem_Click(object sender, EventArgs e)
        {
            var R = from Item in Stu_Info
                        where Item.total >=76 && Item.total <84
                        select Item;
                dataGridView1.DataSource = R.ToArray();
                foreach (StudentInfo Item in R)
                {
                    VeryGood.Push(Item);
                }
            }

        private void goodToolStripMenuItem_Click(object sender, EventArgs e)
        {
             var R = from Item in Stu_Info
                        where Item.total >= 68 && Item.total < 76
                        select Item;
                dataGridView1.DataSource = R.ToArray();
                foreach (StudentInfo Item in R)
                {
                    Good.Push(Item);
                }
        }

        private void countOfVeryGoodToolStripMenuItem_Click(object sender, EventArgs e)
        {
            MessageBox.Show(VeryGood.Count().ToString());
        }

        private void peekOfPeekToolStripMenuItem_Click(object sender, EventArgs e)
        {
            MessageBox.Show(Good.Peek().name);
        }

        private void removeItemFromFailedToolStripMenuItem_Click(object sender, EventArgs e)
        {
            MessageBox.Show(fail.Dequeue().name);
        }

        private void countOfPassedToolStripMenuItem_Click(object sender, EventArgs e)
        {
            MessageBox.Show(pass.Count().ToString());
        }

        private void button1_Click(object sender, EventArgs e)
        {
            Form2 f2 = new Form2();

        }
        }
    }

and in form2 contain textboxes to add new item for these list ! how use this !

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