Re: Need help to show text lines from .txt file in listbox - C# Discussion Boards

Re: Need help to show text lines from .txt file in listbox

13-Dec-12 22:41

I created a text file using StreamWriter in a Form Application.

I can't put a code to show the lines from a text file in a listbox.
How to fix this problem.

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;

namespace AplikacioniStudentet
{
    public partial class FormaStudentet : Form
    {
        public FormaStudentet()
        {
            InitializeComponent();
           
        }

        private void Form1_Load(object sender, EventArgs e)
      
        {

        }

        private void textBoxID_TextChanged(object sender, EventArgs e)
        {
            string pathi = "Studentet.txt";
            using (FileStream fs = new FileStream(pathi, FileMode.Create))
            {
                using (StreamWriter sw = new StreamWriter(fs))
                {
                    foreach (string line in textBoxID.Lines)
                        sw.Write(line + sw.NewLine);
                    sw.Close();
                }
            }
            
        }

        private void button1_Click(object sender, EventArgs e)
        {
            string pathi = "Studentet.txt";
            using (FileStream fs = new FileStream(pathi, FileMode.Create))
            {
                using (StreamWriter sw = new StreamWriter(fs))
                {
                    sw.WriteLine(textBoxID.Text + " " + textBoxEmri.Text + " " + textBoxMbiemri.Text + " ");
                    sw.Flush();
                    sw.Close();
                }
            }
           
            /*TextWriter tw = new StreamWriter (pathi, true);
            tw.ID = ID;
            tw.emri = emri;
            tw.mbiemri = mbiemri;


            tw.Close();*/
            
        }

        private void textBoxEmri_TextChanged(object sender, EventArgs e)
        {
            string pathi = "Studentet.txt";
            using (FileStream fs = new FileStream(pathi, FileMode.Create))
            {
                using (StreamWriter sw = new StreamWriter(fs))
                {
                    foreach (string line in textBoxEmri.Lines)
                        sw.Write(line + sw.NewLine);
                    sw.Close();
                }
            }
        }

        private void textBoxMbiemri_TextChanged(object sender, EventArgs e)
        {
            string pathi = "Studentet.txt";
            using (FileStream fs = new FileStream(pathi, FileMode.Create))
            {
                using (StreamWriter sw = new StreamWriter(fs))
                {
                    foreach (string line in textBoxMbiemri.Lines)
                        sw.Write(line + sw.NewLine);
                    sw.Close();
                }
            }
        }

        private void listBoxStudentat_SelectedIndexChanged(object sender, EventArgs e)
        {
            ListBox lb = new ListBox();
            System.IO.StreamReader sr = new System.IO.StreamReader("Studentet.txt");

            while (!sr.EndOfStream)
            {
                lb.Items.Add(sr.ReadLine());
            }

            sr.Close();
           
        }
    }
}

Richard MacCutchan13-Dec-12 22:48

Richard MacCutchan

13-Dec-12 22:48

You have created a new ListBox in the listBoxStudentat_SelectedIndexChanged method, but at the end of the routine you have let it get lost. You need to create a ListBox on your form and add the items to that. Are you sure you want to add items just at the point that the ListBox selection changes?

One of these days I'm going to think of a really clever signature.

Re: Need help to show text lines from .txt file in listbox

dr_iton14-Dec-12 0:10

Re: Need help to show text lines from .txt file in listbox

14-Dec-12 0:10

Yes, I want that my listbox has to show every line from text file, and also, I don't want that to get lost.

Richard MacCutchan14-Dec-12 1:59

Richard MacCutchan

14-Dec-12 1:59

Then you need to initialise it at the beginning of your program, not in the event handler for selection change. Just read the file and add each item from the file into the listbox that is on your form, don't create a new one that is not connected to anything.

One of these days I'm going to think of a really clever signature.

Re: Need help to show text lines from .txt file in listbox

dr_iton14-Dec-12 4:21

Re: Need help to show text lines from .txt file in listbox

14-Dec-12 4:21

So you are saying that I nedd to create a text file only once, and after that i can manipulate with file details in my form calling them.
Am I right or...

Richard MacCutchan14-Dec-12 4:39

Richard MacCutchan

14-Dec-12 4:39

dr_iton wrote:
So you are saying that I nedd to create a text file only once,

No, I did not discuss creating a file, I was explaining how to fill your ListBox.

dr_iton wrote:
and after that i can manipulate with file details in my form calling them.

Sorry, I have no idea what that is referring to.

Your application should be something like:

During initialisation read your text file and build the ListBox from its contents.
As information is added or changed in the application you may make changes to the ListBox.
At program termination you should save the items from the ListBox if it has been modified.
Other processing as required.

One of these days I'm going to think of a really clever signature.

Re: Need help to show text lines from .txt file in listbox

dr_iton14-Dec-12 6:10

14-Dec-12 6:10

I got the point now.
If I can not post a solution tonight because I'll be busy, tomorrow I'll tell you if I solved my problem.
Once again thank you for your reply.
Cheers.

process

messages13-Dec-12 19:04

messages

13-Dec-12 19:04

Hi everyone
could you explain about GetProcessesByName,I saw MSDN but I could'nt understand
is it true? GetProcessesByName same with GetProcessById ? GetProcessById needs to a PID and return a process but GetProcessesByName needs to process name(its really process name or application name?) and return arrays of processes

Re: process

OriginalGriff13-Dec-12 21:49

The semaphore timeout period has expired.

13-Dec-12 21:49

The MSDN text is pretty clear - you feed it the name of the executable file, and it returns al the processes that are currently running from that file:

Process[] procs = Process.GetProcessesByName("chrome");

Will give you a list of all the processes
running the Chrome browser.

If you get an email telling you that you can catch Swine Flu from tinned pork then just delete it. It's Spam.

C-P-User-313-Dec-12 8:01

C-P-User-3

13-Dec-12 8:01

What are the most likely culprits to this error ?

The semaphore timeout period has expired.

The port in question is a simple BlueTooth USB plug-in thing, which Win'7 sees properly and everything.

This is a "sometimes" error. It is not consistent, and I have not found it documented on MSDN or other sites.

This is the dozen line method I'm using...

public static void Open_ActivePortWeWillUse(String Drone_StringNameFromUser)
    {
        SerialPort TempSerialPort = new SerialPort(Drone_StringNameFromUser, (int) SerialPortSpeed);

        ThePorts.ActivePort1 = TempSerialPort;

        ThePorts.ActivePort1.DataBits = 8;
        ThePorts.ActivePort1.Parity = Parity.None;
        ThePorts.ActivePort1.StopBits = StopBits.One;

        ThePorts.ActivePort1.DataReceived += new SerialDataReceivedEventHandler(OurBackGroundSerialPortReceiver);

        ThePorts.ActivePort1.Open();  // The "Sometimes" Error occurs here
    }

modified 22-Dec-12 22:07pm.

TextChat works fine and users are added to the list but VoiceChat doesn't work

sagarDuwal13-Dec-12 7:47

sagarDuwal

13-Dec-12 7:47

I tried the same sample in VS2012 and the code works but the problem is VoiceChat doesnt work proper. Voice conference ports are not open thus the list of clients for voice conference are not added. No voice conference???? Frown | :(

Thread: Voice and Text Conferencing Library[^]

Re: TextChat works fine and users are added to the list but VoiceChat doesn't work

OriginalGriff13-Dec-12 8:15

13-Dec-12 8:15

Don't post this here - if you got the code from an article, then there is a "new message" button at the bottom of that article, which causes an email to be sent to the author. They are then alerted that you wish to speak to them.
Posting this here relies on them "dropping by" and realising it is for them.

If you get an email telling you that you can catch Swine Flu from tinned pork then just delete it. It's Spam.

txmatt13-Dec-12 5:15

txmatt

13-Dec-12 5:15

I have added a login in screen to my program which works great except if the user just closes the login window it bypasses the the screen and loads the program. Is there a way to prevent this? Here is a sample of the code:

public partial class Login : Form
    {
        public Login()
        {
            InitializeComponent();
        }

        private void Button2_Click(object sender, EventArgs e)
        {
            Application.Exit();
        }

        private void Button1_Click(object sender, EventArgs e)
        {
            if ((textBox1.Text == "puppy") && (textBox2.Text == "password"))
            {
                MessageBox.Show("Welcome Puppy", "Success",
                MessageBoxButtons.OK, MessageBoxIcon.Information);

                this.Dispose();
            }

            else
            {
                MessageBox.Show("Wrong username or password","Invalid Input",
                    MessageBoxButtons.OK,MessageBoxIcon.Error);
            }

OriginalGriff13-Dec-12 5:21

13-Dec-12 5:21

Easiest way is to have a LoggedIn property in your form. Default it to false, then set it to true when he logs in correctly.
When you get back to the main form, check it, and behave appropriately.

BTW: Don't Dispose your form from inside - use Close instead. You can't dispose it until you have finished and the there is still and instance reference in the main form!

If you get an email telling you that you can catch Swine Flu from tinned pork then just delete it. It's Spam.

XsenseX13-Dec-12 5:28

XsenseX

13-Dec-12 5:28

There are many ways to protect this.
However if you want to go for simplicity you can use a global bool for this.

Example:

public bool LoginStatus;

Then an example of usage:
!LoginStatus means basically if false.
LoginStatus means true.

So you set this bool in your login part to true if login is ok.

By using:

LoginStatus = true;

public void YourLogin()
{
   if(!LoginStatus)
   {
     //User is not logged in use a return;
     //Or whatever you wish to do with it.
   }
   else
   {
     //User is logged in continue doing your things.
   }
}

As said above this is a very simple way to do it.
There are lots ways more to do this.

OriginalGriff13-Dec-12 5:42

13-Dec-12 5:42

Never recommend public fields - always use properties. Remember that a lot of the questions (like this one) are from complete beginners, and it is best to get good habits ingrained as early as possible! Laugh | :laugh:

If you get an email telling you that you can catch Swine Flu from tinned pork then just delete it. It's Spam.

Matt U.13-Dec-12 7:00

Matt U.

13-Dec-12 7:00

The way I've always done this is by using a DialogResult from the login form. In the login form's "Button1_Click" event handler, where you have verified that the login info is correct, instead of "this.Close();" use "this.DialogResult = DialogResult.OK;" And then to use it, go to the Program.cs file. In the Main method, remove "Application.Run(new MainForm());" and replace it with:

LoginForm login = new LoginForm();
if (login.ShowDialog() == DialogResult.OK)
{
    Application.Run(new MainForm());
}

That is, unless you want the user to "log out" and see the login screen again. But in this case, if you don't need that functionality, this will show the main form ONLY if the dialog result is "OK". It eliminates the need for a property (global or otherwise). Do you see what I'm doing here or do you need more explanation?

djj55: Nice but may have a permission problem
Pete O'Hanlon: He has my permission to run it.