How do I open data saved in a file from multiple textboxes

Question

1.50/5 (2 votes)

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So my problem is that I want to open data that are previously saved to a file from multiple textboxes, and I want when I open that file in order the data that has been saved in textboxes to show me same thing after I open, e.g. in textbox1 it is written name, textbox2 surname, textbox3 date of birth etc. so if I write in this order I want me to show, my problem is when I open it shows like this:
textbox1: NameSurnameDateofbirth, the exact same thing to all textboxes I get this after opening.

Here are the codes for saveFileDialog and OpenFileDialog:

Open File Dialog

C#

private void openToolStripMenuItem_Click(object sender, EventArgs e)
       {
           if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
           {
               label1.Text = openFileDialog1.FileName;
              textBox1.Text = File.ReadAllText(label1.Text);
               textBox2.Text = File.ReadAllText(label1.Text);
               textBox3.Text = File.ReadAllText(label1.Text);
               textBox4.Text = File.ReadAllText(label1.Text);
               textBox5.Text = File.ReadAllText(label1.Text);
               textBox6.Text = File.ReadAllText(label1.Text);
               textBox7.Text = File.ReadAllText(label1.Text);
               textBox8.Text = File.ReadAllText(label1.Text);
               textBox9.Text = File.ReadAllText(label1.Text);
           }
       }

And Save File Dialog:

C#

private void saveAsToolStripMenuItem_Click(object sender, EventArgs e)
        {
            if (saveFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                File.WriteAllText(saveFileDialog1.FileName, "");
                    File.AppendAllText(saveFileDialog1.FileName, textBox1.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox2.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox3.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox4.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox5.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox6.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox7.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox8.Text);
                    File.AppendAllText(saveFileDialog1.FileName, textBox9.Text);
            }
        }

Posted 8-Sep-15 6:54am

Medin Smaili

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Wendelius · Answer 1 · 2015-09-08T07:11:00

Solution 1

If I understand your question correctly, use ReadAllLines[^] instead of ReadAllText. This would give you an array of strings which you can use to assign correct 'lines' to corresponding text boxes.

So something like

C#

private void openToolStripMenuItem_Click(object sender, EventArgs e)
       {
          string fileName;
          string[] lines;

           if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK)
           {
               fileName = openFileDialog1.FileName;
               label1.Text = fileName;
               lines = File.ReadAllLines(fileName);
               textBox1.Text = lines[0];
               textBox2.Text = lines[1];
               textBox3.Text = lines[2];
               textBox4.Text = lines[3];
               textBox5.Text = lines[4];
               textBox6.Text = lines[5];
               textBox7.Text = lines[6];
               textBox8.Text = lines[7];
               textBox9.Text = lines[8];
           }
       }

Posted 8-Sep-15 7:11am

Wendelius

Updated 9-Sep-15 8:26am

v3

Comments

Medin Smaili 9-Sep-15 13:48pm

the problem is at "lines = File.ReadAllLines(path);" it shows error "Error CS0103 The name 'path' does not exist in the current context"

Wendelius 9-Sep-15 14:26pm

Sorry about the typo. path should be filename, the example is updated.

Medin Smaili 10-Sep-15 7:29am

System.IndexOutOfRangeException was unhandled
HResult=-2146233080
Message=Index was outside the bounds of the array.
Source=Labaratori Ikre
StackTrace:
at Labaratori_Ikre.Form2.openToolStripMenuItem_Click(Object sender, EventArgs e) in C:\Users\Medo\documents\visual studio 2015\Projects\Labaratori Ikre\Labaratori Ikre\Form2.cs:line 77
at System.Windows.Forms.ToolStripItem.RaiseEvent(Object key, EventArgs e)
at System.Windows.Forms.ToolStripMenuItem.OnClick(EventArgs e)
at System.Windows.Forms.ToolStripItem.HandleClick(EventArgs e)
at System.Windows.Forms.ToolStripItem.FireEventInteractive(EventArgs e, ToolStripItemEventType met)
at System.Windows.Forms.ToolStripItem.FireEvent(EventArgs e, ToolStripItemEventType met)
at System.Windows.Forms.ToolStripMenuItem.ProcessCmdKey(Message& m, Keys keyData)
at System.Windows.Forms.ToolStripManager.ProcessShortcut(Message& m, Keys shortcut)
at System.Windows.Forms.ToolStripManager.ProcessCmdKey(Message& m, Keys keyData)
at System.Windows.Forms.ContainerControl.ProcessCmdKey(Message& msg, Keys keyData)
at System.Windows.Forms.Form.ProcessCmdKey(Message& msg, Keys keyData)
at System.Windows.Forms.Control.ProcessCmdKey(Message& msg, Keys keyData)
at System.Windows.Forms.TextBoxBase.ProcessCmdKey(Message& msg, Keys keyData)
at System.Windows.Forms.Control.PreProcessMessage(Message& msg)
at System.Windows.Forms.Control.PreProcessControlMessageInternal(Control target, Message& msg)
at System.Windows.Forms.Application.ThreadContext.PreTranslateMessage(MSG& msg)
at System.Windows.Forms.Application.ThreadContext.System.Windows.Forms.UnsafeNativeMethods.IMsoComponent.FPreTranslateMessage(MSG& msg)
at System.Windows.Forms.Application.ComponentManager.System.Windows.Forms.UnsafeNativeMethods.IMsoComponentManager.FPushMessageLoop(IntPtr dwComponentID, Int32 reason, Int32 pvLoopData)
at System.Windows.Forms.Application.ThreadContext.RunMessageLoopInner(Int32 reason, ApplicationContext context)
at System.Windows.Forms.Application.ThreadContext.RunMessageLoop(Int32 reason, ApplicationContext context)
at System.Windows.Forms.Application.Run(Form mainForm)
at Labaratori_Ikre.Program.Main() in C:\Users\Medo\documents\visual studio 2015\Projects\Labaratori Ikre\Labaratori Ikre\Program.cs:line 19
at System.AppDomain._nExecuteAssembly(RuntimeAssembly assembly, String[] args)
at System.AppDomain.ExecuteAssembly(String assemblyFile, Evidence assemblySecurity, String[] args)
at Microsoft.VisualStudio.HostingProcess.HostProc.RunUsersAssembly()
at System.Threading.ThreadHelper.ThreadStart_Context(Object state)
at System.Threading.ExecutionContext.RunInternal(ExecutionContext executionContext, ContextCallback callback, Object state, Boolean preserveSyncCtx)
at System.Threading.ExecutionContext.Run(ExecutionContext executionContext, ContextCallback callback, Object state, Boolean preserveSyncCtx)
at System.Threading.ExecutionContext.Run(ExecutionContext executionContext, ContextCallback callback, Object state)
at System.Threading.ThreadHelper.ThreadStart()
InnerException:

Now this is what I get at "textBox2.Text = lines[0];"

Wendelius 10-Sep-15 9:25am

If you open the text file for example into Notepad, does it contain the 10 lines?

Based on the error message it seems that the file contains no lines...

Medin Smaili 11-Sep-15 13:56pm

one line and data has not been separated e.g. 123456789

Wendelius 11-Sep-15 23:54pm

Ok, so what I understood from your original post is that the values are on separate lines. I think this would be the easiest solution so why not change the code so that when you write the data, use WriteLine.

Wendelius 19-Sep-15 4:54am

What is the read and write code you currently have?

Medin Smaili 11-Nov-15 16:41pm

exactly, I want my data to be separated in new lines inside text file but after that I must open them in text boxes with those data I inserted before saving e.g. textBox1 = 1; textBox2 = 2; etc, but not show in textBox1 = 123456789;