C# forms: how do I print out a CSV file which is selected using a different button

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Goal: Direct print a CSV file which is selected from another button on my forms application

Problem: I don't know how to tell amy btnPrintFile_click method the FileName coming from another method in form1.cs

Could someone help me? Im new to forms in c#

Code

public void openCVSFile(object sender, EventArgs e)
 {
     OpenFileDialog ofd = new OpenFileDialog();
     ofd.Multiselect = false;
     ofd.Filter = "CSV files (*.csv)|*.csv";
     ofd.FilterIndex = 1;
     if(ofd.ShowDialog() == DialogResult.OK)
     {
         txtAddressCount.Text = ("Address count: "+ ofd.FileName);
     }
 }

 private void btnPrintFile_Click(object sender, EventArgs e)
 {
     try
     {
         streamToPrint = new StreamReader(ofd.FileName);
         try
         {
             printFont = new Font("Arial", 10);
             PrintDocument pd = new PrintDocument();
             pd.PrintPage += new PrintPageEventHandler
                (this.pd_PrintPage);
             pd.Print();
         }
         finally
         {
             streamToPrint.Close();
         }
     }
     catch (Exception ex)
     {
         MessageBox.Show(ex.Message);
     }
 }

What I have tried:

I tried making openCVSFile method return the filename but that doesnt work

Posted 7-Apr-22 19:11pm

Daniel Tsang

Updated 7-Apr-22 19:59pm

v2

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OriginalGriff · Accepted Answer · 2022-04-07T19:59:00

Create a class level private variable, and put the path in there:

private string pathToFile = null;
public void openCVSFile(object sender, EventArgs e)
 {
     OpenFileDialog ofd = new OpenFileDialog();
     ofd.Multiselect = false;
     ofd.Filter = "CSV files (*.csv)|*.csv";
     ofd.FilterIndex = 1;
     if(ofd.ShowDialog() == DialogResult.OK)
     {
         txtAddressCount.Text = ("Address count: "+ ofd.FileName);
         pathToFile = ofd.FileName;
     }
 }

 private void btnPrintFile_Click(object sender, EventArgs e)
 {
     if (pathToFile == null) return;
     try
     {
         streamToPrint = new StreamReader(pathToFile);
         try
         {
             printFont = new Font("Arial", 10);
             PrintDocument pd = new PrintDocument();
             pd.PrintPage += new PrintPageEventHandler
                (this.pd_PrintPage);
             pd.Print();
         }
         finally
         {
             streamToPrint.Close();
         }
     }
     catch (Exception ex)
     {
         MessageBox.Show(ex.Message);
     }
 }