阅读Windows窗体中使用的OpenFileDialog一个文本文件 [英] Reading a text file using OpenFileDialog in windows forms

问题描述

我是新来的的OpenFileDialog功能，但基本想通了。我需要做的就是打开一个文本文件，从文件中读取数据（纯文本），并正确地将数据放置到单独的文本框，在我的应用程序。下面是我在我的打开文件事件处理程序：

I am new to the OpenFileDialog function, but have the basics figured out. What I need to do is open a text file, read the data from the file (text only) and correctly place the data into separate text boxes in my application. Here's what I have in my "open file" event handler:

private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
    OpenFileDialog theDialog = new OpenFileDialog();
    theDialog.Title = "Open Text File";
    theDialog.Filter = "TXT files|*.txt";
    theDialog.InitialDirectory = @"C:\";
    if (theDialog.ShowDialog() == DialogResult.OK)
    {
        MessageBox.Show(theDialog.FileName.ToString());
    }
}

该文本文件我需要阅读这是（对功课，我需要阅读这个确切的文件），它有一个员工号，姓名，地址，工资和工作时间：

The text file I need to read is this (for homework, I need to read this exact file), It has an employee number, name, address, wage, and hours worked:

1
John Merryweather
123 West Main Street
5.00 30

在我得到的文本文件中，有4个员工，在相同的格式后，立即这个信息。你可以看到，员工的工资和工时都在同一行，而不是一个错字。

In the text file I was given, there are 4 more employees with info immediately after this in the same format. You can see that the employee wage and hours are on the same line, not a typo.

我有一个Employee类在这里：

I have an employee class here:

public class Employee
{
    //get and set properties for each 
    public int EmployeeNum { get; set; }
    public string Name { get; set; }
    public string Address { get; set; }
    public double Wage { get; set; }
    public double Hours { get; set; }

    public void employeeConst() //constructor method
    {
        EmployeeNum = 0;
        Name = "";
        Address = "";
        Wage = 0.0;
        Hours = 0.0;
    }

    //Method prologue
    //calculates employee earnings
    //parameters: 2 doubles, hours and wages
    //returns: a double, the calculated salary
    public static double calcSalary(double h, double w)
    {
        int OT = 40;
        double timeandahalf = 1.5;
        double FED = .20;
        double STATE = .075;
        double OThours = 0;
        double OTwage = 0;
        double OTpay = 0;
        double gross = 0; ;
        double net = 0;
        double net1 = 0;
        double net2 = 0;
        if (h > OT)
        {
            OThours = h - OT;
            OTwage = w * timeandahalf;
            OTpay = OThours * OTwage;
            gross = w * h;
            net = gross + OTpay;
        }
        else
        {
            net = w * h;
        }

        net1 = net * FED; //the net after federal taxes
        net2 = net * STATE; // the net after state taxes

        net = net - (net1 + net2);
        return net; //total net
    }
}

所以我需要拉文从该文件到我的Employee类，然后将数据输出到正确的文本框的Windows窗体应用程序。我无法理解如何做到这一点的权利。我需要用一个StreamReader？还是有另一种，在这种情况下更好的办法？谢谢

So I need to pull the text from that file into my Employee class, then output the data to the correct textbox in the windows forms application. I am having trouble understanding how to do this right. Do I need to use a streamreader? Or is there another, better way in this instance? Thank you.

推荐答案

下面是一种方法：

Stream myStream = null;
OpenFileDialog theDialog = new OpenFileDialog();
theDialog.Title = "Open Text File";
theDialog.Filter = "TXT files|*.txt";
theDialog.InitialDirectory = @"C:\";
if (theDialog.ShowDialog() == DialogResult.OK)
{
    try
    {
        if ((myStream = theDialog.OpenFile()) != null)
        {
            using (myStream)
            {
                // Insert code to read the stream here.
            }
        }
    }
    catch (Exception ex)
    {
        MessageBox.Show("Error: Could not read file from disk. Original error: " + ex.Message);
    }
}

从这里修改：的MSDN OpenFileDialog.OpenFile

修改这里是你的需要另一种方式更适合：

EDIT Here's another way more suited to your needs:

private void openToolStripMenuItem_Click(object sender, EventArgs e)
{
    OpenFileDialog theDialog = new OpenFileDialog();
    theDialog.Title = "Open Text File";
    theDialog.Filter = "TXT files|*.txt";
    theDialog.InitialDirectory = @"C:\";
    if (theDialog.ShowDialog() == DialogResult.OK)
    {
        string filename = theDialog.FileName;

        string[] filelines = File.ReadAllLines(filename);

        List<Employee> employeeList = new List<Employee>();
        int linesPerEmployee = 4;
        int currEmployeeLine = 0;
        //parse line by line into instance of employee class
        Employee employee = new Employee();
        for (int a = 0; a < filelines.Length; a++)
        {

            //check if to move to next employee
            if (a != 0 && a % linesPerEmployee == 0)
            {
                employeeList.Add(employee);
                employee = new Employee();
                currEmployeeLine = 1;
            }

            else
            {
                currEmployeeLine++;
            }
            switch (currEmployeeLine)
            {
                case 1:
                    employee.EmployeeNum = Convert.ToInt32(filelines[a].Trim());
                    break;
                case 2:
                    employee.Name = filelines[a].Trim();
                    break;
                case 3:
                    employee.Address = filelines[a].Trim();
                    break;
                case 4:
                    string[] splitLines = filelines[a].Split(' ');

                    employee.Wage = Convert.ToDouble(splitLines[0].Trim());
                    employee.Hours = Convert.ToDouble(splitLines[1].Trim());
                    break;


            }

        }
        //Test to see if it works
        foreach (Employee emp in employeeList)
        {
            MessageBox.Show(emp.EmployeeNum + Environment.NewLine +
                emp.Name + Environment.NewLine +
                emp.Address + Environment.NewLine +
                emp.Wage + Environment.NewLine +
                emp.Hours + Environment.NewLine);
        }
    }
}

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