Question

C# OpenDialog如何获取打开文件的路径？

请问下面这个程序如何获取打开文件的全路径？谢谢！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;namespace MostlyUsedDialog{    public partial class Form1 : Form    {        OpenFileDialog openFileDialog;        public Form1()        {            InitializeComponent();        }        private void openFile_Click(object sender, EventArgs e)        {            //声明打开文件对话框            openFileDialog = new OpenFileDialog();            /**             *FileInfo myFile=new FileInfo(openFileDialog1.FileName);             *FileName= myFile.FileName;//myFile.FileName为所需无路径文件名             */            //指出文件筛选列表            openFileDialog.Filter = "文本文件(*.txt)|*.*|所有文件(*.*)|*.*";            //指出默认使用的筛选器            openFileDialog.FilterIndex = 1;            //指出恢复路径            openFileDialog.RestoreDirectory = true;            //openFileDialog.InitialDirectory = @"F:\123";            //如果点击对话框上的“打开”按钮（如果用户取消表示没必要下一步操作）            if (openFileDialog.ShowDialog(this) == DialogResult.OK)            {                string content = System.IO.File.ReadAllText(openFileDialog.FileName, Encoding.Default);                richTextBox1.AppendText(content);            }        }    }}

Answer 1

openFileDialog.FileName获取文件路径