打开特定文件夹，然后从该文件夹加载选定文件 [英] Open specific folder and than load a selcted file from that folder

问题描述

嗨



i想要打开一个特定文件夹，点击打开按钮并在该文件夹中.XML扩展文件应该只过滤并显示，之后我可以选择任何.xml文件和加载。



请帮助这个..

i需要获胜形式。

hi

i want to open a specifc folder clicking on open button and in that folder .XML extension file should be filter and show only and after than i can select any of the .xml file and load.

plase help on this..
i need thi in win form.

推荐答案

using System;
using System.Windows.Forms;

namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
        // Show the dialog and get result.
        DialogResult result = openFileDialog1.ShowDialog();
        if (result == DialogResult.OK) // Test result.
        {
        }
        Console.WriteLine(result); // <-- For debugging use.
    }
    }
}

和diaolog box过滤器物业

过滤=XML | * .xml

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