如何制作OpenFileLocation按钮 [英] How to make OpenFileLocation button

问题描述

嘿

我想知道如何构建OpenFileLocation按钮.
我的意思是，选择浏览"后，ofd会显示出来.我选择.exe或其他内容，然后将其打印到textbox1.text

Hey

I want to know how to build OpenFileLocation button.
I mean , after I select browse , ofd shows up. I select the .exe or whatever , then it prints to textbox1.text

OpenFileDialog ofd = new OpenFileDialog();
            
            
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                this.textBox1.Text = ofd.FileName;
                
            }

我创建了这个

and I created this

Process.Start(textbox1.Text)

但是，我想创建一个按钮来打开文件夹，该文件夹找到我从浏览按钮中选择的exe.
像，打开文件位置按钮
我试过了

but , I want to create a button to open the folder which locates the exe i selected from the browse button.
Like , Open file location button
I tried this

if (ofd.ShowDialog() == DialogResult.OK)
            {
                textBox1.Text = OpenFileDialog;
            }

但是，它没有用.

But , it didnt worked.

推荐答案

使用FolderBrowserDialog类

Use the FolderBrowserDialog class

if (ofd.ShowDialog() == DialogResult.OK)
{
   FolderBrowserDialog dialog = new FolderBrowserDialog();
   dialog.SelectedPath = Path.GetDirectoryName(ofd.FileName);
   dialog.ShowDialog();
   textBox1.Text = dialog.SelectedPath;
}

这篇关于如何制作OpenFileLocation按钮的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！