从OpenFileDialog c#中选择它后启动进程 [英] Start process after select it from OpenFileDialog c#

问题描述

嘿
我想知道.
从OpenFileDialog中选择它后如何启动过程

Hey
I want to know .
How to start process after select it from OpenFileDialog

OpenFileDialog ofd = new OpenFileDialog();
            ofd.Title = "XXXXXXXXXXX";
            ofd.Filter = "xxxxxxxxxxxx|xxxxxx";
            if (ofd.ShowDialog() == System.Windows.Forms.DialogResult.OK)
            {
                this.textBox1.Text = ofd.FileName;
            }

它应该将Ive选择的文件的行路径放入textbox1
现在，我要创建一个按钮以启动我从openfiledialog中选择的项目.
例如:
浏览.->选定的firefox.exe
名为开始"的按钮以启动我选择的firefox.exe


有人可以帮助我吗?

It should put the line path of the file Ive selected to textbox1
Now , I want to make a button to launch that item I;ve selected from openfiledialog.
For example:
browse.. -> selected firefox.exe
button named ''Start'' to start firefox.exe that I''ve selected


can anyone help me?

推荐答案

请使用

Process.start(Firefox.exe);

正如Dominic所说的，您应该使用Process.Start方法.

您可以对按钮单击事件进行编码，例如:

As Dominic said you should Process.Start method.

You can code to your button click event such as:

Process.Start(textbox1.Text);

以下是一些适合您的详细链接:
MSDN Process.Start方法 [ ^ ]
如何使用此方法的示例 [ ^ ]

祝你好运，

OI

Here are some detailed links for you:
MSDN Process.Start Method[^]
Examples of how you can use this method[^]

Good luck,

OI

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