单击列表框中的项目时如何打开文件 [英] How to open a file on click of item in List Box

问题描述

我想在列表框中单击项目时打开一个特定文件..谢谢...

i want to open a particular file on click of item in List box..thanks...

推荐答案

我建​​议您在.

做类似的事情，

I would suggest you to do it on DoubleClick.

Do something like,

switch (ListBox1.SelectedIndex) 
{
	case 0:
		//open file 1
		//Process.Start("notepad");
		break;
	case 1:
		//open file 2
		//Process.Start("calc");
		break;
	default:
		break;
	//do nothing
}

这样做:

正如Prera​​k所说，请在DoubleClick事件中执行以下操作.

将此添加到您的构造函数中:

Do it like this :

As said by Prerak, do the below in a DoubleClick event.

Add this to your constructor :

listBox2.MouseDoubleClick += new MouseButtonEventHandler(listBox2_MouseDoubleClick);

这是实现它的功能:

This is the function to implement it:

void listBox2_MouseDoubleClick(object sender, MouseButtonEventArgs e)
{
    ListBoxItem item =(ListBoxItem)listBox2.SelectedItem;
    Process.Start(item.Content.ToString());
}

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