打开文件所在位置 [英] Open file location

问题描述

在Windows资源管理器中搜索文件时，右键单击搜索结果中的文件；有一个选项:打开文件位置".我想在我的C#WinForm中实现相同的功能.我是这样做的:

When searching a file in Windows Explorer and right-click a file from the search results; there is an option: "Open file location". I want to implement the same in my C# WinForm. I did this:

if (File.Exists(filePath)
{
    openFileDialog1.InitialDirectory = new FileInfo(filePath).DirectoryName;
    openFileDialog1.ShowDialog();
}

还有更好的方法吗?

推荐答案

如果openFileDialog_View是

If openFileDialog_View is an OpenFileDialog then you'll just get a dialog prompting a user to open a file. I assume you want to actually open the location in explorer.

您可以这样做:

if (File.Exists(filePath))
{
    Process.Start("explorer.exe", filePath);
}

---

要选择，文件explorer.exe采用这样的/select参数:

---

To select a file explorer.exe takes a /select argument like this:

explorer.exe /select, <filelist>

我是从SO帖子中获得的:在资源管理器中打开一个文件夹并选择一个文件

I got this from an SO post: Opening a folder in explorer and selecting a file

因此您的代码应为:

if (File.Exists(filePath))
{
    Process.Start("explorer.exe", "/select, " + filePath);
}

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