C# 打开文件所在位置
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Open file location
提问by Haroon A.
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:
在 Windows 资源管理器中搜索文件并右键单击搜索结果中的文件时；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。我这样做了：
if (File.Exists(filePath)
{
openFileDialog1.InitialDirectory = new FileInfo(filePath).DirectoryName;
openFileDialog1.ShowDialog();
}
Is there any better way to do it?
有没有更好的方法来做到这一点？
采纳答案by gideon
If openFileDialog_Viewis an OpenFileDialogthen you'll just get a dialog prompting a user to open a file. I assume you want to actually openthe location in explorer.
如果openFileDialog_View是OpenFileDialog，那么您只会得到一个对话框，提示用户打开文件。我假设您想在资源管理器中实际打开该位置。
You would do this:
你会这样做：
if (File.Exists(filePath))
{
Process.Start("explorer.exe", filePath);
}
To selecta file explorer.exetakes a /selectargument like this:
要选择一个文件explorer.exe需要一个/select像这样的说法：
explorer.exe /select, <filelist>
I got this from an SO post: Opening a folder in explorer and selecting a file
我从 SO 帖子中得到了这个：在资源管理器中打开一个文件夹并选择一个文件
So your code would be:
所以你的代码是：
if (File.Exists(filePath))
{
Process.Start("explorer.exe", "/select, " + filePath);
}
回答by Chibueze Opata
This is how I do it in my code. This will open the file directory in explorer and select the specified file just the way windows explorer does it.
这就是我在代码中的做法。这将在资源管理器中打开文件目录，并按照 Windows 资源管理器的方式选择指定的文件。
if (File.Exists(path))
{
Process.Start(new ProcessStartInfo("explorer.exe", " /select, " + path);
}
