从打开文件对话框路径/文件名路径提取 [英] Extracting Path from OpenFileDialog path/filename

问题描述

我正在写一个小工具，以选择一个文件开始，然后我需要选择一个文件夹。我想默认的文件夹选定的文件在何处

I'm writing a little utility that starts with selecting a file, and then I need to select a folder. I'd like to default the folder to where the selected file was.

OpenFileDialog.FileName返回的完整路径和放大器;文件名 - 我想要的是获得人的路部分（没有文件名），这样我就可以使用它作为初始选择的文件夹

OpenFileDialog.FileName returns the full path & filename - what I want is to obtain just the path portion (sans filename), so I can use that as the initial selected folder

    private System.Windows.Forms.OpenFileDialog ofd;
    private System.Windows.Forms.FolderBrowserDialog fbd;
    ...
    if (ofd.ShowDialog() == DialogResult.OK)
    {
        string sourceFile = ofd.FileName;
        string sourceFolder = ???;
    }
    ...
    fbd.SelectedPath = sourceFolder; // set initial fbd.ShowDialog() folder
    if (fbd.ShowDialog() == DialogResult.OK)
    {
       ...
    }

是否有任何.NET方法要做到这一点，或者我需要使用正则表达式，拆分，修剪等??

Are there any .NET methods to do this, or do I need to use regex, split, trim, etc??

推荐答案

使用的 路径 从的 System.IO 。它包含了操作文件路径，包括 GetDirectoryName <有用的电话/ code> 这你想要做什么，返回文件路径的目录部分。

Use the Path class from System.IO. It contains useful calls for manipulating file paths, including GetDirectoryName which does what you want, returning the directory portion of the file path.

用法很简单。

string directoryPath = Path.GetDirectoryName(filePath);

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