如何将openFileDialog.FileNames转换为FileInfo []

Question

I have a OpenFileDialog and I want to convert selected file names to a FileInfo[] variable.

But I don't know how to convert all the selected files in one line code.

This obviously doesn't work:

FileInfo[] files = openFileDialog.FileNames;

Thank you.

Answer 1

The FileInfo class offers a constructor that expects a filename . Therefore, to get a FileInfo instance for a single filename string , simply call that constructor:

FileInfo file = new FileInfo(openFileDialog.FileName);

In your case, you want to get an array and have several filename strings, therefore you can use the LINQ extension methods from the Enumerable class :

FileInfo[] files = openFileDialog.FileNames.Select(fn => new FileInfo(fn)).ToArray();

Note the additional call to ToArray in the end, as Select will return an IEnumerable<FileInfo> .

Answer 2

使用LINQ：

FileInfo[] files = openFileDialog.FileNames.Select(f => new FileInfo(f)).ToArray();