OpenFileDialogue with Multiselction enabaled

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4.00/5 (1 vote)

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HI guys,

I have a question here.

The situation is:
I use OpenFileDialogue to select some files (e.g TextOne.txt,TextTwo.txt, TextThree.txt).
Now I want to access send the TextTwo.txt to a stream object to be read.
How to do that?
Because;

C#

stream = OpenFileDialog.OpenFile();

is getting the TextOne.txt.

C++

FileNames[i]

or

C++

FileName[i]

just returns the string but not the stream.

Need help on this?

Posted 21-Feb-11 21:10pm

skunkhead

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OriginalGriff · Accepted Answer · 2011-02-21T21:18:00

Solution 1

OpenFileDialog always returns only the file name: It can't return the stream because it does not know what you are going to do with it: Open it as text or binary.
If you are using it with

this.openFileDialog1.Multiselect = true;

Then you need to loop through all names and open them yourself:

C#

foreach (String file in openFileDialog1.FileNames)
   {
   FileStream fileStream = new FileStream(file, FileMode.Open);

   ... // Do something with the stream.

   }

Posted 21-Feb-11 21:18pm

OriginalGriff

Comments

Olivier Levrey 22-Feb-11 4:02am

Voted 5. Perfect answer.

Sergey Alexandrovich Kryukov 22-Feb-11 4:13am

Correct, 5.
--SA

skunkhead 28-Feb-11 22:55pm

thanks :)