Opening a file with a command button

[sistinus]

What I'm trying to do is this. I have table called Videos that lists video
files on my computer. Two of the fields are Filename and FilePath. I'm
trying to make a command button in a form to play the video that corresponds
to the record that's open. If anyone could help me with this it would be a
big help. Thanks.

 

[Dymondjack]

Try the Shell command

Shell(FilePath & Filename)

If the video file has an associated program in the win registry it should
automatically execute that program (usually medai player).

Make sure the argument is a full path including extension.

"C:\Vids\MyVid.wmv"

Hopefully that helps

Jack

 

[Dymondjack]

I forgot: This command would go in the Button_Click procedure

 

[sistinus]

That still doesn't tell me how to get the filename and filepath from the
record. Just typing the path\name in vba defeats the purpose. I have over
350 videos and I really don't want to type a different module for each video.

 

[Keith Wilby]

That still doesn't tell me how to get the filename and filepath from the
record. Just typing the path\name in vba defeats the purpose. I have
over
350 videos and I really don't want to type a different module for each
video.

So just substitute control names to build the path string:

Dim strFilePath as String
strFilePath = Me.txtPath & "\" & Me.txtFilename
Shell(strFilePath)

Keith.
www.keithwilby.co.uk

 

[Daniel Pineault]

The easiest way to open any type of file is to use the FollowHyperlink method
as it will open the file in the default application. So try something like:

Application.FollowHyperlink FilePath & FileName
--
Hope this helps,

Daniel Pineault
http://www.cardaconsultants.com/
For Access Tips and Examples: http://www.devhut.net
Please rate this post using the vote buttons if it was helpful.

 

[papazar]

What I'm trying to do is this. I have table called Videos that lists
video
files on my computer. Two of the fields are Filename and FilePath. I'm
trying to make a command button in a form to play the video that
corresponds
to the record that's open. If anyone could help me with this it would be
a
big help. Thanks.

 

[sistinus]

Thanks. This worked like a charm.

The easiest way to open any type of file is to use the FollowHyperlink method
as it will open the file in the default application. So try something like:

Application.FollowHyperlink FilePath & FileName