How to make open any name mpeg file.

[roma123]

How to make Video Object to open any name mpeg file in root cdrom.
to open *.mpg file in root cdrom.
If you have solution please attach simple example project file.
I learning better from examples.

Thanks

[roma123]

Basically when I insert video object to
the project must have name of the file.

But name video file in my project the always change, but it in same location
as project.
How to make video object to run file witout knowing the name of the file?

Or how to make this action:


find name of the file mpeg file in root cd-rom
and put name of mpeg file to "video object" and play.

Or

play \\*.mpg file

[sside]

You must find the video file first then play it.

result = File.Find("AutoPlay\\Videos", "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

[yosik]

If you are talking about runtime finding of your mpg file, then SSIDE gave you the answer, but from your post, it looks as if you want to do it at BUILD time.
If so, it is much simpler. 2 ways:
1. use a dummy name in your video object for the file and then, put the mpeg file in the dir, renaming it to the name of your dummy file.
2. you CAN have a video object without a name at design time. When you run the project, it will give you an error though. So what you can do is build sort of a template project abd for each project, you just insert the mpg file name in the video object field.

I hope that between SSIDE and this, you found what you were looking for.

Yossi

[roma123]

Thanks to SSIDE and Thanks(תודה) to the Yossi.
The project will be defaut mpeg player with logo of my company,
so when burn CD-ROM I do need change name of mpeg file,
to default name in the project.

So SSIDE gave right solution.

But I have tiny problem, when I change path of search to SourceDrive
or SourceFolder its not working, Its working if I change path to folder.

I doing something wrong? :

result = File.Find("{_SourceFolder}", "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

or

result = File.Find("SourceFolder", "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

Thanks

[sside]

result = File.Find(_SourceFolder.."\\AutoPlay\\Videos", "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

[roma123]

Thanks, but its not working.
Take look in the project with mpeg file.

[sside]

In this occasion it would be:

result = File.Find(_SourceFolder, "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

You had to place the video file in the video folder. AutoPlay\Videos

[roma123]

In this occasion it would be:

result = File.Find(_SourceFolder, "*.mpg", false, false, nil, nil);
Video.Load("Video1", result[1], true, false);

You had to place the video file in the video folder. AutoPlay\Videos

Its WORKING from root cdrom!!! THANKS!

[roma123]

I change it to:

result = File.Find(_SourceFolder, "*.mpg", false, false, nil, nil);
MediaPlayer.Load("Plugin1", result[1]);

Now its look like Windows Media Player interface!

[Derek]

I thought roma123 was gonna tell us all how to open any mpeg file. The title suggests so

[roma123]

I thought roma123 was gonna tell us all how to open any mpeg file. The title suggests so

You missing one word: " How to make open any name mpeg file"

[Derek]

Same thing!

[roma123]

open any mpeg files-say it any type of mpeg files (mpeg1,mpeg2,mpeg4)

open any name mpeg files - say it any file name mpeg files.

Aniway Derek you don't explanation for this, with 809 posting

[Derek]

I read your 'statement' .. job done!