View Full Version : What do I do if the part where I put disks doesn't work
Meg93
Jun 29, 2010, 10:28 AM
Technology
ScottGem
Jun 29, 2010, 10:45 AM
Assuming we are talking about a CD/DVD burner, the first this is to define what you mean by "doesn't work". But if the drive has gone bad, then you replace it.
Meg93
Jun 29, 2010, 10:47 AM
Like when I put anything in there it doesn't play music, do anything, or make any response.
ScottGem
Jun 29, 2010, 10:48 AM
Have you tried accessing the drive directly like through Windows Explorer?
Meg93
Jun 29, 2010, 10:52 AM
No, what's that how do you do that?
ScottGem
Jun 29, 2010, 11:16 AM
If you go into Windows Explorer, you will see a drive letter (usually D or E) see if you can open that and run files from there.
Meg93
Jun 29, 2010, 11:31 AM
Oh, OK I'll try that, but if that doesn't work that part in the computer will need to be replaced?
ScottGem
Jun 29, 2010, 12:15 PM
Its hard to tell. There is a software feature in your OS that reads when a disk is put in and tries to do something with that disk. You may have this autoplay feature disabled.
Do a lookup in Help for Autoplay to check if its turned on.
Meg93
Jun 30, 2010, 10:05 AM
I did the lookup, and everything is turned in autoplay, so I guess that means this part of my computer needs to be replaced, about how much would someone charge if they had to replace it.
ScottGem
Jun 30, 2010, 10:19 AM
The drive itself should cost you under $50. Maybe another $50 to install it.
NeedKarma
Jun 30, 2010, 10:25 AM
Did you recently uninstall any CD burning programs?
Meg93
Jun 30, 2010, 10:25 AM
Oh. But I did what you said, and it said this device is working properly so I don't know what's going on. I mean, is there another way to play the disk once it's in there.
Meg93
Jun 30, 2010, 12:07 PM
Oh. I did everything you said, and it said this device is working properly. Is there a way to manually see or listen to your music with steps through start, of what disk you put in?
Meg93
Jun 30, 2010, 12:08 PM
No, I didn't. The hardrive isn't working.