Hello,

I have a desktop that has Windows 7 and Windows XP installed on separate hard drives. Hard drive 1 contains Windows 7, and that's what the computer boots each time. Hard drive 2 has XP, but the only way I'm able to boot into XP is to disable hard drive 1 in the BIOS. I'd like to be able to add XP into Windows 7's boot menu so I can select which OS to boot when the computer starts up. I tried running these commands (from Windows 7):

bcdedit /create {ntldr} /d "Windows XP"bcdedit /set {ntldr} device partition=F: path \ntldrbcdedit /displayorder {ntldr} /addlast

I rebooted, but whenever I tried selecting "Windows XP" from the boot menu, the computer just restarts. How do I configure this correctly? I am aware of third-party tools like EasyBCD, but I'm looking for a way to do this using bcdedit on the command-line.

 

Another thing to note is that XP labels its partition (hard drive 2) as C:, but Windows 7 labels hard drive 2 as F: (which is why I specified partition=F). Do I have to do anything extra to make sure that when XP boots, it correctly identifies its own partition as C: and not F:?

Automatic way:

You can try Dual-boot Repair tool for Windows 8.1 update.

It fixes not only boot manager (for Windows 7/8) but also ntldr (for Windows XP) and boot.ini.

What the tool does is to place correct(corrected) boot files on "System partition".

 

Manual way:

You can do this manually - for XP place ntldr, ntdetect.com and boot.ini on System partition,

check boot.ini entry for pointing to correct device - you can use ArcPaths utility to display correct

ARC Path for XP system folder like "multi(i)disk(k)rdisk(m)partition(n)\WINDOWS".