Windows 7: Grub question for any Linux users.

I am a complete newbie - I know zero about Linux.

What does this mean - it is complete nonsense:



First it says change X to the partition number - then it says do not enter any numbers.

I would hazard a guess you enter the number to mount the partition - but you don't enter the number to install grub to it?




No wonder the general public won't go near it.

They're a bit confusing there in talking about partitions but with the intentions of installing grub to the root of the drive, regardless of partitions. In the example where they talk about "/dev/sda6", the take home point is that they are identifying that the Mint partition (/dev/sda6, since the number 6 denotes which partition it is on the drive) is on the drive mounted as /dev/sda. What matters here is that you are installing grub to the root of the drive with the Mint partition on it. Thus, the drive containing the Mint partition in this case is /dev/sda. That's all you really care about and that's the identifier you would use, as noted above in "--root-directory=/mnt/ /dev/sda". If there was a second drive, it would be called /dev/sdb, and so forth. They are just trying to make sure you install Grub on the drive where you have Mint installed. If you only have one drive in the system, you don't need to identify which partition Mint is installed onto.

Thanks for the response - but it is still not clear to me.

I want to install it to the Mint partition.

Does it understand that is what I am trying to do - or will it install grub to the Hard drive Mint is on - instead of just the partition Mint is on.

( I will add it to Windows boot menu after. )

The Grub will replace the windows boot loader, so you won't be able to use the windows boot loader afterwords. You will have the choice to load windows from the grub as long as windows was installed before you installed Mint. The root of the drive in the example given here would be sda and not sda6.

I don't want to do that.

There has to be a way to get grub directly onto Mint partition .

I'd do it the other way round.
Install all of your Windows operating systems and then install Linux Mint.
That's the way I installed Ubuntu and I've never had any boot problems.

Note:
I installed Windows without that useless 100 MB boot partition.

When I boot my PC:

• I see the GRUB menu (showing Windows 7 and Ubuntu options)
• If I select Windows 7, I see the Windows 7 menu (showing Windows 7, W8 & XP)

I haven't physically installed Linux Mint, but I don't see any reason why it would be different.

^^ That is the easiest way to do it if you're multi-booting.

If you're just set on installing GRUB to the Mint Partition itself, you can do that but it doesn't make much sense. You'd still have to have a way for the Windows bootloader to recognize that there is another OS installed, otherwise you'll never be able to boot to it. In that case, you'd have to use EasyBCD to modify the Windows bootloader to add that Mint partition to your options of which OS to load at startup. At that point, having GRUB on the Mint partition just adds another bootloader you don't need.

Got it working - you can add a bootsector loader to bcd -you just need to point it at a bootsector file it can find. Like D:\MY\grub.mbr

Real-mode Boot Sector
---------------------
identifier {58638d6d-bc11-11e1-a5d9-001966657855}
device partition=D:
path \MY\Grub.mbr
description Mint Maya

Hi, Si. I don't know whether this is needed but it's the best way I've found to Dual boot Ubuntu-Win7 without interfering with Windows 7. I've been linking it since Barman first posted it and only gotten good feedback.

LOL - I got it working already - see my post just before yours.

bcdedit /create /d "Linux Whatever" /application bootsector