GRUB loads Windows 7 instead of Bootmenu with default Menu.lst

[terriwild]

Binary package hint: grub

Ubuntu Release: 9.04 Desktop
GRUB Package Version: 0.97-29ubuntu53

I've recently tries out several Windows 7 Versions ( Beta, RC1, MSDN RTM Version ).
After installation i had to reinstall GRUB ( like common if one installs Windows ).
Installation worked fine. Before the Win7 tests i had Vista and used the Vista entries in GRUB's menu.lst.
GRUB loaded Windows 7 but after reboot the Menu wasn't shown.

I've determined that GRUB Activated the Windows 7 Partition.
So i took my Live CD and changed the active flag back to my Linux partition.
And everything worked fine until next Windows 7 boot.

The Problem was the assumption i've could use my Menu.lst with the Vista entry unchanged.

The Vista entry looked this:

title Windows Vista
root (hd0,0)
savedefault
makeactive
chainloader +1
quiet

I had to disable the makeactive command to get GRUB working as expected.

So my Win7 entry looks like

title Windows 7
root (hd0,0)
savedefault
#makeactive
chainloader +1
quiet

This solved the problem.

ProblemType: Bug
Architecture: i386
DistroRelease: Ubuntu 9.04
NonfreeKernelModules: nvidia
Package: grub 0.97-29ubuntu53
ProcEnviron:
 PATH=(custom, user)
 LANG=de_DE.UTF-8
 SHELL=/bin/bash
SourcePackage: grub
Uname: Linux 2.6.28-15-generic i686

[dino99 (9d9)]

outdated report & no more maintained distro; please send a new one if that issue still exist (using ubuntu-bug)