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Re: grub 'if' expression with 'or' operation

From: Lennart Sorensen
Subject: Re: grub 'if' expression with 'or' operation
Date: Tue, 21 Aug 2012 12:41:14 -0400
User-agent: Mutt/1.5.20 (2009-06-14)

On Tue, Aug 21, 2012 at 08:18:57AM -0700, Loving, Kent wrote:
> I'm trying to write an expression that will be true if either one of two 
> files are present. The expression must return true if either ntldr or bootmgr 
> is present on the first partition of the first drive. If either one is 
> present I want to boot to windows on that partition.
> So I tried:
> if [ -e (hd0,msdos1)/ntldr -o -e (hd0,msdos1)/bootmgr ] ; then
>    #do stuff to boot windows
> This works great if ntldr is present. But if ntldr is NOT present, and 
> bootmgr is present, then the test fails. So I reversed the order:
> if [ -e (hd0,msdos1)/bootmgr -o -e (hd0,msdos1)/ntldr ] ; then
>    #do stuff to boot windows
> Which works if bootmgr is present, but not for ntldr.
> In other words, it appears that combining expressions with or operation does 
> not work. Only the first operand is used. Is the or operation implemented?
> I guess I can change separate the expression into two separate if statements.

It seems the whole use of [ -e ... isn't actually covered in the grub
manual at all, other than having an example that uses it.

Poking a bit at the source code I suspect this might work:

if [ ( -e (hd0,msdos1)/ntldr ) -o ( -e (hd0,msdos1)/bootmgr ) ] ; then
   #do stuff to boot windows

Len Sorensen

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