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 _______________________________________________ Grub-devel mailing list Grub-devel@gnu.org https://lists.gnu.org/mailman/listinfo/grub-devel
