Overwriting /boot/grub/menu.lst?

Before overwriting I have copied /boot/grub/menu.lst to /home/san. I am running sed on the /home/san/menu.lst just for testing.

How can i overwrite

default 0

to

default 1

with the help of sed. I used following commands but none worked. It's most probably because i don't how many spaces are there in between "default" and "0". I thought there were two spaces and one tab but I was wrong.

sed -e 's/default  \t0/default  \t1/' /home/san/menu.lst
sed -e 's/default\t0/default\t1/' /home/san/menu.lst
sed -e 's/default \t0/defaul开发者_开发百科t \t1/' /home/san/menu.lst
sed -e 's/default  0/default 1/' /home/san/menu.lst

I actually want to write a script that may see if 'default 0' is written in menu.lst then replace it with 'default 1' and if 'default 1' is written then replace it with 'default 2'.

Thanks

Update:

How can used a conditional statement to see if the line starting with 'default' has "0" or "1" written after it? If "0" replace with "1" and if "1" replace it with "0"?

This works for me (with another file, of course ;)

sed -re 's/^default([ \t]+)0$/default\11/' /home/san/menu.lst

How it works:

1. Passing -r to sed allows us to use extended regular expressions, thus no need to escape the parentheses and plus sign.
2. [ \t]+ matches one or more tabs or spaces, in any order.
3. Putting parentheses around this, that is ([ \t]+), turns this into a group, to which we can refer.
4. This is the only group in this case, thus \1. That is what happens in the replacement string.
5. We don't want to replace default 0 as part of a larger string. Thus we use ^ and $, which match the start and end of the line, respectively.

Note that:

1. Using a group is not strictly necessary. You can also opt to replace all those tabs and spaces by a single tab or space. In that case just omit the parentheses and replace \1 with a space:

   sed -re 's/^default[ \t]+0$/default 1/' /home/san/menu.lst
   

2. This will fail if there is trailing whitespace after default 0. If that is a concern, then you can match [ \t]* (zero or more tabs and spaces) just before the EOL:

   sed -re 's/^default([ \t]+)0[ \t]*$/default\11/' /home/san/menu.lst
   

Firstly, you could use several lines like this:

 sed -re 's/default([ \t]*)0/default\11/' /home/san/menu.lst.

However, you might be better off with one line like this:

 awk '/^default/ { if ($2 == 1) print "default  0" ; else print "default 1" } !/^default/ {print}' /boot/grub/menu.lst

This substitutes 1 for 0, 0 for 1:

sed -r '/^default[ \t]+[01]$/{s/0/1/;t;s/1/0/}'

This substitutes 1 for 0, 2 for 1, 0 for 2:

sed -r '/^default[ \t]+[012]$/{s/0/1/;t;s/1/2/;t;s/2/0/}'

Briefly:

• /^default[ \t]+[01]$/ uses addressing to select lines that consist only of "default 0" or "default 1" (or "default 2" in the second form) with any non-zero amount of spaces and/or tabs between "default" and the number
• s/0/1/ - substitute a "1" for a "0"
• t - branch to the end if a substitution was made, if not then continue with the next command
• followed by more substitution(s) (and branches)

Edit:

Here's another way to do it:

sed -r '/^default[ \t]+[012]$/y/012/120/'