I need help following this script:

#!/bin/sh
file=$1
shift
vi `grep -l "$file" "$@"`

when i run it i get nothing. does the 'shift' before the 'vi' command
remove the value of $1, thereby not allowing the script to finish or am
i missing something.
i removed the 'shift' line from the above script and i get an empty vi page.
i thought this script would open the file which matches the 1st argument
in the vi editor??
any help is appreciated.

Works for me, could you post the output with -xv?

sets the variable file to the 1st param supplied.
gets rid of this first param and just leaves the rest of the params in $@
grep searches for a string in a file. The -l option just lists the file names it's in. The odd thing is the string to search for is in the variable $file and the file name it is searching is in the variable $@, if this contains more than 1 value (which it might if you typed in say 3 params) it will cause an error.
The backwards single quotes will run the command just mentioned and then pass those file names to vi for you to edit.

It seems a complicated way of not doing anything very interesting. Why were you running it anyway? what were you trying to do?

Mike

Same as