Shell Programming and Scripting

aigles,

Works perfectly!! Very quick! This is exactly what I needed -- Thanks!

Now for the really difficult question how does it work?
About all that I can figure out from the man page is -n will "Suppresses the default output". I know the standard sed functions but I can't look at this and tell what it is doing.

$PKG=XXXpkg8
VER=$( sed -n "/XXXpkg8/s/.*XXXpkg8\([^)]*).*/\1/p' )

Code:

sed -n                      Suppress automatic print, output must be explicit (p command)
/XXXpkg8/                   Select lines containing XXXpkg8
s/.*XXXpkg8\([^)]*).\).*/   Substitute all the line, memorize version XXXpkg8(version)
/\1/                        with memorized version
p                           Print the result if a substitution have been made

Jean-Pierre.

Thanks JP!

grep "XXXpkg8(5.7.0.10)" /tmp/trash
echo $?
if its = 0 then u found the pattern and the file.

Is this is what u r looking for?

No! If file exist and if pattern is found, the exit status is 0.

Jean-Pierre.

Aigles,

Thanks for providing the breakdown but I still can't translate this to not look for the version in XXXpkg8(version)....
IE how would I do this same task if all I wanted in the output was XXpkg8(version) ....

Thanks!

The parentheses \(...\) select what is being memorized as \1. The parentheses don't affect what is being matched (in this particular case) so you can move them around in the expression. If you have multiple parenthesized expressions, they will be remembered in \1 \2 etc according to the order of the opening parentheses.

The "select" part of the script is actually redundant, since the substitution obviously can only take place on lines where the left-hand side of the substitution expression matches.