using shell variables in awk??? 1

[RobJordan]

This is from a k shell script.

echo " $LINE\c"|awk '{printf("%-20s",$0)}' # print menu item

As you can see, in the second line of code,
I am trying to make the statement dynamic
by replacing the hard coded 20 with a shell variable.

echo " $LINE\c"|awk '{printf("%-${MENU_ITEM_WIDTH}s",$0)}'

Anyone know the correct syntax for the line above?

Thanks in advance,

Robert

Robert G. Jordan

Robert G. Jordan

Robert@JORDAN2000.com

http://www.JORDAN2000.com

Unix Sys Admin
Chicago, Illinois U.S.A.

 

[dickiebird]

I reckon you need to pass the variable to awk ie
echo " $LINE\c"|awk -v MIW=${MENU_ITEM_WIDTH} '{printf("%-MIWs",$0)}'
I didn't test it, though
DickieBird

 

[RobJordan]

Thanks! I got it to work by adding double quotes
around the variable MIW.

echo " $LINE\c"|awk -v MIW=${MENU_ITEM_WIDTH} '{printf("%-"MIW"s",$0)}' Robert G. Jordan

Robert@JORDAN2000.com

http://www.JORDAN2000.com

Unix Sys Admin
Chicago, Illinois U.S.A.