Var substitution in awk - not working as expected

videsh77 · #1 (**permalink**) 01-13-2006

countA=`awk '/X/''{print substr($0,38,1)}' fName | wc -l`
countB=`wc -l fName | awk '{print int($1)}'`
echo > temp
ratio=`awk -va=$countA -vc=$countB '{printf "%.4f", a/c}' temp`

After running script for above I am getting an error as :

awk: 0602-533 Cannot find or open file -vc=25.
The source line number is 1.

Same kind of script is working elsewhere. Not sure why its going wrong.

blowtorch · #2 (**permalink**) 01-13-2006

This might have something to do with which awk you are using. Try using nawk instead of awk in your script - especially if it is Solaris.

videsh77 · #3 (**permalink**) 01-13-2006

I am using AIX environment.

linuxpenguin · #4 (**permalink**) 01-13-2006

did nawk work ? also try gawk

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