Script Help


 
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# 8  
Old 07-12-2008
Crystal after changing command as below its working as expected.

file_name=`cat standardfilecleanup.lst|awk -F: '{print '$var_value'}'`

Thanks for your valuable input.

Thanks,
Prakash
# 9  
Old 07-13-2008
You can reduce the number of lines of code by using a for loop and directly printing output from command substitution.

The following works for both bash and ksh93
Code:
for ((i=3;i<=5;i++))
do
     print $(awk -F: -v var=$i '{print $var}' standardfilecleanup.lst)
done

 
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