I am trying to pass an argument to

[shaokat]

I am trying to pass an argument to a script and use that argument in an awk statement. Is there something missing in the code? Its not working! It is creating a file with blank extension.

TIA
-KAT


#!/usr/bin/ksh

if [ $# != 2 ] ;then
echo "usage: $0 argument"
exit
fi

ch_ind="$1" ;export ch_ind
ind_pos="$2" ;export ind_pos


ProcessFiles (){
echo GetFiles

for i in *.svd
do

awk 'BEGIN { count=0;}
{
print_ind=substr($0,'$ind_pos', 1);
if (print_ind=="Y&quot {
if (count > 0) {
printf("\r\n&quot > "file."'ch_ind'
}
count=1;
printf("%s", $0) > "file."'$ch_ind'
}
}
' < &quot;$i&quot;
done

}

ProcessFiles
echo done

 

[olded]

Kat:

You're shell variables within the awk script has to be protected by two double quotes:

Check out this FAQ on the 3 methods for using variables in awk.

&quot;&quot;$ch_ind&quot;&quot;

faq271-1281

Regards,

Ed

 

[vgersh99]

or:

nawk -n ind_pos=${ind_pos} 'BEGIN { count=0;}
{
print_ind=substr($0,ind_pos, 1);
........
vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[vgersh99]

ooops - sorry 'bout that:

nawk -v ind_pos=${ind_pos} 'BEGIN { count=0;}
{
print_ind=substr($0,ind_pos, 1); vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+