Hi, I have a script as below and I would like to replace the first field of record 5th of the file parameter.PAR by the file name that I read as line. #!/bin/bash for line in *.mod; do awk '{ if (NR == 5) { $1=$line } }' parameter.PAR > out done But this seems does not work. Any comments please yacob

There are several ways of using shell parameters in awk scripts. Check out this link: http://www.tek-tips.com/faqs.cfm?fi...

when I used this modified script for i in *.mod do awk '{ if (NR == 5) { $1=$i ; print } else print $0}' out1 > out done the "out" output file is not modified by the file name "$i". I got the same as the original fie only the second column of the 5th recored repeated. Any help, please Yacob

by defining a variable as: arg=$i awk -v var=$arg '{ if (NR == 5) { $1=var ; print } else print $0}' out1 > out it seems that I have solved the problems. But if you have more elegant ways please comment.

Three things: 1) I see no reason for arg=$i. var="$i" will work as well. 2) Note the double quotes around "$i" above. If your argument contains white space, it won't pass properly to the awk script without the quotes. 3) Under Linux, your use of the -v option is correct. Most Linux distros use gawk which is POSIX compliant. If you tried to run that on older versions of AIX and solaris it would fail. The method below is more portable: awk '{ if (NR == 5) { $1=var ; print } else print $0}'var="$i" out1