Shell Programming and Scripting

I have a nawk that reads in a log file and outputs a file that matches my search.

IFS=" "
while read record
do
`echo $record | nawk 'BEGIN {
FS=" "
}
{
type_record=substr($0, 1, 1);

if ( type_record == "-" )
printf ("%s" , $9) >>"filelist.dat";
else
printf ("\n");
}
'`

done < x.log
OFS=$IFS

the file that comes out of the nawk has carriage return on the end of the two lines:

nawk_output - filelist.dat

abc.csvOD
def.csvOD

i want to set the filenames in a variable i.e.

files_found=`cat x.dat`

the problem is that when i call the variable $files_found it shows as this:

abc.csvf.csv

I want it to show:

abc.csv
def.csv

Its the carriage return that is the problem but can any one help??

You can use quotes to preserve the newline.

i.e.

Code:

echo "$files_found"

thanks for the reply but even with double quotes i get:

abc.csvf.csv

????

Try the following awk script :

Code:

awk -F '[ \r]*' '
/^-/ { print $9 }
' x.log > filelist.dat

Jean-Pierre.

this code produced a syntax error

bailing out at line 1

Works fine on my AIX box.
Try to replace awk by nawk or gawk.

Jean-Pierre.

Hi Jean-Pierre,

replaced with nawk but got this output:

abc.csvdef.csv

I need them seperating, like below?

abc.csv
def.csv