Shell Programming and Scripting

Hi,

I want to make a script using shell and awk together, the objective is----
Using for loop I am reading data from a file and then want to print (using awk) the first and fifth field from a directory.
This script is not running as it should........
Please help
Thanks in advance.........

for i in `cat file1` ;
do
grep $i /user2/INAUTO/LOG/OCS*24092008* | awk '/Success/ { print $1,$5}' *24092008*;
done

Is the grep supposed to tell you which files contain a pattern from file1, and run awk on those which do? You need grep -l to list file names, and awk should run with xargs or something to get only the files returned by grep.

Code:

awk '/Success/ { print $1, $5 }' $(grep -lf file1 /usr2/INAUTO/LOG/OCS*24092008*)

If your shell is old, you might need to use `grep ...` with grave accents (aka backticks, ASCII 96 -- not straight quotes!) instead of $(grep ...)

Another change is to use grep -f file1 to read all patterns from file1 in one go. If your grep doesn't support the -f option, or if file1 contains too many patterns for your grep, you will obviously have to undo that part.

You could probably refactor this further to a single awk script which collects output from one file at a time, and only prints the Success lines once it has determined a match on one of the patterns in file1.

What does "not running as it should" mean? Describe what you want to happen and what actually happens. Post any error messages you get (copy and paste, do not retype them).

DON'T DO THAT!

That may or may not read the file line by line.

If you want to read a file line by line, use a while loop:

Code:

while IFS= read -r line
do
 : do whatever with "$line"
done < FILE

grep -f is not working..
Is there any alternate for this or a total different awk script to achieve the objective.......

What does "not working" mean? Please explain what happens and why that is not what you want.

If you get error messages, post them exactly as they appear; do not retype them.