script help

[roger_114]

I have a file that contains records like this record/:

8,388,616,192 B 02/13/05 12:18:23 /u37/oradata/PRODODS/clm_hdr_dt06.dbf 03/20/05

What is the easiest way to retrieve just the full file name out these records?
(in this record , I only need /u37/oradata/PRODODS/clm_hdr_dt06.dbf)

Thanks

[Chris Vail · Accepted solution]

I've counted only 3 spaces in the line you posted. Therefore awk '{ print $3 }' $FILE would return the 3rd field--the data you're looking for.


Chris

[A. Clay Stephenson]

If yoiu need to do some processing:

INFILE="xxx/yyy.txt"
cat ${INFILE} | awk '{print $(NF -1)}' | while read FNAME
do
echo "${FNAME}"
done

The simplest:
awk '{print $(NF -1}}' xxx/yyy.txt

[Rodney Hills]

data="8,388,... /u37/oradata/..."
file=${data##* }

HTH

-- Rod Hills

[Robert Bennett_3]

Try this:

cat | awk -F \ '{print $4}'

This should give you the part you need.

[Rodney Hills]

Oops. Ignore my previous suggestion. I thought the filename was the last field on the line.

You could use-
awk '{print $5}' inputfile
or
cut -f5-5 inputfile

HTH

-- Rod Hills