Shell Programming and Scripting

I'm trying the following as i wish to get the last field in a report containing several fields, each line can have more fields than the previous, so I tried

rvwords=`echo $rvstatus |wc -w`
rvstat=`echo $rvstatus |cut -f $rvwords`

but it prints the whole variable contents of $rvstatus rather than just the seventh field, there are no delimeters (other than spaces / tabs) in the line, which would read along the lines of

0 1111 my file 102.00 1003.00 2.99
1 2222 my own file 111.00 112.98 200.10
...

Thanks in anticipation

rvstat=`echo $rvstatus | nawk '{print $NF}'`

or with sed:

rvstat=`echo $rvstatus | sed -e 's/.* \([^ ][^ ]*\)[ ]*$/\1/'`

Try this ....

Code:

$ echo "0 1111 my file 102.00 1003.00 2.99" | cut -d" " -f 7
2.99

Try this if we have more than a one embedded space between the fields

Code:

$ echo "0 1111 my file 102.00 1003.00 2.99" | tr -s " " | cut -d" " -f 7
2.99

Try as with awk as,

Code:

awk ' { print $NF }' <filename>

or 

sed 's/.*[ ]\(.*\)$/\1/' <filename>

IT will do too.

Thanks the awk method works fine, however this now raises another issue, how about if I now wanted to print the field fourth and/or fifth from the right ?

Code:

# 4th AND 5th from last
rvstat=`echo $rvstatus | nawk '{print $(NF-4), $(NF-5)}'`

# 4th from last
rvstat=`echo $rvstatus | nawk '{print $(NF-4)}'`

Thanks again, that works as required, I had tried something similar but missed off the brackets !.