How to awk or grep the last column in file when date on column contains spaces?

Shell Programming and Scripting
1

Hi have a large spreadsheet which has 4 columns

APM00111803814  server_2        96085   Corp IT Desktop and Apps
APM00111803814  server_2         96085   Corp IT Desktop and Apps
APM00111803814  server_2        96034   Storage Mgmt Team
APM00111803814  server_2        96152   GWP Implementations UK
APM00111803814  server_2        96138   GWP Production-UK
APM00111803814  server_2        96109   GWP Production - US

I need to be able to read this file and define a variable for each entry in each column. The last column has spaces but I want to be able to read it a one variable. I am trying to put it into a while loop but the last column will only read the first word/characters (for example in the first line Corp)

For example

while read c1 c2 c3 c4
do
echo "$c1 $c2 $c3 $c4"
done < file.txt

Obviously this does not work. Can anybody help me out with this?

2

Why would it not work with a while read loop? Have you tried it?

3

My apologies. It actually does work. My bad. There are other columns after C4 which I did not paste and it picks these up as well. I just need to make sure that the column with the spaces is the last column and the while loop will work. Thanks for you reply.

4

If you specify less parameters than there are fields then any leftover data is assigned to the last parameter.

EDIT: As you've noticed :).

If there's more data afterwards (that you don't want) then you could pre-process the file with man cut ("posix").

5 
cut -c40- file1
6

Try this:

nawk '{ f4="";
        for ( i=4; i<= NF; i++)
                f4=f4 $i;
        print $1, $2, $3, "<", f4, ">"; }'
7

Can you post or attach the actual input file and specify exactly which columns you want in the output...