UNIX for Dummies Questions & Answers

Is there a way to count the no. of fields (columns) in a file?
Actually I need to cut some fields starting from the middle to the end. How can I specify to cut till last field?

thanks in advance

You can use awk to do this. It's automatically stored in "NF":

ll |tail -1 |awk '{print NF}'
9

I bet you could do the whole thing in awk, but I'm not too good with awk, so maybe someone else can help with an example...

file1:
aaa bbb ccc ddd
aaa bbb ccc ddd
aaa bbb ccc ddd
aaa bbb ccc ddd

If you are dealing with a fixed-column format, you can pull the columns you want with:

cut -c1-3,18-21 file1 > file2

cut command also works with columns (fields), but if the field delimiter is space, for example, multiple spaces in a row are seen as multiple field delimiters, thus making the field numbers very inconsistent. On the other hand, by default, awk considers any amount of white space as a single field delimiter. Following awk command will output fields 1 and 4 with one space between:

awk '{print $1,$4}' file1 > file2

For both solutions, file2 looks like:

aaa ddd
aaa ddd
aaa ddd
aaa ddd

awk can apply logic as to which lines to process, or process certain lines differently from others based on content, etc.

im thinking something like

Code:

(host):/home0/  cat me
1c 2 3b 4aE2
(host):/home0/ sed 's/[^ ]*$//' me
1c 2 3b
(host):/home0/

I fixed it.

As clarification, the sed solution, as coded, will drop the last two characters of each line that ends with two characters:
first character must be space, letter or digit
second character can be anything
otherwise will leave the line unchanged.