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# 1  
Old 12-06-2001
count fields

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
# 2  
Old 12-07-2001
You can use awk to do this. It's automatically stored in "NF":

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

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...
# 3  
Old 12-07-2001
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.
# 4  
Old 12-07-2001
im thinking something like
(host):/home0/  cat me
1c 2 3b 4aE2
(host):/home0/ sed 's/[^ ]*$//' me
1c 2 3b

I fixed it.

Last edited by Optimus_P; 12-07-2001 at 01:28 PM..
# 5  
Old 12-07-2001
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.
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