UNIX for Dummies Questions & Answers

how do i count column in a file.

test1,test2,test3,,, = 5 columns

If i am not mistaken having 5 separators in a line gives not 5 fields but 6:

Code:

field1 , field2 , field3 , field4 , field5 , field6

Anyway, you can count the numbers of fields:

Code:

echo "field1 , field2 , field3 , field4 , field5 , field6" | awk '{FS=",";print NF}'

I hope this helps.

bakunin

my goal is see output of # of columns such as wc -lc.

I have understood that. I suggest you first try the command i wrote above, then modify the input (the string behind "echo") and see how that changes the output. Especially the line:

Code:

echo ",,,,," | awk '{FS=",";print NF}'

might tell you something.

I hope this helps.

bakunin

thank you. worked perfectly.

I'm not sure the FS is getting set in the right place. I get the following:-

Code:

# echo "field1 , field2 , field3 , field4 , field5 , field6" | awk '{FS=",";print NF}'
11
# echo "field1,field2,field3,field4,field5,field6" | awk '{FS=",";print NF}'          
1
#

The only difference is that I've removed the spaces. The FS="," seemingly has no effect:-

Code:

# echo "field1 , field2 , field3 , field4 , field5 , field6" | awk '{print NF}'       
11
# echo "field1,field2,field3,field4,field5,field6" | awk '{print NF}'          
1
#

Not being too experienced in awk, I did get more sucess with these:-

Code:

# echo "field1,field2,field3,field4,field5,field6" | awk -F, '{print NF}'
6
# echo "field1 , field2 , field3 , field4 , field5 , field6" | awk -F, '{print NF}'
6
#

Have I missed something important in your solution? I'm using RHEL 6


Kind regards,
Robin

If this is your result your awk is definitely working different than mine. I used AIX (7.1.3 SP3) awk.

This is in line with my documentation. Per default the field separator (internal variable "FS") is set to a blank (which is translated to any whitespace in the current locale by awk). You got 11 fields because it treated each word and each comma as a separate field. You got 1 on the second try, because with blanks used as field separators there is only one field once you remove all the blanks. So it seems your awk just ignored the FS-declaration somehow and fell back to its defaults.


According to my documentation the "-F" parameter sets the value for the internal FS variable the same way i did and i suppose that means the two ways ought to produce the same result.

Addendum: i just checked with a Ubuntu system and the awk there works the same way you described it. The version is:

Code:

# awk -W version
mawk 1.3.3 Nov 1996, Copyright (C) Michael D. Brennan

Its documentation says that the variable FS can only be set to a ERE, so i tried

Code:

# echo "f1,f2,f3,f4" | awk 'BEGIN {FS="[,]";} {print NF;}'
4

which finally did the trick. Seems like mawk is more picky about where FS is set and only allows EREs whereas AIX's awk allows it to be set anywhere and allows single characters, blanks or EREs.

I hope this helps.

bakunin