awk count fields not working

Shell Programming and Scripting
1

Hi,

i am trying to count the fields in a file.
Input:

100,1000,,2000,3000,10/26/2012 12:12:30
200,3000,,1000,01/28/2012 17:12:30
300,5000,,5000,7000,09/06/2012 16:12:30

output:
Cout of the fileds for each row

6
5
6

awk -F"," '{print $NF}' file1.txt

When i try with above awk command i am getting the below output and it is not working as expected.

12:12:30
17:12:30
16:12:30

Any suggestion using awk command for couting the fields.

2

Use NF instead of $NF.

But your output with $NF is strange, given the field separator you've used.

3

thanks for swift responce.
i can find the total fileds in the row which is not equal to 6

awk -F"," '{print NF","$0}' |grep -v "^6"

Input

100,1000,,2000,3000,10/26/2012 12:12:30
200,3000,,1000,01/28/2012 17:12:30
300,5000,,5000,7000,09/06/2012 16:12:30

Output

5,200,3000,,1000,01/28/2012 17:12:30

i unable to print the only values not include the count

required output : fileds not having 6 and print only values and not include 
200,3000,,1000,01/28/2012 17:12:30

i want to print only data not count in the output.

4

Please use code tags for data as well!
Try

awk -F"," 'NF!=6'
200,3000,,1000,01/28/2012 17:12:30

No need for grep . awk will check the field count, and, if eq. 6, print the line, otherwise, not.