Shell Programming and Scripting

Hi,
Could someone please let me know me how i can use an awk command to print the No. of fields of each file present in a Directory?Suppose the Directory has 5 text files in which the first record in each file contains fields separated by a '|'. I need to use a awk command to display the No of fields for each of these txt files as follows:

Name1.txt 30
Name2.txt 12
Name3.txt 10
Name4.txt 21
Name5.txt 0

Also some of these txt files may be empty and i need to print '0' for the No. of fields in this case as shown above. Is this Possible.Please suggest.

Thanks .

Try:

Code:

awk -F'|'  'FNR == 1 { print FILENAME, NF}'  DIRECTORY/*

Thanks Yazu. Your command worked well except it does not print the filename.txt and '0' for a file which is empty.Is there any way to get this?

From GNU awk manual:
"All known awk implementations silently skip over zero-length files. This is a by-product of awk's implicit read-a-record-and-match-against-the-rules loop: when awk tries to read a record from an empty file, it immediately receives an end of file indication, closes the file, and proceeds on to the next command-line data file, without executing any user-level awk program code."

GNU awk can deal with this but it's very specific. I believe it's better to write a shell script.

Code:

for i in *.txt; do wc -l $i; done

tha will print the number of lines, not the number of fields, as the OP wanted.

You could do with awk, with a little hack -- use a shell loop around it and getline in BEGIN section, that is before the processing of file starts:

Code:

for i in *.txt ; do 
  awk  -F"|" 'BEGIN{
     if(getline < "'$i'") 
       {print "'$i'", NF} 
     else print "'$i' 0"}' 
 done

Try:

Code:

awk -F\| 'BEGIN{for(i=1;i<ARGC;i++){f=ARGV[i];print f,(getline<f)?NF:0}}' *.txt