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Help with grep & count in ksh

 
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# 1  
Old 10-05-2011
Help with grep & count in ksh
Hello,

I have input file with queue names
HTML Code:
  No.esprd.Queue|No.esdev.Queue|||n|120|No_User||No_User|
  No.esdev.Queue|No.esdev.Queue|||n|120|No_User||No_User|
I have to check if the input file contains word "esprd" and the number of times it occurs. I will have to do further processing based on the no of occurances of "esprd". Can you please help me out with this.

Thanks.
# 2  
Old 10-05-2011
depending on how elaborate you want to be....
Code:
grep -c 'esprd' myFile

# 3  
Old 10-05-2011
thanks vgersh99
Quote:
Originally Posted by vgersh99
depending on how elaborate you want to be....
Code:
grep -c 'esprd' myFile

My problem is it does not give me the count of words. It gives me the count of lines. In production region "esprd" should occur only 4 times in the queues. It test region it should occur 0 times. So I want to get the count of words and then process depending on the count.
# 4  
Old 10-05-2011
how 'bout this:
Code:
nawk '{n+=gsub("esprd","")}END{print n}' myFile

# 5  
Old 10-05-2011
Thanks vgersh99. It seems working and i will include this code in my scripts to see if i get required results. Can you please detail me on the code(ofcourse if u have time or else i will figure out myself)
# 6  
Old 10-05-2011
gsub replaces "esrpd" for "" in the line and returns the count of the number of times it did so. This is run for each and every line. n keeps a running total. when it runs out of lines, it prints n.
# 7  
Old 10-05-2011
from 'man awk':
Code:
       sub(r, t, s)
              substitutes t for the first occurrence of the regular expression
              r in the string s.  If s is not given, $0 is used.

       gsub   same as sub except that all occurrences of the  regular  expres-
              sion  are  replaced;  sub and gsub return the number of replace-
              ments.

'n' accumulates the number of replacements and gets printed at the 'END' after all the file(s) have be processed.
 
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