Shell Programming and Scripting

Dears,

I would like to count the number of "(" and ")" that occur in a file.
(syntax checking script). I tried to use "grep -c" and this works fine as long as there is only one character (for which I do a search) on a line.
Has anyone an idea how I can count the number of specific characters in a file ?

(I prefer not to use awk or sed but if there's no other option ...)

I don't understand your aversion to awk in this context because it's so simple:

Code:

echo '(a)(bc)d)' | awk '-F[()]' { t += NF - 1 } END { print t }'

produces "5" on the output.

Thanks, works fine.
My aversion to awk or sed
1) I got a good book about it but, no team to practise/learn it
2) the people who will be using/changing my script don't know anything of awk/sed => if they want to change it they will always come to me.

But anyway I think I'll stick to awk cause it is indeed simple.

I don't know what shell you're using, but this works in ksh93:

Code:

#! /bin/ksh

cnt=0
while read -n 1 char; do
        [[ "$char" == "(" || "$char" == ")" ]] && {
        ((cnt++))
        }
done

print There were $cnt characters found!

$ echo "this (by that I mean this) is a \"paren\": )" | ./count.ksh
There were 3 characters found!
$

Code:

awk '{\
workline=$0
lcnt=gsub("\(","",workline)
rcnt=gsub("\)","",workline)
ocnt=ocnt+lcnt-rcnt
printf "%3d %3d %3d %-s\n",lcnt,rcnt,ocnt,$0
}' myscript

Sorry, another awk solution. This one prints the script being checked with each line preceded by 3 counts: left parens on the line, right parens on the line, and a running count of open parens after the line. Sample output:

1 1 0 function chkforEnd() {
2 2 0 if (match($0,"=$"))
0 0 0 SOstat="C"
0 0 0 else
0 0 0 SOstat="O"}

But the third column is the important one, so I would go with:

Code:

awk '{\
workline=$0
lcnt=gsub("\(","",workline)
rcnt=gsub("\)","",workline)
ocnt=ocnt+lcnt-rcnt
printf "%3d %-s\n",ocnt,$0
}' myscript

As long as paren sets open and close on the same line, the count stays zero. When paren sets remain open across lines, the count will go non-zero for those few lines. When I omit a right paren, the open count remains positive for remainder of script:

0 {if (phase==1)
0 if (SOstat=="C")
0 if (NF>0)
0 chkforEnd()
0 else
0 phase=2
0 else
1 if (NF>1 && match($1,"^[A-Z]")
1 {holdline=holdline "="
1 corrected++
1 chkforEnd()}
1 else
1 if (NF>0 && match($1,"^[0-9][0-9]*$"))
1 chkforEnd()

and when I omit the left paren instead of the right paren:

0 {if (phase==1)
0 if (SOstat=="C")
0 if (NF>0)
0 chkforEnd()
0 else
0 phase=2
0 else
-1 if NF>1 && match($1,"^[A-Z]"))
-1 {holdline=holdline "="
-1 corrected++
-1 chkforEnd()}
-1 else
-1 if (NF>0 && match($1,"^[0-9][0-9]*$"))
-1 chkforEnd()

Hi,
Anyone interested in Perl?
Perl can do that faster than awk where we call it pattern matching.

When it match, it can count the character. Say ur filename is Myfile.txt

open (FILE, "<Myfile.txt")
{
while (<FILE>)
{
if /(\W+\sand\s\W+)/)
{
my $data = $1;
my $data1 =length ($data) -2;
#$data1 contains the value of the length of the word (" and ")
print $data1;
}
}

I think you misunderstood the problem: The OP wants to count parentheses, not occurrences of "and" between parentheses. One variation in awk might go like this (untested code):

Code:

#!/usr/bin/awk
BEGIN { t = 0 }
 { len1 = length
   gsub("[()]", "", $0)
   len2 = length
   t += len1 - len2
 }
END { print sum }

Translation into perl is left as an exercise for the reader. My original awk solution is simpler, and can be easily incorporated on a command line.