Counter increment in Bash loop not working

Question

I have the following simple script where I am running a loop and want to maintain a COUNTER. I am unable to figure out why the counter is not updating. Is it du

Answer 1

count=0   
base=1
(( count += base ))

Answer 2

Try to use

COUNTER=$((COUNTER+1))

instead of

COUNTER=$((COUNTER))

Answer 3

I think this single awk call is equivalent to your grep|grep|awk|awk pipeline: please test it. Your last awk command appears to change nothing at all.

The problem with COUNTER is that the while loop is running in a subshell, so any changes to the variable vanish when the subshell exits. You need to access the value of COUNTER in that same subshell. Or take @DennisWilliamson's advice, use a process substitution, and avoid the subshell altogether.

awk '
  /GET \/log_/ && /upstream timed out/ {
    split($0, a, ", ")
    split(a[2] FS a[4] FS $0, b)
    print "http://example.com" b[5] "&ip=" b[2] "&date=" b[7] "&time=" b[8] "&end=1"
  }
' | {
    while read WFY_URL
    do
        echo $WFY_URL #Some more action
        (( COUNTER++ ))
    done
    echo $COUNTER
}

Answer 4

minimalist

counter=0
((counter++))
echo $counter

Answer 5

This is all you need to do:

$((COUNTER++))

Here's an excerpt from Learning the bash Shell, 3rd Edition, pp. 147, 148:

bash arithmetic expressions are equivalent to their counterparts in the Java and C languages.[9] Precedence and associativity are the same as in C. Table 6-2 shows the arithmetic operators that are supported. Although some of these are (or contain) special characters, there is no need to backslash-escape them, because they are within the $((...)) syntax.

..........................

The ++ and - operators are useful when you want to increment or decrement a value by one.[11] They work the same as in Java and C, e.g., value++ increments value by 1. This is called post-increment; there is also a pre-increment: ++value. The difference becomes evident with an example:

$ i=0
$ echo $i
0
$ echo $((i++))
0
$ echo $i
1
$ echo $((++i))
2
$ echo $i
2

See http://www.safaribooksonline.com/a/learning-the-bash/7572399/

Answer 6

Instead of using a temporary file, you can avoid creating a subshell around the while loop by using process substitution.

while ...
do
   ...
done < <(grep ...)

By the way, you should be able to transform all that grep, grep, awk, awk, awk into a single awk.

Starting with Bash 4.2, there is a lastpipe option that

runs the last command of a pipeline in the current shell context. The lastpipe option has no effect if job control is enabled.

bash -c 'echo foo | while read -r s; do c=3; done; echo "$c"'

bash -c 'shopt -s lastpipe; echo foo | while read -r s; do c=3; done; echo "$c"'
3