Bash - increment a counter in a while loop if something is true

Question

I want to increment a counter in a while loop, and once that loop is complete, echo out the counter. The $COUNTER should only increment if $DATE >= $SINCE

   last -f /var/log/btmp |
while read line; do
    set -- $line
    [[ $1 == "$CURRENTUSER" ]] || continue
    [[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $SINCE ]] && echo "FAILED LOGIN $line" && ((COUNTER+=1))
    echo $COUNTER //echos 1, 2 etc
done

echo "Top counter:"$COUNTER //echos 0

Tagged as bash, so the next approach is not suitable as an answer: Start your script with #!/bin/ksh — Walter A
– Walter A, Commented Jul 25, 2014 at 12:36

David C. Rankin · Accepted Answer · 2014-07-24 08:13:15Z

1

Your problem is with this line:

[[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $since ]] && echo "$line && COUNTER=$[$COUNTER +1]"

You will want something like:

[[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $since ]] && echo "$line" && ((COUNTER+=1))

If you want to display and increment, then:

[[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $since ]] && echo "$line $((COUNTER+=1))"

answered Jul 24, 2014 at 8:13

David C. Rankin

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7 Comments

Kiksy Over a year ago

That works for my echo in the while loop, but the echo of $COUNTER afterwards still prints 0

David C. Rankin Over a year ago

Hmm. that's strange. I tested here and it works fine: ( COUNTER=0; line="dog"; for i in $(seq 1 3); do echo "$line $((COUNTER+=1))"; done; echo "COUNTER $COUNTER" ) output is: dog 1 dog 2 dog 3 COUNTER 3

David C. Rankin Over a year ago

@Kiksy Where is the input for your while loop? You should have something line done < file_to_read.txt Also, you don't really have the // for comments in the file do you? That is not a valid bash comment. Use # instead.

David C. Rankin Over a year ago

Try: echo "Top counter: $COUNTER" and get rid of the // change to # for comments.

Kiksy Over a year ago

outputs 2,2,2,2,2 TOPCOUNTER:0

|

Kiksy · Accepted Answer · 2014-07-24 14:13:00Z

0

Issue was one of scope. Piping into the while means the $COUNTER goes out of scope. A good explanation is http://mywiki.wooledge.org/BashFAQ/024

Solution was from Chris's comment on this post:

http://www.edwardawebb.com/linux/scope-issue-bash-loops enter link description here

COUNTER=0

STDOUT=$(last -f /var/log/btmp)

        while read line; do
            set -- $line
            [[ $1 == "$CURRENTUSER" ]] || continue
            [[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $SINCE ]] && echo "FAILED LOGIN $line" && ((COUNTER+=1))
        done <<EOF

        $STDOUT
        EOF

        echo "counter:$COUNTER"

answered Jul 24, 2014 at 14:13

Kiksy

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Comments

Sylvain Leroux · Accepted Answer · 2014-07-25 10:34:19Z

0

Pipes commands are executed each in their own sub-shell. Variables modifications are confined in that sub-shell, and will not propagate back to the parent shell.

Using bash, a common workaround to avoid that pitfall is to use process substitution :

while read line; do
    set -- $line
    [[ $1 == "$CURRENTUSER" ]] || continue
    [[ $(date -d "$4 $5 $6 $7" +%Y%m%d%H%M%S) -ge $SINCE ]] && echo "FAILED LOGIN $line" && ((COUNTER+=1))
    echo $COUNTER //echos 1, 2 etc
done < <(last -f /var/log/btmp)
#    ^^^^^^^^^^^^^^^^^^^^^^^^^^

answered Jul 25, 2014 at 10:34

Sylvain Leroux

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