CODEWITHSUNDEEP
bashshellloopsvariablesfor-loop
Tarun
Tarun

Reputation: 75

Using variable in for Loop using seq -w option

I am trying to run a for loop as below without success

hour=`date +%H -d "1 Hour Ago"`
cd /Log1/
for i in `seq -w 00 "$hour"`;  // $hour not working
do
zgrep -a "Packet" LOG1.txt| grep "#SUCCESS#" |wc -l >>success_chour.txt
zgrep -a "Packet" LOG2.txt| grep "#FAIL#" |wc -l >>fail_chour.txt
done

I have tried $hour, "$hour", and '$hour' without success.

Upvotes: 0

Views: 171

Answers (2)

Charles Duffy
Charles Duffy

Reputation: 295443

It's not clear what "without success" means in this context, since you're not even trying to expand $i anywhere inside the body of your loop.

That said, using seq is actually not ideal -- it's a nonstandard tool not built into bash. Better to use functionality guaranteed to be available everywhere the shell is:

# Calculate both day and hour at once
log_scan_end_time=$(date -d "1 hour ago" '+%Y-%M-%d-%H')

# ...then create log_scan_prefix by removing the hour
log_scan_prefix=${log_scan_end_time%-*}

# and extract the hour alone
max_hour=${log_scan_end_time##*-}

for ((cur_hour=0; cur_hour<max_hour; cur_hour++)); do
  printf -v cur_hour_str '%02d' "$cur_hour" # generate a 0-padded string
  logs=( Packet*"${log_scan_prefix}-${cur_hour_str}"*.gz )
  [[ -e ${logs[0]} ]] || {
    echo "No logs found matching pattern ${logs[0]}" >&2
    continue
  }

  # print a header line in the output file to allow diagnosing results
  for f in success_chour.txt fail_chour.txt; do
    {
      printf '# '
      printf '%q ' "${logs[@]}"
      printf '\n'
    } >>"$f"
  done

  zgrep -E -a --count \
    -e '(Packet.*#SUCCESS#)|(#SUCCESS#.*Packet)' \
    -- /dev/null "${logs[@]}" >>success_chour.txt
  zgrep -E -a --count \
    -e '(Packet.*#FAIL#)|(#FAIL#.*Packet)' \
    -- /dev/null "${logs[@]}" >>fail_chour.txt
done

Upvotes: 0

Jiaping Zhu
Jiaping Zhu

Reputation: 56

jiapingzjp@jiapingzjp-Dell:~$ cat my.sh 

hour=`date +%H -d "1 Hour Ago"`
for i in `seq -w 00 "$hour"`;
do
  echo $i
done

jiapingzjp@jiapingzjp-Dell:~$ bash my.sh 

00
01
02
03
04
05
06
07
08
09
10
11
12

I think the code is OK, I try both "bash test.sh" "dash test.sh" in ubuntu system, it works fine. Maybe you can show what the error message you get and what the command you run.

Upvotes: 2

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