BASH Script - Check if consecutive numbers in a string are above a value

Question

I am echoing some data from an Oracle DB cluster, via a bash script. Currently, my output into a variable in the script from SQLPlus is:

11/12 0 0 0 0 0 0 1 0 1 0 5 4 1 0 0 0 0 0 0 0 0 0 0 0

What I'd like to be able to do is evaluate that string of numbers, excluding the first one (the date), to see if any consecutive 6 of the numbers are above a certain value, lets say 10.

I only want the logic to return true if all 6 consecutive values were above "10".

So for example, if the output was:

11/12 0 0 8 10 5 1 1 0 8 10 25 40 6 2 0 0 0 0 0 0 0 0 0 0

The logic should return false/null/zero, anything I can handle negatively.

But if the string looked like this:

11/12 0 0 0 0 5 9 1 0 1 10 28 10 12 19 15 11 6 7 0 0 0 0

Then it would return true/1 etc..

Is there any bash component that I can make use of to do this? I've been stuck on this part for a while now.

Answer 1

For variety, here is a solution not depending on awk:

#!/usr/bin/env bash

contains() {
   local nums=$* count=0 threshold=10 limit=6 i
   for i in ${nums#* }; do
      if (( i >= threshold )); then
         (( ++count >= limit )) && return 0
      else
         count=0
      fi
   done
   return 1
}

output="11/12 0 0 0 0 5 9 1 0 1 10 28 10 12 19 15 11 6 7 0 0 0 0"
if contains "$output"; then
   echo "Yaaay!"
else
   echo "Noooo!"
fi