CODEWITHSUNDEEP
bash
awenis
awenis

Reputation: 11

Bash script: select variable from array with highest value and output variable name, not value

I have 50 variables that changes value all the time. Their values are integer e.g. 1-9999, but I don't know what the values are until the script runs.

I need to display the name of the variable with the highest value. The actual value is of no importance to me.

The variables are called Q1 - Q50.

Example:

Q34=22
Q1=23
Q45=3
Q15=99

Output must just say Q15

Could you please help me in the right direction?

Upvotes: 1

Views: 506

Answers (4)

Walter A
Walter A

Reputation: 20002

Ask for all vars and grep yours:

set | egrep "^Q[0-9]=|^Q[1-4][0-9]=|^Q50=" | sort -nt= -k2 | tail -1 | cut -d= -f1

Upvotes: 1

Jose Ricardo Bustos M.
Jose Ricardo Bustos M.

Reputation: 8164

this works for me, for example Q1-Q5

Q1=22
Q2=23
Q3=3
Q4=99
Q5=16

for i in $(seq 1 5); do
    name=Q$i
    value=${!name}
    echo "$name $value"
done  | sort -k2n | tail -1 | awk '{print $1}'

you get

Q4

Upvotes: 0

chepner
chepner

Reputation: 531215

Run through each of the 50 variables using indirect variable expansion, then save the name whose value is the greatest.

for ((i=1; i<= 50; i++)); do
    name=Q$i
    value=${!name}
    if ((value > biggest_value)); then
        biggest_value=$value
        biggest_name=name
    fi
done

printf "%s has the largest value\n" "$biggest_name"

Upvotes: 0

Josh Jolly
Josh Jolly

Reputation: 11786

You can use variable indirection for this:

for var in Q{1..50}; do
    [[ ${!var} -gt $max ]] && max=$var
done
echo \$$max

Upvotes: 3

Related Questions