[SOLVED] bash script to calculate age difference with loop

shayno90 · 05-25-2012, 08:13 AM

I essentially want to calculate this:
"calculate how much beer an 18+ user has drunk statistically (100 liters/year) and print this information for the user."

Not sure how to calculate it i.e. use a loop or other to use the difference in specified age versus user input age, multiply the result, I made an attempt in the last section of code.

Code:

#!/bin/bash
 
# This script check's the user's age and if they can legally drink alcohol

age=16

lage=18

echo "What age are you?"

read uage

years=$[$age - $uage]

beer=$[$years ge $lage]

if [ "$uage" -ge "$age" ] ; then
    echo -e "You are $uage and are of legal age to drink alcohol\n" && 
    echo -e "You have drank $ litres of beer since you turned 18 years of age."
    exit 1
else
    echo -e "You are $uage and are not of legal age to drink alcohol\n" &&
    echo -e "You have to wait $years years before you can legally drink alcohol"
fi 

while [$beer]; do
        echo " "
    let $beer = $beer+1 * 100
    done

if [ "$uage" == "$lage" ] ; then
    echo " You have drank $beer litres of beer since you were born"
else
    echo "You have not drank much beer"
fi

uhelp · 05-25-2012, 08:26 AM

Code:

lage=18
read -p "What is your age? " uage
(( beeryears = age - lage  ))
(( beer = $beeryears * 100 ))
echo $beer

shayno90 · 05-25-2012, 09:05 AM

Quote:

Originally Posted by uhelp

Code:

lage=18
read -p "What is your age? " uage
(( beeryears = age - lage  ))
(( beer = $beeryears * 100 ))
echo $beer

It only works if the user age is 18 and not if older. I ran this code separately without adding to my own.

What is your age? 19
-1800

What is your age? 78
-1800

What is your age? 23
-1800

Nominal Animal · 05-25-2012, 02:40 PM

Code:

beerperyear=100
legalage=18

read -p "What is your age? " age

beer=$(( (age - legalage) * beerperyear ))

if (( beer > 0 )); then
    echo "You've drank $beer liters of beer."
else
    beer=0
    echo "You're not old enough to have drunk beer."
fi

For fractional years (but integral liters of beer) it's easiest to use awk:

Code:

beer=$( echo "$age $legalage $beerperyear" | awk '{ printf("%.0f\n", ($1-$2)*$3) }' )

grail · 05-27-2012, 09:07 AM

The bash code example from uhelp has mixed up uage and age variables, a simple change will fix it:

Code:

(( beeryears = uage - lage  ))

shayno90 · 05-28-2012, 06:58 AM

Great that worked!