#!/bin/bash
for ((i=1 ;i<=3;i++))
do
echo "Enter gallon used(gal):"
read gal
echo "Enter Miles Obtained(mil):"
read mil
mileage=`echo $mil / $gal |bc`
echo "scale=4; $mileage " | bc
c=`echo $c + $mileage | bc`
echo "$c + $mileage = $c"
echo
done
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And what's the reported error? – kos Jun 29 '15 at 13:53
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For one you are trying to use a variable in it's own declaration and then echoing that the variable + another equals itself. – 123 Jun 29 '15 at 13:56
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Yes, the problem is in this line: `c=\`echo $c + $mileage | bc\``; `echo $c + $mileage | bc` is (obviously) run before the shell assigns its output to `$c`, so `$c` is undeclared while running `echo $c + $mileage | bc` – kos Jun 29 '15 at 14:02
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I am trying to calculate the mileage for 3 input and get the average. plus the mileage only come out as integer. I need decimals. – okuru joe Jun 29 '15 at 14:06
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1By the way, your conceptual algorithm is flawed — the average of a series of mileage values is generally not the average mileage. For example, if you use 4 gallons driving 80 miles up a mountain (→ 80/4 = 20 mpg) and then you use 1 gallon while driving/coasting 80 miles back down the mountain (→ 80/1 = 80 mpg), your average mileage is not (20+80)/2=50, it is (80+80)/(4+1)=160/5=32. – G-Man Says 'Reinstate Monica' Oct 12 '16 at 19:10
2 Answers
1
Is c your accumulator? Set it to zero to start with, then you will not get a syntax error in line 10.
You get an integer result because there is no operation in line 9. Merge lines 8 and 9 to
mileage=$(echo "scale=4; $mil / $gal" | bc)
and then mileage will have a decimal result.
You do not do anything useful with $c and fail to print it after the loop.
G-Man Says 'Reinstate Monica'
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X Tian
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Thanks that was helpful, now i have decimal result. I need to work on accumulating. – okuru joe Jun 29 '15 at 14:24
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1
#!/usr/bin/env bash
# Above, get the path to BASH from the environment.
# Below, you could just set the total mileage here.
total_mileage=0
# Below, start from zero and count up for the three loops.
for ((i=0; i<3; i++)); do
# Below, use `-n` to prevent the new line.
# It's ok to use descriptive variable names.
# echo -n "Enter gallons used: "
# Below, quote variables.
# read "gallons"
# Using the suggestion for `read` from @roaima :
read -p "Enter gallons used : " "gallons"
# Use a regular expression (regex). Here, a number with optional decimal:
while [[ ! $gallons =~ ^[+-]?[0-9]+\.?[0-9]*$ ]]; do
echo "Please enter a number or [CTRL]+[C] to exit."
read -p "Enter gallons used : " "gallons"
done
# echo -n "Enter miles obtained: "
# read "miles"
# Using the suggestion for `read` from @roaima :
read -p "Enter miles obtained: " "miles"
while [[ ! $miles =~ ^[+-]?[0-9]+\.?[0-9]*$ ]]; do
echo "Please enter a number or [CTRL]+[C] to exit."
read -p "Enter miles obtained: " "miles"
done
# Below, backticks are antiquated.
mileage=$(echo "scale=4; ($miles) / ($gallons)" | bc)
echo "Mileage: $mileage"
total_mileage=$(echo "scale=4; $total_mileage + $mileage" | bc)
done
average_mileage=$(echo "scale=4; ($total_mileage) / ($i)" | bc)
echo "Average mileage is $average_mileage"
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Christopher
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#### I did this### #!/bin/bash c=0 for ((i=1 ;i<=3;i++)) do echo "Enter gallon used(gal):" read gal echo "Enter Miles Obtained(mil):" read mil mileage=`echo "scale=6; $mil / $gal" |bc` echo "The miles/gallon of a tankful was: $mileage" c=`echo "scale=6; $c + $mileage" |bc` done echo average=`echo "scale=6; $c / 3" |bc` echo "The overall Average miles/gallon attained was :$average" – okuru joe Jun 29 '15 at 15:15
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You're running bash. Consider `read -p "Enter miles obtained: " miles`, etc. – roaima Jun 29 '15 at 15:46
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thanks, that part is working well now. But I still having difficulty terminating the program when user input -1 – okuru joe Jun 29 '15 at 15:54
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Yes, you should always quote all *references* to shell variables (e.g., `"$gallons"` and `"$miles"`) unless you have a good reason not to, and you’re sure you know what you’re doing. But you don’t need to quote their *names* when you’re setting them; `read gallons` is perfectly fine and safe. … … … … … … P.S. Why do you divide by 1 in `bc`? – G-Man Says 'Reinstate Monica' Oct 12 '16 at 19:14