help required for maths in bash script

[needee]

#!/bin/bash
a=19
b=17
s1=$(echo "scale=1;($a/31)*100" | bc -l)
s2=$(echo "scale=1;($b/31)*100" | bc -l)
echo "s1 = $s1 and s2 = $s2"
if [ $s1 -gt $s2 ]; then echo "Holiday"; else echo "More work"; fi

output:
s1 = 60.0 and s2 = 50.0
n: line 7: [: 60.0: integer expression expected
More work

and when I tried
s1=$(echo "scale=0;($a/31)*100" | bc -l)
s2=$(echo "scale=0;($b/31)*100" | bc -l)
echo "s1 = $s1 and s2 = $s2"

got the following output
s1 = 0 and s2 = 0
More work


Regards

[GNU-Fan]

Linux tip: Bash test and comparison functions

[vsemaska]

The only type of numbers that bash can handle are integers. You're trying to compare 2 reals. The decimal point in the numbers caused the error 'integer expression expected'.

You'd have to compare two reals using bc, something like this:

rslt=$(echo "scale=1; if ($s1 > $s2) print 1 else print 0" | bc -l)
if [ $rslt = 1 ]; then echo "Holiday"; else echo "More work"; fi

[drl]

Hi.

If you needed the values in bash, and you were keeping track of the scaling yourself, you could use:

#!/bin/bash -

# @(#) s1       Demonstrate integer arithmetic by arranging operations.

echo
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version "=o" $(_eat $0 $1) bc
set -o nounset

echo
echo " Results:"
a=19
b=17
# s1=$(echo "scale=0;($a/31)*100" | bc -l)
# s2=$(echo "scale=0;($b/31)*100" | bc -l)
s1=$(echo "scale=0;($a*100)/31" | bc -l)
s2=$(echo "scale=0;($b*100)/31" | bc -l)
echo "s1 = $s1 and s2 = $s2"

exit 0

Producing:

% ./s1

(Versions displayed with local utility "version")
Linux 2.6.11-x1
GNU bash 2.05b.0
bc 1.06

 Results:
s1 = 61 and s2 = 54

cheers, drl

[burschik]

Since you are already using bc, you might as well use its relational expressions.