Sorry about bits and snippit of information
So I am writing an average shell script program
so if use inputs
echo 1 3, .... | sh get_number
I would have to pull the numbers seperated by spaces from echo to be
var1 = 1, var2= 3, etc.
I tried
#!/bin/sh
sum=0
for i in $*
do
sum=`expr $sum + $i`
done
avg=`expr $sum / $n`
echo Average=$avg
but doesnt work....
do I include a read here?
also how would I do
sh get_number <file1>, <file2>... to grab numbers in them and sum them
in shell script?
Thanks
Sounds like you are looking for the read shell builtin:
% echo "1 2 3 4" | read a b stuff
% echo $b
2
% echo $stuff
3 4
To fix up your code:
for i in $*; do
sum=$(( sum + i ))
n=$(( n + 1 ))
done
echo "Average=$(( sum / n ))"
#!/bin/sh
while [ $# -gt 0 ]; do
(( i++ ))
(( sum += $1 ))
shift
done
echo "Average=$(( sum/i ))"
Note: This fails in dash which is the closest shell I could find to a real sh.
An example of reading values from files passed as command line arguments or from lines read from stdin:
add_to_sum() {
set $*
while [ $# -gt 0 ]; do
I=`expr $I + 1`
SUM=`expr $SUM + $1`
shift
done
}
I=0
SUM=0
if [ $# -gt 0 ]; then
# process any arguments on the command line
while [ $# -gt 0 ]; do
FILE=$1
shift
while read LINE; do
add_to_sum "$LINE"
done < "$FILE"
done
else
# if no arguments on the command line, read from stdin
while read LINE; do
add_to_sum "$LINE"
done
fi
# be sure not to divide by zero
[ $I -gt 0 ] && echo Average=`expr $SUM / $I`
Related
n=20
x=3
count=0
flag=0
i=1
declare -a arr[n+1]
for (( j=0;j<=n;j++ ))
do
arr+=(0)
done
#echo "${arr[#]}"
while [[ $count -ne $n ]]
do
if [[ $i -le $n ]]
then
if [[ ${arr[$i]} -eq '0' ]]
then
echo "Value is ${arr[$i]}"
#${arr[$(i-1)]}= (( ${arr[$i-1]++} ))
${arr[$i]}+=${arr[$i]}
echo " "
#echo -n "${arr[$i]}"
echo -n " $i"
count=$(( count+1 ))
i=$(( i+1+x ))
else
i=$(( i+1 ))
fi
else
i=$(( i-n ))
flag=$(( flag+1 ))
fi
done
echo " "
echo "No of round : $flag"
This is the whole code, I've tried to print numbers that follows this: n=20 is the number of elements and x=3 is the number that we have to avoid. For example,
20
3
1,5,9,13,17,2,6,10,14,18,3,7,11,15,19,4,8,12,16,20,
3
But, the problem is that my second if condition is not fulfilling, if ignores the condition. Above example is for the C++, but in bash script, 2nd if statement isn't working. This can be because syntax is wrong. So can you please help me to find the mistakes.
Output of the above code:
output
${arr[$i]}+=${arr[$i]}
This is incorrect. $ should not be used when you assign the value.
If you want to double the value, replace this string with the following:
arr[$i]=$(( ${arr[$i]} + ${arr[$i]} ))
Or what you want to do there?
the code below cannot make the fibonacci sequence more than 93 sequences, how can i solve this? I would like you to do with any number
#!/bin/bash
clear
echo "Program to Find Fibonacci Series"
echo "How many number of terms to be generated ?"
read n
x=0
y=1
i=2
echo "Fibonacci Series up to $n terms :"
echo "$x"
echo "$y"
while [ $i -lt $n ]
do
i=`expr $i + 1 `
z=`expr $x + $y `
echo "$z"
x=$y
y=$z
done
You can use the "bc" command (an interactive algebraic language with arbitrary precision) to get past numeric limits of the shell. Here is the re-write of your while loop:
while [[ $i -lt $n ]]
do
i=$(( $i + 1 ))
z=$( bc <<< "$x + $y" )
echo "$z"
x=$y
y=$z
done
On Debian/Ubuntu/RHEL/CentOS systems, install the optional "bc" package.
I am a newbie .Please help me with this
The output says syntax error near unexpected token 'do'
The code is
if [ $# -eq 0 ];
then
echo "Command line arguments are missing."
else
n=$1
sum=0
while[ $n -gt 0 ]
do
rem=$(( $n % 10 ))
sum=$(( $sum + $rem ))
n=$(( $n / 10 ))
done
echo "Sum of digit for given number is $sum "
fi
A whitespace after while. Try this out:
if [ $# -eq 0 ];
then
echo Command line arguments are missing.
else
n=$1
sum=0
while [ $n -gt 0 ];
do
rem=$(( $n % 10 ))
sum=$(( $sum + $rem ))
n=$(( $n / 10 ))
done
echo "Sum of digit for given number is $sum"
fi
Put spaces before and after [ and ].
You can replace the while-loop with
sum=$(( $(echo $1| sed 's/./&+/g; s/+$//' ) ))
I have loop which writes to file, but I want to write each 0.5 value to the file. I tried with let count+=0.5 but that didn't work somehow. Is this possible?
Script:
#!/bin/bash
COUNTER=50
count=0
until [ $COUNTER -lt 20 ]; do
echo $count >> value.txt
echo COUNTER $COUNTER
let COUNTER-=1
let count+=0.5
sleep 1
done
bash doesn't do floating-point arithmetic natively; you need to use an external tool. -= is also not a supported operator.
until [ "$COUNTER" -lt 20 ]; do
printf "%0.1f\n" "$count"
echo "COUNTER $COUNTER"
count=$(bc <<< "$count + 0.5")
COUNTER=$((COUNTER - 1))
sleep 1
done > value.txt
I need to perform a calculation(addition/multiplication) using the command line input.
For an example: I'm executing the below ./calculation.sh 1 2 3 4 5. It has to sum up the output as 15. Any idea to this ? I've tried with the below logic but couldn't make it.
set -x
while [ $# -gt 0 ]
do
expr $1 + 1
shift
done
OUTPUT=0
for i in $*; do
OUTPUT=$(($OUTPUT + $i))
done
echo $OUTPUT
You need to make use of a variable to save the result of expr. Moreover, +1 doesn't seem to make much sense. You probably wanted to replace that with the variable itself.
You need to print the variable at the end.
Try:
set -x
res=0
while [ $# -gt 0 ]
do
res=`expr $1 + $res`
shift
done
echo $res
Try
set -x
sum=0
while [ $# -gt 0 ]
do
sum=$(expr "$sum" + "$1")
shift
done
echo "sum: $sum"
And it's simpler in bash:
sum=0
for i; do
(( sum += i ))
done
echo "sum: $sum"