#!/bin/bash
n=1
while (( $n <= 5 ))
do
num$n=`echo "$n"`
n=$(( n+1 ))
done
echo "$num1"
ok so what I am trying to do is create a while loop that will create variables and just put something into it in this case its just the value of n but i cant get it to do this!
so basically it will create num1, num2, num3 and so on
echo "$num1"
echo "$num2"
echo "$num3"
should display
1
2
3
but i keep getting an error am i missing something here cause it shouldnt be anything crazy to do this...
Try with
#!/bin/bash
n=1
while (( $n <= 5 ))
do
eval num$n=`echo "$n"`
n=$(( n+1 ))
done
echo "$num1"
echo "$num2"
echo "$num3"
The problem here is that bash is trying to evaluate num$n as a command, which does not exist, so the error.
Don't dynamically create numbered variable names like this; use an array.
n=1
while (( $n <= 5 )); do
nums[$n]=$n # No need to execute $(echo ...)
n=$((n+1))
done
echo "${num[1]}"
echo "${num[2]}"
echo "${num[3]}"
Related
n=20
x=3
count=0
flag=0
i=1
declare -a arr[n+1]
for (( j=0;j<=n;j++ ))
do
arr+=(0)
done
#echo "${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?
Requirement: Based upon IF condition in the called function: myfunc, echo hello in the for loop should not get executed and control should go to the next iteration.
In the below script, when the value of k becomes 2 and 3, echo hello should not get executed.
This is the script that I am trying to develop but no success.
#!/usr/bin/env bash
myfunc() {
if [[ $k -gt 1 ]]; then
echo "in the loop"
return
else
echo continue
fi
}
for (( k=1; k<=3; k++ ))
do
myfunc
echo hello
done
Please help.
Your loop is all wrong and I don't know why you have an if / else if you're just interested in one output:
#!/usr/bin/env bash
myfunc() {
if [[ $k -lt 2 ]]; then
echo "hello my value is $k"
fi
}
for (( k=1; k<=3; k++ ))
do
myfunc
echo "$k just to prove it is looping" # this is always run regardless of what's in the function
done
output:
hello my value is 1
1 just to prove it is looping
2 just to prove it is looping
3 just to prove it is looping
#!/usr/bin/env bash
myfunc() {
if [[ $k -gt 1 ]]; then
echo "in the loop"
x=1
else
echo Welcome
fi
}
for (( k=1; k<=3; k++ ))
do
myfunc
if [[ $x -eq 1 ]];then
continue
fi
echo hello
done
I'm struggling with this script, I have tried different thing but I can't get it to work.
I tried with if ,doesn't work, I tried with nested while also without success. I don't know what to do any more...
If the value imax has reached 10 i want it to start over with 1
this is the code:
folder="f"
ps=20000
name="test"
i=1
imax=1
while [ $i -le 20 ]
do
cd
cd "$folder$((i))"
sed -i 1s/.*/$name$((imax))/ file.txt
sed -i 2s/.*/$ps/ file.txt
(( i++ ))
(( imax++ ))
done
Thank you
Something like this works:
i=1
imax=1
while [ $i -le 20 ]
do
echo $i $imax
(( i++ ))
(( imax++ ))
if [ $imax -eq 10 ]; then imax=1; fi
done
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`
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"