Printing a Variable Which Contains $ Sign
I have a bash script which contains awscli as well. I am trying to print a variable which is created in a for loop. The variable that I am trying to print contains $ sign because of for loop. I couldn't print the value. Below I ams sharing the script. The output of this script is only numbers which is generated in the for loop. I want to print the value which is generated in the command.
!/bin/bash
declare -i counter=11
declare -i counter2=14
for i in {1..2}
do
declare v1$i=$(aws iam get-group --group-name VideoEditors | awk -v counter1=$counter 'NR==counter1' | awk -F" '{print $4}')
counter=$counter+7
declare v2$i=$(aws iam get-group --group-name VideoEditors | awk -v counter3=$counter2 'NR==counter3' | awk -F" '{print $4}')
counter2=$counter2+7
echo $v1$i
echo $v2$i
done
linux bash shell-script variable
asked 5 mins ago
user335828user335828
1
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I have a bash script which contains awscli as well. I am trying to print a variable which is created in a for loop. The variable that I am trying to print contains $ sign because of for loop. I couldn't print the value. Below I ams sharing the script. The output of this script is only numbers which is generated in the for loop. I want to print the value which is generated in the command.
!/bin/bash
declare -i counter=11
declare -i counter2=14
for i in {1..2}
do
declare v1$i=$(aws iam get-group --group-name VideoEditors | awk -v counter1=$counter 'NR==counter1' | awk -F" '{print $4}')
counter=$counter+7
declare v2$i=$(aws iam get-group --group-name VideoEditors | awk -v counter3=$counter2 'NR==counter3' | awk -F" '{print $4}')
counter2=$counter2+7
echo $v1$i
echo $v2$i
done
linux bash shell-script variable
asked 5 mins ago
user335828user335828
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Use"${var}"instead of just$var. See unix.stackexchange.com/questions/4899/…
– Kenneth B. Jensen
1 min ago
add a comment |
I have a bash script which contains awscli as well. I am trying to print a variable which is created in a for loop. The variable that I am trying to print contains $ sign because of for loop. I couldn't print the value. Below I ams sharing the script. The output of this script is only numbers which is generated in the for loop. I want to print the value which is generated in the command.
!/bin/bash
declare -i counter=11
declare -i counter2=14
for i in {1..2}
do
declare v1$i=$(aws iam get-group --group-name VideoEditors | awk -v counter1=$counter 'NR==counter1' | awk -F" '{print $4}')
counter=$counter+7
declare v2$i=$(aws iam get-group --group-name VideoEditors | awk -v counter3=$counter2 'NR==counter3' | awk -F" '{print $4}')
counter2=$counter2+7
echo $v1$i
echo $v2$i
done
linux bash shell-script variable
asked 5 mins ago
user335828user335828
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have a bash script which contains awscli as well. I am trying to print a variable which is created in a for loop. The variable that I am trying to print contains $ sign because of for loop. I couldn't print the value. Below I ams sharing the script. The output of this script is only numbers which is generated in the for loop. I want to print the value which is generated in the command.
!/bin/bash
declare -i counter=11
declare -i counter2=14
for i in {1..2}
do
declare v1$i=$(aws iam get-group --group-name VideoEditors | awk -v counter1=$counter 'NR==counter1' | awk -F" '{print $4}')
counter=$counter+7
declare v2$i=$(aws iam get-group --group-name VideoEditors | awk -v counter3=$counter2 'NR==counter3' | awk -F" '{print $4}')
counter2=$counter2+7
echo $v1$i
echo $v2$i
done
linux bash shell-script variable
linux bash shell-script variable
asked 5 mins ago
user335828user335828
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
user335828user335828
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
user335828user335828
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
user335828user335828
1
asked 5 mins ago
user335828user335828
1
1
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user335828 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Use"${var}"instead of just$var. See unix.stackexchange.com/questions/4899/…
– Kenneth B. Jensen
1 min ago
add a comment |
Use"${var}"instead of just$var. See unix.stackexchange.com/questions/4899/…
– Kenneth B. Jensen
1 min ago
Use
"${var}" instead of just $var. See unix.stackexchange.com/questions/4899/…– Kenneth B. Jensen
1 min ago
Use
"${var}" instead of just $var. See unix.stackexchange.com/questions/4899/…– Kenneth B. Jensen
1 min ago
add a comment |
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Use
"${var}"instead of just$var. See unix.stackexchange.com/questions/4899/…– Kenneth B. Jensen
1 min ago