This question already has answers here:
Dynamic variable names in Bash
(19 answers)
how to use variable variable names in bash script [duplicate]
(1 answer)
Closed 6 years ago.
How do I assign a value to variable that has a variable in its name?
var1="file"
var2_$var1="folder"
The code above gives me the error -bash: var2_file=folder: command not found. I was curious to know how to assign to a variable with another variable in its name.
Version of Bash is "GNU bash, version 4.1.2"
With bash you can use declare:
declare var2_$var1="123"
How about using another variable to hold the dynamic name and use it for retrieving the value after setting?
new_var=var2_$var1
declare var2_$var1="123"
echo "${!new_var}" # => 123
Unfortunately, Bash doesn't allow declare $new_var="123" - that would have made this a little prettier.
Related
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 12 months ago.
I want to execute a string command and then get the result of it to assign a variable in shell script bash.
for example :
strCode="scontrol show jodid --dd $VALUE"
eval $strCode
The point here is this codepart worked but how can I get the results from the `
eval
` command and assign it to a variable, because I need it to use.
when I type the theResult=eval $strCode it does not work.
thanks for the help
The solution is:
strCode=$(scontrol show jodid --dd $VALUE)
echo "${strCode}"
It's easier than you are doing:
theResult=$(scontrol show jodid --dd $VALUE)
You should follow some bash tutorial, as this is a quite basic question.
This question already has answers here:
Replace a string in shell script using a variable
(12 answers)
sed substitution with Bash variables
(6 answers)
Closed 4 years ago.
I need to replace current value in configuration file with new value which is assigned to variable ,
like
file_name=abc.txt
needs to be replaced like
file_name=xyz.txt
where $file=xyz.txt
I tried
sed -i 's/file_name=.*/file_name=$file/g' conf_file.conf
however the variable is not getting expanded,
it comes like file_name=$file.
any pointers?
This should work,assuming that variable file has value:xyz.txt assigned to it:
sed "s/file_name=.*/file_name=${file}/g" file_name
Output:
file_name=xyz.txt
This question already has answers here:
Lookup shell variables by name, indirectly [duplicate]
(5 answers)
How to get a variable value if variable name is stored as string?
(10 answers)
Bash - variable variables [duplicate]
(4 answers)
Closed 4 years ago.
I have a set of directories listed inside a text file as
DIR_A= (name of directory 1)
DIR_B= (name of directory 2)
....
I have a second script to which I would like to pass an argument like sh Scriptname varname where varname could be A, B, ...
Scriptname sources the initial text file and I would like this script to accept the passed varname (using $1) to echo DIR_varname.
Any ideas? TIA
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Variable variable assignment error -"command not found"
(1 answer)
Why does a space in a variable assignment give an error in Bash? [duplicate]
(3 answers)
Closed 4 years ago.
So I've got a shell script to do some lazy stuff for if the directory isn't changing for a user. It's below. Essentially, it should be an if statement that if the user enters "default" for the directory, it'll pull them to the default directory for the files. However, I'm getting a command not found on line 16, which is the reassignment statement.
The entire if statement below:
if [ $directory = "default" ];
then
echo Enter your ldap:
read $ldap
$directory = "/usr/local/home/google/${ldap}/Downloads"
fi
I've tried doing it without the dollar sign too...nothing. What's going on here? New to shell, couldn't find this question asked before either.
This is how you should assign a value to a variable in shell:
directory="/usr/local/home/google/${ldap}/Downloads"
No dollar ($) sign.
No space around equal (=) sign.
Also, you should wrap your variables inside double quotes ("). This way, you avoid errors with undefined variables, arguments with spaces, etc.
That gives us:
if [ "$directory" = "default" ]
then
echo "Enter your ldap:"
read $ldap
directory="/usr/local/home/google/${ldap}/Downloads"
fi
This question already has answers here:
Why does a space in a variable assignment give an error in Bash? [duplicate]
(3 answers)
Closed 4 years ago.
I'm newer on bash scripting ,I have a global variable that I want to change his value insead a loop in my script but still get an error that commande not found
this my script :
SCRIPT_BASE = "/home/scripts/test-Scripts"
CURRENT_SCRIPT_PATH = ""
declare -a arr=("A" "B" "C" "D")
for i in "${arr[#]}"
do
if [ $i == "A" ]; then
CURRENT_SCRIPT_PATH = $SCRIPT_BASE
echo -e "Current Path : $CURRENT_SCRIPT_PATH"
fi
done
when I run this script I get that CURRENT_SCRIPT_PATH commande not found
Thanks in advance for any help
In bash you should be really cautious about spaces in if conditions but also when you assign a value to a variable.
Replace in your code the following tree lines:
SCRIPT_BASE="/home/scripts/test-Scripts"
CURRENT_SCRIPT_PATH=""
CURRENT_SCRIPT_PATH=$SCRIPT_BASE
If you keep a space after the variable name bash will interpret it as a command and as you do not have commands SCRIPT_BASE, CURRENT_SCRIPT_PATH, CURRENT_SCRIPT_PATH in your current $PATH you have the error command not found that is produced.