I want to declare a variable, the name of which comes from the value of another variable, and I wrote the following piece of code:
a="bbb"
$a="ccc"
but it didn't work. What's the right way to get this job done?
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If you are on Bash 5+ the better solution is nearly always an associative array. The duplicate has several demonstrations of this.tripleee– tripleee2023-01-04 07:18:18 +00:00Commented Jan 4, 2023 at 7:18
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6 Answers
In highly portable shell code, eval is used for this, but if you do it naively, there are going to be nasty escaping issues. This sort of thing is generally safe:
name_of_variable=abc
eval $name_of_variable="simpleword" # abc set to simpleword
This breaks:
eval $name_of_variable="word splitting occurs"
The fix:
eval $name_of_variable="\"word splitting occurs\"" # not anymore
The ultimate fix: put the text you want to assign into a variable. Let's call it safevariable. Then you can do this:
eval $name_of_variable=\$safevariable # note escaped dollar sign
Escaping the dollar sign solves all escape issues. The dollar sign survives verbatim into the eval function, which will effectively perform this:
eval 'abc=$safevariable' # dollar sign now comes to life inside eval!
And of course this assignment is immune to everything. safevariable can contain *, spaces, $, etc. (The caveat being that we're assuming name_of_variable contains nothing but a valid variable name, and one we are free to use: not something special.)
The Korn shell introduces something called namerefs which Bash also implements. With namerefs we can do this directly:
declare -n foo=bar
Now foo is a nameref which holds the name bar. The nameref behaves as a variable. When we assign to foo, we are actually assinging to bar, and $foo produces the value $bar. It's like a symbolic link for variables.
Namerefs provide a much cleaner and safer solution for indirecting on variables than eval at the loss of maximal portability.
3 Comments
inetphantom
So the
$safevariable is the $safe_target_value to assign the variable name saved in $name_of_variable ?
Stony
eval $name_of_variable="word\ splitting\ occurs" also works
Charles Duffy
It works only if the backslashes remain part of the string and aren't eaten during the initial parsing pass. That's not documented/guaranteed/version-independent behavior.
You can use declare and !, like this:
John="nice guy"
programmer=John
echo ${!programmer} # echos nice guy
Second example:
programmer=Ines
declare $programmer="nice gal"
echo $Ines # echos nice gal
6 Comments
Golar Ramblar
@leaf: Maybe the
-a-option of declare helps? See declare --help, and test for yourself.
Mr. Developerdude
It is nice gesture to say such nice things about your colleagues :D:D
Mr. Developerdude
Just a note, this will not work inside a function, for that I found KAz' eval solution to work.
Michael Burr
Bash docs call the
${!var_name} form "indirect expansion": gnu.org/software/bash/manual/html_node/… |
This might work for you:
foo=bar
declare $foo=baz
echo $bar
baz
or this:
foo=bar
read $foo <<<"baz"
echo $bar
baz
6 Comments
meso_2600
correct, better to use declare over eval (security reasons)
niieani
@meso_2600 declare does not behave the same way as assignment, because it will change the scope of the variable to the local one (you won't be able to modify a variable from a parent scope with declare, unless you declare it as a global variable).
bhu Boue vidya
your second answer is great. no subshell spawned. cheers.
Golar Ramblar
@niieani: According to
declare --help, the -g option to declare always declares a global variable.niieani
@GolarRamblar that's correct. I didn't state otherwise, unless you're referring to something else?
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You could make use of eval for this.
Example:
$ a="bbb"
$ eval $a="ccc"
$ echo $bbb
ccc
Hope this helps!
Comments
If you want to get the value of the variable instead of setting it you can do this
var_name1="var_name2"
var_name2=value_you_want
eval temp_var=\$$var_name1
echo "$temp_var"
You can read about it here indirect references.
Comments
You can assign a value to a variable using simple assignment using a value from another variable like so:
#!/usr/bin/bash
#variable one
a="one"
echo "Variable a is $a"
#variable two with a's variable
b="$a"
echo "Variable b is $b"
#change a
a="two"
echo "Variable a is $a"
echo "Variable b is $b"
The output of that is this:
Variable a is one
Variable b is one
Variable a is two
Variable b is one
So just be sure to assign it like this b="$a" and you should be good.
1 Comment
Kaz
The question is how to indirect upon computed variable names, now how to assign expressions fixed variables.