I am using a bash script in an azure devops pipeline where a variable is created dynamically from one of the pipeline tasks.
I need to use the variable's value in subsequent scripts, I can formulate the string which is used for the variable name, however cannot get the value of the variable using this string. I hope the below example makes it clear on what I need.
Thanks in advance for your help.
PartAPartB="This is my text"
echo "$PartAPartB" #Shows "This is my text" as expected
#HOW DO I GET BELOW TO PRINT "This is my text"
#without using the PartAPartB variable and
#using VarAB value to become the variable name
VarAB="PartAPartB"
VARNAME="$VarAB"
echo $("$VARNAME") #INCORRECT
You can use eval to do this
$ PartAPartB="This is my text"
$ VarAB="PartAPartB"
$ eval "echo \${${VarAB}}"
This is my text
You have two choices with bash. Using a nameref with declare -n (which is the preferred approach for Bash >= 4.26) or using variable indirection. In your case, examples of both would be:
#!/bin/bash
VarAB="PartAPartB"
## using a nameref
declare -n VARNAME=VarAB
echo "$VARNAME" # CORRECT
## using indirection
othervar=VarAB
echo "${!othervar}" # Also Correct
(note: do not use ALLCAPS variable names, those a generally reserved for environment variables or system variables)
Related
I am trying to create an environment variable in bash script, user will input the name of environment variable to be created and will input its value as well.
this is a hard coded way just to elaborate my question :
#!/bin/bash
echo Hello
export varName="nameX" #
echo $varName
export "$varName"="val" #here I am trying to create an environment
#variable whose name is nameX and assigning it value val
echo $nameX
it works fine
it's output is :
Hello
nameX
val
But, I want a generic code. So I am trying to take input from user the name of variable and its value but I am having trouble in it. I don't know how to echo variable whose name is user-defined
echo "enter the environment variable name"
read varName
echo "enter the value to be assigned to env variable"
read value
export "$varName"=$value
Now, I don't know how to echo environment variable
if I do like this :
echo "$varName"
it outputs the name that user has given to environment variable not the value that is assigned to it. how to echo value in it?
Thanks
To get closure: the OP's question boils down to this:
How can I get the value of a variable whose name is stored in another variable in bash?
var='value' # the target variable
varName='var' # the variable storing $var's *name*
gniourf_gniourf provided the solution in a comment:
Use bash's indirection expansion feature:
echo "${!varName}" # -> 'value'
The ! preceding varName tells bash not to return the value of $varName, but the value of the variable whose name is the value of $varName.
The enclosing curly braces ({ and }) are required, unlike with direct variable references (typically).
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
The page above also describes the forms ${!prefix#} and ${!prefix*}, which return a list of variable names that start with prefix.
bash 4.3+ supports a more flexible mechanism: namerefs, via declare -n or, inside functions, local -n:
Note: For the specific use case at hand, indirect expansion is the simpler solution.
var='value'
declare -n varAlias='var' # $varAlias is now another name for $var
echo "$varAlias" # -> 'value' - same as $var
The advantage of this approach is that the nameref is effectively just an another name for the original variable (storage location), so you can also assign to the nameref to update the original variable:
varAlias='new value' # assign a new value to the nameref
echo "$var" # -> 'new value' - the original variable has been updated
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameters.html
Compatibility note:
Indirect expansion and namerefs are NOT POSIX-compliant; a strictly POSIX-compliant shell will have neither feature.
ksh and zsh have comparable features, but with different syntax.
I am trying to create an environment variable in bash script, user will input the name of environment variable to be created and will input its value as well.
this is a hard coded way just to elaborate my question :
#!/bin/bash
echo Hello
export varName="nameX" #
echo $varName
export "$varName"="val" #here I am trying to create an environment
#variable whose name is nameX and assigning it value val
echo $nameX
it works fine
it's output is :
Hello
nameX
val
But, I want a generic code. So I am trying to take input from user the name of variable and its value but I am having trouble in it. I don't know how to echo variable whose name is user-defined
echo "enter the environment variable name"
read varName
echo "enter the value to be assigned to env variable"
read value
export "$varName"=$value
Now, I don't know how to echo environment variable
if I do like this :
echo "$varName"
it outputs the name that user has given to environment variable not the value that is assigned to it. how to echo value in it?
Thanks
To get closure: the OP's question boils down to this:
How can I get the value of a variable whose name is stored in another variable in bash?
var='value' # the target variable
varName='var' # the variable storing $var's *name*
gniourf_gniourf provided the solution in a comment:
Use bash's indirection expansion feature:
echo "${!varName}" # -> 'value'
The ! preceding varName tells bash not to return the value of $varName, but the value of the variable whose name is the value of $varName.
The enclosing curly braces ({ and }) are required, unlike with direct variable references (typically).
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
The page above also describes the forms ${!prefix#} and ${!prefix*}, which return a list of variable names that start with prefix.
bash 4.3+ supports a more flexible mechanism: namerefs, via declare -n or, inside functions, local -n:
Note: For the specific use case at hand, indirect expansion is the simpler solution.
var='value'
declare -n varAlias='var' # $varAlias is now another name for $var
echo "$varAlias" # -> 'value' - same as $var
The advantage of this approach is that the nameref is effectively just an another name for the original variable (storage location), so you can also assign to the nameref to update the original variable:
varAlias='new value' # assign a new value to the nameref
echo "$var" # -> 'new value' - the original variable has been updated
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameters.html
Compatibility note:
Indirect expansion and namerefs are NOT POSIX-compliant; a strictly POSIX-compliant shell will have neither feature.
ksh and zsh have comparable features, but with different syntax.
I am looking for a way to add a string to a variable name in bash and evaluate this as a new variable. Example:
#!/bin/bash
file="filename"
declare ${file}_info="about a file"
declare ${file}_status="status of file"
declare ${file}_number="29083451"
echo ${${file}_info} # <-- This fails with error: "bad substitution"
Is there a way to do this?
I'm not actually implementing this in any production code anywhere. I just want to know if there is some way to create a variable name using a variable and a string. It seemed like an interesting problem.
Also note that I am not asking about bash indirection, which is the use of ${!x} to evaluate the contents of a variable as a variable.
You aren't asking about indirection, but that's what can help you:
info=$file\_info
echo ${!info}
Hi I would like to set a variable by concatenating two other variables.
Example
A=1
B=2
12=C
echo $A$B
desired result being C
however the answer I get is always 12
Is it possible?
UPDATED
Example
A=X
B=Y
D=$A$B
xy=test
echo $D
desired result being "test"
It looks like you want indirect variable references.
BASH allows you to expand a parameter indirectly -- that is, one variable may contain the name of another variable:
# Bash
realvariable=contents
ref=realvariable
echo "${!ref}" # prints the contents of the real variable
But as Pieter21 indicates in his comment 12 is not a valid variable name.
Since 12 is not a valid variable name, here's an example with string variables:
> a='hello'
> b='world'
> declare my_$a_$b='my string'
> echo $my_hello_world
my string
What you are trying to do is (almost) called indirection: http://wiki.bash-hackers.org/syntax/pe#indirection
...I did some quick tests, but it does not seem logical to do this without a third variable - you cannot do concatenated indirection directly as the variables/parts being concatenated do not evaluate to the result on their own - you would have to do another evaluation. I think concatenating them first might be the easiest. That said, there is a chance you could rethink what you're doing.
Oh, and you cannot use numbers (alone or as the starting character) for variable names.
Here we go:
cake="cheese"
var1="ca"
var2="ke"
# this does not work as the indirection sees "ca" and "ke", not "cake". No output.
echo ${!var1}${!var2}
# there might be some other ways of tricking it to do this, but they don't look to sensible as indirection probably needs to work on a real variable.
# ...this works, though:
var3=${var1}${var2}
echo ${!var3}
I have some code that creates a variable of some name automatically and assigns some value to it. The code is something like the following:
myVariableName="zappo"
eval "${myVariableName}=zappo_value"
How would I access the value of this variable using the automatically generated name of the variable? So, I'm looking for some code a bit like the following (but working):
eval "echo ${${myVariableName}}"
(... which may be used in something such as myVariableValue="$(eval "echo ${${myVariableName}}")"...).
Thanks muchly for any assistance
If you think this approach is madness and want offer more general advice, the general idea I'm working on is having variables defined in functions in a library with such names as ${usage} and ${prerequisiteFunctions}. These variables that are defined within functions would be accessed by an interrogation function that can, for instance, ensure that prerequisites etc. are installed. So a loop within this interrogation function is something like this:
for currentFunction in ${functionList}; do
echo "function: ${currentFunction}"
${currentFunction} -interrogate # (This puts the function variables into memory.)
currentInterrogationVariables="${interrogationVariables}" # The variable interrogationVariables contains a list of all function variables available for interrogation.
for currentInterrogationVariable in ${currentInterrogationVariables}; do
echo "content of ${currentInterrogationVariable}:"
eval "echo ${${currentInterrogationVariable}}"
done
done
Thanks again for any ideas!
IIRC, indirection in bash is by !, so try ${!myVariableName}
Try:
echo ${!myVariableName}
It will echo the variable who's name is contained in $myVariableName
For example:
#!/bin/bash
VAR1="ONE"
VAR2="TWO"
VARx="VAR1"
echo ${VARx} # prints "VAR1"
echo ${!VARx} # prints "ONE"