I have a line of code
eval echo \$$var
which prints a string. How can I store this string into a variable?
Like this:
eval c=\$$var
a better, safer way is to use indirection:
c=${!var}
newVariable=$(eval echo \$$var)
You can try this: It worked for me on bash.
c=$(eval echo \${$var})
Example: If HOME is the environment variable that contains your home directory like /export/users/john, then embedding a variable within another variable and then unpacking that combination would work like this:
var=HOME
c=$(eval echo \${$var})
echo $c
/export/users/john
While this might look unnecessarily convoluted it will be useful in certain scenarios.
This can achieved by using the back quote syntax : ` `
c=`$var`
Related
I'm trying to use a variable as part of another variable's name.
For example, I have some variables in a CONFIG_ scope.
echo $CONFIG_systema_value > "testconfig"
echo $CONFIG_systemb_value > "testconfig2"
and then I have a "deciding" variable like WANTEDSYSTEM
echo $WANTEDSYSTEM > "systema"
I need to be able to "dynamically" switch the config depending on the wantedsystem variable.
My initial attempts I tried something like this echo $CONFIG_$WANTEDSYSTEM_value which I would like to have output testconfig as if I was asking it to echo $CONFIG_systema_value
You can use the ${!name} syntax to indirectly reference a variable, like this:
CONFIG_systema_value=foo
CONFIG_systemb_value=bar
WANTEDSYSTEM=systema
target_var_name="CONFIG_${WANTEDSYSTEM}_value"
echo ${!target_var_name}
(Note that this is bash-specific syntax, but it sounds like that's what you're looking for.)
I would like to know if I can substitute a variable twice.
For example:
#global variable
TEST_SERV_EXT=""
#variables become from myconf.sh
TEST_SERV_EXT_FO='foo01'
TEST_SERV_EXT_BR='bar01'
I want dynamically construct those last two and assign them in TEST_SERV_EXT.
I tried something like this ${$TEST_SERV_COMP} but I'm getting "bad substitution" message.
I need something like php's feature "$$" or tcl's subst command.
Regards,
thandem
TEST_SERV_COMP=TEST_SERV_EXT_FO
TEST_SERV_EXT=${!TEST_SERV_COMP}
Look for indirect expansion in the bash manual.
I have some code that creates a variable of some name automatically and assigns some value to it. The code is something like the following:
myVariableName="zappo"
eval "${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"
I have multiple variables in a shell script; i was trying to save some code duplication and wanted to do something like following
# variables
FLAG=SIM
SIM_ICR_KEY_VAL="http://www.example.com/simi/icr"
REAL_ICR_KEY_VAL="http://www.example.com/real"
Based on the FLAG value i want to access the correct variable (without using IF's)
When i try this it echos the variable name & not the value itself.
echo $(echo ${FLAG}_ICR_KEY_VAL)
On further note; i need to use these substitutions inline in a sed statememt:
sed "s!${ISTR_KEY}=.*!${ISTR_KEY}=${SIM_ISTR_KEY_VAL}!" > tmp.file
... i am not sure its possible or not, please suggest
Reflection can be achieved with the infamous eval:
eval thisvar=\$${FLAG}_INC_KEY_VAL;
echo "We are using $thisvar"
Whenever you find yourself dynamically synthesizing a variable name, though, you are probably Doing It Wrong. You should consider alternatives like arrays:
ICR_KEY_VAL[0]="http://www.example.com/simi/icr"
ICR_KEY_VAL[1]="http://www.example.com/real"
SIM=0
echo ${ICR_KEY_VAL[$SIM]}
I don't know how to do it directly, but in bash you can do it indirectly:
FLAG=SIM
SIM_ICR_KEY_VAL="http://www.example.com/simi/icr"
REAL_ICR_KEY_VAL="http://www.example.com/real"
FLAG_ICR_KEY_VAL=${FLAG}_ICR_KEY_VAL
sed "s!${ISTR_KEY}=.*!${ISTR_KEY}=${!FLAG_ISTR_KEY_VAL}!" > tmp.file
I'm trying to write a shell script to automate a job for me. But i'm currently stuck.
Here's the problem :
I have a variable named var1 (a decreasing number from 25 to 0
and another variable named
var${var1} and this equals to some string.
then when i try to call var${var1} in anywhere in script via echo it fails.
I have tried $[var$var1], ${var$var} and many others but everytime it fails and gives the value of var1 or says operand expected error.
Thanks for your help
It's probably better if you use an array, but you can use indirection:
var25="some string"
var1=25
indirect_var="var$var1"
echo ${!indirect_var} # echoes "some string"
There's only one round of variable expansion, so you can't do it directly. You could use eval:
eval echo \${var$var1}
A better solution is to use an array:
i=5
var[$i]='foo'
echo ${var[$i]}
It sounds like you need bash variable indirection. Take a look at the link below.
http://mywiki.wooledge.org/BashFAQ/006