I have a additionaly script help_script.sh:
....
result="example"
and the
main.bash
var=`sh help_script.sh`
echo $var
Is there any way to assign the result from help_script.sh to the var value in the main.sh? I know solution with "echo", but I'm looking for another solution.
Use the source keyword to inline the help_script.sh
This will run the script in the same context so any variables should remain in scope.
See this Super User question for more details https://superuser.com/q/46139/149483
Related
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 ${!info}
I have some code that creates a variable of some name automatically and assigns some value to it. The code is something like the following:
myVariableName="zappo"
eval "${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 ${!VARx} # prints "ONE"
I have a do while loop where I am adding a variable to itself
while read line
do
let variable=$variable+$someOtherVariable
done
return $variable
When I echo the value of $variable I get no output ...
Is this the correct way to add some value back to the variable itself (i.e. i = i+j)
Also, in the context of bash scripting what is the scope in this case..
return returns an "exit" code, a number, not what you are looking for. You should do an echo.
The problem is that the variable is not visible outside of the scope (the assignment is not propagated outside the loop).
The first way that comes to mind is to run the command in a subshell and forcing the loop to emit the variable:
variable=$(variable=0; while read line; do variable=$((variable+someOtherVariable)); done; echo $variable)
I'm trying to write a shell script to automate a job for me. But i'm currently stuck.
Here's the problem :
I have a variable named var1 (a decreasing number from 25 to 0
and another variable named
var${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
This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
Let's say I have a variable's name stored in another variable:
myvar=123
varname=myvar
Now, I'd like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:
function var { v="\$$1"; eval "echo "$v; }
so
var $varname # gives 123
Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious.
From the man page of bash:
${!varname}
If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.
There isn't a direct Posix-conforming syntax, only a bashism. I usually do this:
eval t="\$$varname"
This will work on any Posix shell, including those systems where bash is the login shell and /bin/sh is something smaller and faster like ash. I like bash and use it for my login shell but I avoid bashisms in command files.
Note: One problem with writing bash-specific scripts is that even if you can count on bash being installed, it could be anywhere on the path. It might be a good idea in that case to use the fully general /usr/bin/env shebang style, but note that this is still not 100% portable and has security issues.
${!varname} should do the trick
$ var="content"
$ myvar=var
$ echo ${!myvar}
content
I usually look at Advance Bash-Scripting Guide when I need to freshen up my Bash skills.
Regarding your question look at Indirect References
Notation is:
Version < 2
\$$var
Version >= 2
${!varname}
# bmuSetIndirectVar()
# TO DOUBLE CHECK THIS COMMENT AND DEMO
# This function is an helper to read indirect variables.
# i.e. get the content of a variable whose name is saved
# within an other variable. Like:
# MYDIR="/tmp"
# WHICHDIR="MYDIR"
# bmuSetIndirectVar "WHICHDIR" "$MYDIR"
#
bmuSetIndirectVar(){
tmpVarName=$1
locVarName=$1
extVarName=$2
#echo "debug Ind Input >$1< >$2<"
eval tmpVarName=\$$extVarName
#echo "debug Ind Output >$tmpVarName< >$extVarName<"
export $locVarName="${tmpVarName}"
}
I am currently using this little function. I am not fully happy with it, and I have seen different solutions on the web (if I could recall I would write them here), but it seems to work. Within these few lines there is already some redundancy and extra data but it was helpful for debugging.
If you want to see it in place, i.e. where I am using it, check:
https://github.com/mariotti/bmu/blob/master/bin/backmeup.shellfunctions.sh
Of course it is not the best solution, but made me going on with the work, in
the hope I can replace it with something a bit more general soon.