I have a config file something like below.
_ispip=$_octet.129
_octet=10.89.2
_rxpip=$_octet.132
And when i try to echo the value its not printing the full values for the 1st variable. Is there a easy way to fix this?
# source test.cfg
# echo $_ispip
.129
# echo $_octet
10.89.2
# echo $_rxpip
10.89.2.132
bash doesn't have lazy evaluation, it will try to replace $_octet when you refer to it. If you do this before the assignment, you get an empty string.
You need to put the _octet assignment before _ispip.
_octet=10.89.2
_ispip=$_octet.129
_rxpip=$_octet.132
Related
I am trying to read Xcode build settings via shell script, i.e. if there is a build setting called PRODUCT_NAME, I read its value by writing echo ${PRODUCT_NAME} in shell script.
Now, I get this PRODUCT_NAME in an array, lets call myArr having just one element PRODUCT_NAME. I loop over that array as
for i in "${myarr[#]}"
do
:
echo $i
done
echo $i would simply output PRODUCT_NAME. However, I want to write something that would evaluate ${PRODUCT_NAME} and give me results.
I have also tried eval echo $i but that also outputs PRODUCT_NAME only
Solved it by using
echo "${!i}"
The second line gives the output as desired. I have taken this answer from here: How to get a variable value if variable name is stored as string?
As I interpret your problem, You hold the array of variables, and while traversing, you want to use the actual values of the variables.
I tried the below code which is almost similar to which you have written and seems working.
#!/bin/bash
PRODUCT_NAME="Calendar"
PROCUCT_VER="2.3"
PROCUCT_OWNER="Sam"
PRODUCT_DETAILS=( $PRODUCT_NAME $PROCUCT_VER $PROCUCT_OWNER )
for PRODUCT_PROPERTY in "${PRODUCT_DETAILS[#]}"
do
echo "PRODUCT_PROPERTY: $PRODUCT_PROPERTY"
done
Output:
PRODUCT_PROPERTY: Calendar
PRODUCT_PROPERTY: 2.3
PRODUCT_PROPERTY: Sam
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 ${!VARx} # prints "ONE"
I'm trying to write a shell script to automate a job for me. But i'm currently stuck.
Here's the problem :
I have a variable named var1 (a decreasing number from 25 to 0
and another variable named
var${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