if suppose I have an local variable input <- 100 Can You suggest me How to use the local varaibles in RS.eval(c1,xx <-input) please comment
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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)
I've seen that some people when writing bash script they define local variables inside an if else statement like example 1
Example 1:
#!/bin/bash
function ok() {
local animal
if [ ${A} ]; then
animal="zebra"
fi
echo "$animal"
}
A=true
ok
For another example, this is the same:
Example 2:
#!/bin/bash
function ok() {
if [ ${A} ]; then
local animal
animal="zebra"
fi
echo "$animal"
}
A=true
ok
So, the example above printed the same result but which one is the best practice to follow. I prefer the example 2 but I've seen a lot people declaring local variable inside a function like example 1. Would it be better to declare all local variables on top like below:
function ok() {
# all local variable declaration must be here
# Next statement
}
the best practice to follow
Check your scripts with https://shellcheck.net .
Quote variable expansions. Don't $var, do "$var". https://mywiki.wooledge.org/Quotes
For script local variables, prefer to use lowercase variable names. For exported variables, use upper case and unique variable names.
Do not use function name(). Use name(). https://wiki.bash-hackers.org/scripting/obsolete
Document the usage of global variables a=true. Or add local before using variables local a; then a=true. https://google.github.io/styleguide/shellguide.html#s4.2-function-comments
scope best practice
Generally, use the smallest scope possible. Keep stuff close to each other. Put local close to the variable usage. (This is like the rule from C or C++, to define a variable close to its usage, but unlike in C or C++, in shell declaration and assignment should be on separate lines).
Note that your examples are not the same. In the case variable A (or a) is an empty string, the first version will print an empty line (the local animal variable is empty), the second version will print the value of the global variable animal (there was no local). Although the scope should be as smallest, animal is used outside of if - so local should also be outside.
The local command constrains the variables declared to the function scope.
With that said, you can deduce that doing so inside an if block will be the same as if you did outside of it, as long as it's inside of a function.
vv=1
cc() { local vv=2; echo $vv; unset vv; echo "${vv}3"; }
cc
echo $vv
Gives:
2
3
1
I was expecting:
2
13
1
How can I access the global variable once a variable with the same name has been set local in a function?
I don't think you can. If it's an exported environment variable you can find it by reading the environment, but as far as a global variable masked by a local one, AFAIK you're out of luck. Check the contents of, and copy as necessary, before declaring your local variable.
I'm trying to determine the existing HDDs in each system using a for loop as show below, the problem is when I try to set the variable using the below code i get sda=true: command not found. What is the proper way to do this?
#!/bin/bash
for i in a b c d e f
do
grep -q sd$i /proc/partitions
if [ $? == 0 ]
then
sd$i=true
else
sd$i=false
fi
done
You need to use an array or declare:
declare sd$i=true
I would use an array in this case. For example:
$ i=a
$ sd[$i]=true
$ echo ${sd[a]}
true
As another poster stated, if you want to do this without an array, you can instead make a local variable by using syntax like declare sd$i=true. If you want to make a global variable, use export sd$i=true.
BASH FAQ entry #6: "How can I use variable variables (indirect variables, pointers, references) or associative arrays?": "Assigning indirect/reference variables"
Consider the following Lua code:
local var1, var2;
Is var2 a local variable here? Or is only var1 a local?
Both are local.
Both variables are local, and both are given a value of nil.
To assign them to 2 different values, simply:
local var1,var2 = 1,2