Bash, variables and arrays - arrays

I crawled through lots of boards but didn't find the final solution for my problem.
I have got an array, named "array0", the name is stored to a variable called arrayname. Also I've got a logged IP address, let's say 127.0.0.1, also stored in a variable, called ip.
Now I want to assign the IP to index 3 in the array like that:
"$arrayname"[3]="$ip"
So, this didn't work. I tried lots of ways how to solve that but none worked.
Is anyone out there who can tell me the final solution for this case?
Update: The given opportunities to handle the problem are great! But I forgot to mention that the array I'm working with is just sourced from another file (also written in bash). My goal is now to edit the array in the sourced file itself. Any more ideas for that?

Try
read ${arrayname}[3] <<<"$ip"

You'll need to use the declare command and indirect parameter expansion, but it's a little tricky to use with array names. It helps if you think of the index as part of the variable name, instead of an operator applied to the array name, like in most languages.
array0=(1 2 3 4 5)
arrayname=array0
name=$arrayname[3]
declare "$name=$ip"
echo "${!name}

And yet another way to do it, this time using the versatile printf.
printf -v "$arrayname[3]" %s "$ip"
demo
#!/bin/bash
array0=(a b c d e)
echo "${array0[#]}"
arrayname='array0'
ip='127.0.0.1'
printf -v "$arrayname[3]" %s "$ip"
echo "${array0[#]}"
output
a b c d e
a b c 127.0.0.1 e

See this:
# declare -a arrayname=(element1 element2 element3)
# echo ${arrayname[0]}
element1
# arrayname[4]="Yellow"
# echo ${arrayname[4]}
Yellow
# export ip="192.168.190.23"
# arrayname[5]=$ip
# echo ${arrayname[5]}
192.168.190.23
You don't have to use quotes.

After initializing the arrays, you can access the array elements using their indices as follows.
Access as:
${arrayname[3]}

Related

How to set arrays with variables with loop in bash [duplicate]

I am confused about a bash script.
I have the following code:
function grep_search() {
magic_way_to_define_magic_variable_$1=`ls | tail -1`
echo $magic_variable_$1
}
I want to be able to create a variable name containing the first argument of the command and bearing the value of e.g. the last line of ls.
So to illustrate what I want:
$ ls | tail -1
stack-overflow.txt
$ grep_search() open_box
stack-overflow.txt
So, how should I define/declare $magic_way_to_define_magic_variable_$1 and how should I call it within the script?
I have tried eval, ${...}, \$${...}, but I am still confused.
I've been looking for better way of doing it recently. Associative array sounded like overkill for me. Look what I found:
suffix=bzz
declare prefix_$suffix=mystr
...and then...
varname=prefix_$suffix
echo ${!varname}
From the docs:
The ‘$’ character introduces parameter expansion, command substitution, or arithmetic expansion. ...
The basic form of parameter expansion is ${parameter}. The value of parameter is substituted. ...
If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of indirection. Bash uses the value formed by expanding the rest of parameter as the new parameter; this is then expanded and that value is used in the rest of the expansion, rather than the expansion of the original parameter. This is known as indirect expansion. The value is subject to tilde expansion, parameter expansion, command substitution, and arithmetic expansion. ...
Use an associative array, with command names as keys.
# Requires bash 4, though
declare -A magic_variable=()
function grep_search() {
magic_variable[$1]=$( ls | tail -1 )
echo ${magic_variable[$1]}
}
If you can't use associative arrays (e.g., you must support bash 3), you can use declare to create dynamic variable names:
declare "magic_variable_$1=$(ls | tail -1)"
and use indirect parameter expansion to access the value.
var="magic_variable_$1"
echo "${!var}"
See BashFAQ: Indirection - Evaluating indirect/reference variables.
Beyond associative arrays, there are several ways of achieving dynamic variables in Bash. Note that all these techniques present risks, which are discussed at the end of this answer.
In the following examples I will assume that i=37 and that you want to alias the variable named var_37 whose initial value is lolilol.
Method 1. Using a “pointer” variable
You can simply store the name of the variable in an indirection variable, not unlike a C pointer. Bash then has a syntax for reading the aliased variable: ${!name} expands to the value of the variable whose name is the value of the variable name. You can think of it as a two-stage expansion: ${!name} expands to $var_37, which expands to lolilol.
name="var_$i"
echo "$name" # outputs “var_37”
echo "${!name}" # outputs “lolilol”
echo "${!name%lol}" # outputs “loli”
# etc.
Unfortunately, there is no counterpart syntax for modifying the aliased variable. Instead, you can achieve assignment with one of the following tricks.
1a. Assigning with eval
eval is evil, but is also the simplest and most portable way of achieving our goal. You have to carefully escape the right-hand side of the assignment, as it will be evaluated twice. An easy and systematic way of doing this is to evaluate the right-hand side beforehand (or to use printf %q).
And you should check manually that the left-hand side is a valid variable name, or a name with index (what if it was evil_code # ?). By contrast, all other methods below enforce it automatically.
# check that name is a valid variable name:
# note: this code does not support variable_name[index]
shopt -s globasciiranges
[[ "$name" == [a-zA-Z_]*([a-zA-Z_0-9]) ]] || exit
value='babibab'
eval "$name"='$value' # carefully escape the right-hand side!
echo "$var_37" # outputs “babibab”
Downsides:
does not check the validity of the variable name.
eval is evil.
eval is evil.
eval is evil.
1b. Assigning with read
The read builtin lets you assign values to a variable of which you give the name, a fact which can be exploited in conjunction with here-strings:
IFS= read -r -d '' "$name" <<< 'babibab'
echo "$var_37" # outputs “babibab\n”
The IFS part and the option -r make sure that the value is assigned as-is, while the option -d '' allows to assign multi-line values. Because of this last option, the command returns with an non-zero exit code.
Note that, since we are using a here-string, a newline character is appended to the value.
Downsides:
somewhat obscure;
returns with a non-zero exit code;
appends a newline to the value.
1c. Assigning with printf
Since Bash 3.1 (released 2005), the printf builtin can also assign its result to a variable whose name is given. By contrast with the previous solutions, it just works, no extra effort is needed to escape things, to prevent splitting and so on.
printf -v "$name" '%s' 'babibab'
echo "$var_37" # outputs “babibab”
Downsides:
Less portable (but, well).
Method 2. Using a “reference” variable
Since Bash 4.3 (released 2014), the declare builtin has an option -n for creating a variable which is a “name reference” to another variable, much like C++ references. Just as in Method 1, the reference stores the name of the aliased variable, but each time the reference is accessed (either for reading or assigning), Bash automatically resolves the indirection.
In addition, Bash has a special and very confusing syntax for getting the value of the reference itself, judge by yourself: ${!ref}.
declare -n ref="var_$i"
echo "${!ref}" # outputs “var_37”
echo "$ref" # outputs “lolilol”
ref='babibab'
echo "$var_37" # outputs “babibab”
This does not avoid the pitfalls explained below, but at least it makes the syntax straightforward.
Downsides:
Not portable.
Risks
All these aliasing techniques present several risks. The first one is executing arbitrary code each time you resolve the indirection (either for reading or for assigning). Indeed, instead of a scalar variable name, like var_37, you may as well alias an array subscript, like arr[42]. But Bash evaluates the contents of the square brackets each time it is needed, so aliasing arr[$(do_evil)] will have unexpected effects… As a consequence, only use these techniques when you control the provenance of the alias.
function guillemots {
declare -n var="$1"
var="«${var}»"
}
arr=( aaa bbb ccc )
guillemots 'arr[1]' # modifies the second cell of the array, as expected
guillemots 'arr[$(date>>date.out)1]' # writes twice into date.out
# (once when expanding var, once when assigning to it)
The second risk is creating a cyclic alias. As Bash variables are identified by their name and not by their scope, you may inadvertently create an alias to itself (while thinking it would alias a variable from an enclosing scope). This may happen in particular when using common variable names (like var). As a consequence, only use these techniques when you control the name of the aliased variable.
function guillemots {
# var is intended to be local to the function,
# aliasing a variable which comes from outside
declare -n var="$1"
var="«${var}»"
}
var='lolilol'
guillemots var # Bash warnings: “var: circular name reference”
echo "$var" # outputs anything!
Source:
BashFaq/006: How can I use variable variables (indirect variables, pointers, references) or associative arrays?
BashFAQ/048: eval command and security issues
Example below returns value of $name_of_var
var=name_of_var
echo $(eval echo "\$$var")
Use declare
There is no need on using prefixes like on other answers, neither arrays. Use just declare, double quotes, and parameter expansion.
I often use the following trick to parse argument lists contanining one to n arguments formatted as key=value otherkey=othervalue etc=etc, Like:
# brace expansion just to exemplify
for variable in {one=foo,two=bar,ninja=tip}
do
declare "${variable%=*}=${variable#*=}"
done
echo $one $two $ninja
# foo bar tip
But expanding the argv list like
for v in "$#"; do declare "${v%=*}=${v#*=}"; done
Extra tips
# parse argv's leading key=value parameters
for v in "$#"; do
case "$v" in ?*=?*) declare "${v%=*}=${v#*=}";; *) break;; esac
done
# consume argv's leading key=value parameters
while test $# -gt 0; do
case "$1" in ?*=?*) declare "${1%=*}=${1#*=}";; *) break;; esac
shift
done
Combining two highly rated answers here into a complete example that is hopefully useful and self-explanatory:
#!/bin/bash
intro="You know what,"
pet1="cat"
pet2="chicken"
pet3="cow"
pet4="dog"
pet5="pig"
# Setting and reading dynamic variables
for i in {1..5}; do
pet="pet$i"
declare "sentence$i=$intro I have a pet ${!pet} at home"
done
# Just reading dynamic variables
for i in {1..5}; do
sentence="sentence$i"
echo "${!sentence}"
done
echo
echo "Again, but reading regular variables:"
echo $sentence1
echo $sentence2
echo $sentence3
echo $sentence4
echo $sentence5
Output:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
Again, but reading regular variables:
You know what, I have a pet cat at home
You know what, I have a pet chicken at home
You know what, I have a pet cow at home
You know what, I have a pet dog at home
You know what, I have a pet pig at home
This will work too
my_country_code="green"
x="country"
eval z='$'my_"$x"_code
echo $z ## o/p: green
In your case
eval final_val='$'magic_way_to_define_magic_variable_"$1"
echo $final_val
This should work:
function grep_search() {
declare magic_variable_$1="$(ls | tail -1)"
echo "$(tmpvar=magic_variable_$1 && echo ${!tmpvar})"
}
grep_search var # calling grep_search with argument "var"
An extra method that doesn't rely on which shell/bash version you have is by using envsubst. For example:
newvar=$(echo '$magic_variable_'"${dynamic_part}" | envsubst)
For zsh (newers mac os versions), you should use
real_var="holaaaa"
aux_var="real_var"
echo ${(P)aux_var}
holaaaa
Instead of "!"
As per BashFAQ/006, you can use read with here string syntax for assigning indirect variables:
function grep_search() {
read "$1" <<<$(ls | tail -1);
}
Usage:
$ grep_search open_box
$ echo $open_box
stack-overflow.txt
Even though it's an old question, I still had some hard time with fetching dynamic variables names, while avoiding the eval (evil) command.
Solved it with declare -n which creates a reference to a dynamic value, this is especially useful in CI/CD processes, where the required secret names of the CI/CD service are not known until runtime. Here's how:
# Bash v4.3+
# -----------------------------------------------------------
# Secerts in CI/CD service, injected as environment variables
# AWS_ACCESS_KEY_ID_DEV, AWS_SECRET_ACCESS_KEY_DEV
# AWS_ACCESS_KEY_ID_STG, AWS_SECRET_ACCESS_KEY_STG
# -----------------------------------------------------------
# Environment variables injected by CI/CD service
# BRANCH_NAME="DEV"
# -----------------------------------------------------------
declare -n _AWS_ACCESS_KEY_ID_REF=AWS_ACCESS_KEY_ID_${BRANCH_NAME}
declare -n _AWS_SECRET_ACCESS_KEY_REF=AWS_SECRET_ACCESS_KEY_${BRANCH_NAME}
export AWS_ACCESS_KEY_ID=${_AWS_ACCESS_KEY_ID_REF}
export AWS_SECRET_ACCESS_KEY=${_AWS_SECRET_ACCESS_KEY_REF}
echo $AWS_ACCESS_KEY_ID $AWS_SECRET_ACCESS_KEY
aws s3 ls
Wow, most of the syntax is horrible! Here is one solution with some simpler syntax if you need to indirectly reference arrays:
#!/bin/bash
foo_1=(fff ddd) ;
foo_2=(ggg ccc) ;
for i in 1 2 ;
do
eval mine=( \${foo_$i[#]} ) ;
echo ${mine[#]}" " ;
done ;
For simpler use cases I recommend the syntax described in the Advanced Bash-Scripting Guide.
KISS approach:
a=1
c="bam"
let "$c$a"=4
echo $bam1
results in 4
I want to be able to create a variable name containing the first argument of the command
script.sh file:
#!/usr/bin/env bash
function grep_search() {
eval $1=$(ls | tail -1)
}
Test:
$ source script.sh
$ grep_search open_box
$ echo $open_box
script.sh
As per help eval:
Execute arguments as a shell command.
You may also use Bash ${!var} indirect expansion, as already mentioned, however it doesn't support retrieving of array indices.
For further read or examples, check BashFAQ/006 about Indirection.
We are not aware of any trick that can duplicate that functionality in POSIX or Bourne shells without eval, which can be difficult to do securely. So, consider this a use at your own risk hack.
However, you should re-consider using indirection as per the following notes.
Normally, in bash scripting, you won't need indirect references at all. Generally, people look at this for a solution when they don't understand or know about Bash Arrays or haven't fully considered other Bash features such as functions.
Putting variable names or any other bash syntax inside parameters is frequently done incorrectly and in inappropriate situations to solve problems that have better solutions. It violates the separation between code and data, and as such puts you on a slippery slope toward bugs and security issues. Indirection can make your code less transparent and harder to follow.
For indexed arrays, you can reference them like so:
foo=(a b c)
bar=(d e f)
for arr_var in 'foo' 'bar'; do
declare -a 'arr=("${'"$arr_var"'[#]}")'
# do something with $arr
echo "\$$arr_var contains:"
for char in "${arr[#]}"; do
echo "$char"
done
done
Associative arrays can be referenced similarly but need the -A switch on declare instead of -a.
POSIX compliant answer
For this solution you'll need to have r/w permissions to the /tmp folder.
We create a temporary file holding our variables and leverage the -a flag of the set built-in:
$ man set
...
-a Each variable or function that is created or modified is given the export attribute and marked for export to the environment of subsequent commands.
Therefore, if we create a file holding our dynamic variables, we can use set to bring them to life inside our script.
The implementation
#!/bin/sh
# Give the temp file a unique name so you don't mess with any other files in there
ENV_FILE="/tmp/$(date +%s)"
MY_KEY=foo
MY_VALUE=bar
echo "$MY_KEY=$MY_VALUE" >> "$ENV_FILE"
# Now that our env file is created and populated, we can use "set"
set -a; . "$ENV_FILE"; set +a
rm "$ENV_FILE"
echo "$foo"
# Output is "bar" (without quotes)
Explaining the steps above:
# Enables the -a behavior
set -a
# Sources the env file
. "$ENV_FILE"
# Disables the -a behavior
set +a
While I think declare -n is still the best way to do it there is another way nobody mentioned it, very useful in CI/CD
function dynamic(){
export a_$1="bla"
}
dynamic 2
echo $a_2
This function will not support spaces so dynamic "2 3" will return an error.
for varname=$prefix_suffix format, just use:
varname=${prefix}_suffix

Bash-Scripting: what does ${VARIABLE[number]} mean if VARIABLE is not an array?

I know that there are arrays in bash-scripting. For example:
JOBS=("JOB1", "JOB2", "JOB3")
Then one can refer to, say, JOB2 as follows:
${JOBS[1]}
In a bash script I recently encountered a normal (non-array) variable:
JOB="XYZ"
Later in the script, this variable was referred to as follows:
${JOB[0]}
and:
${JOB[1]}
Since JOB is not an array, I do not understand what ${JOB[number]} is expanded to.
Is this just a programming mistake? Or is there some construct regarding normal variables I am not aware of?
bash doesn't have array values at all; it provides array syntax to use with names that have an array attribute set:
$ a=()
$ declare -p a
declare -a a=()
The -a in the output indicates that the array attribute is set on a, having been set implicitly by the preceding array assignment.
For such variables, the name-plus-index acts as a sort of "virtual" name, and like any other name, expands to the empty string if the name doesn't actually exist, as is the case with ${JOB[1]}. A side effect of the implementation is the $foo and ${foo[0]} are generally equivalent, whether or not foo has its array attribute set.
I believe newer versions of Bash support one-dimensional arrays and the syntax can be seen in the following link as follows :
Advanced Bash-Scripting Guide
If Variable is not an array, ${Variable[Number]} will work only for Number=0 but not for any other number as ${Variable[0]} and $Variable - both are same.

How can I reference an existing bash array using a 2nd variable containing the name of the array?

My closest most helpful matches when I searched for an answer ahead of posting:
Iterate over array in shell whose name is stored in a variable
How to use an argument/parameter name as a variable in a bash script
How to iterate over an array using indirect reference?
My attempt with partial success:
#!/bin/bash
declare -a large_furry_mammals
declare -a array_reference
# I tried both declaring array_reference as an array and
# not declaring it as an array. no change in behavior.
large_furry_mammals=(horse zebra gorilla)
size=large
category=mammals
tmp="${size}_furry_${category}"
eval array_reference='$'$tmp
echo tmp=$tmp
echo array_reference[0]=${array_reference[0]}
echo array_reference[1]=${array_reference[1]}
Output
tmp=large_furry_mammals
array_reference[0]=horse
array_reference[1]=
Expectation
I would have expected to get the output zebra when I echoed array_reference[1].
...but I'm missing some subtlety...
Why can I not access elements of the index array beyond index 0?
This suggests that array_reference is not actually being treated as an array.
I'm not looking to make a copy of the array. I want to reference (what will be) a static array based on a variable pointing to that array, i.e., ${size}_furry_${category} -> large_furry_mammals.
I've been successful with the general idea here using the links I've posted but only as long as its not an array. When it's an array, it's falling down for me.
Addendum Dec 5, 2018
bash 4.3 is not available in this case. #benjamin's answer does work on under 4.3.
I'll be needing to loop over the resulting array variable's contents. This kinda dumb example I gave involving mammals was just to describe the concept. There's actually a real world case around this. I have set of static reference arrays and an input string would be parsed to select which array was relevant and then I will loop over the array that was selected. I could do a case statement but with more than 100 reference arrays that would be the direct but overly verbose way to do it.
This pseudo code is probably better example of what I'm going after.
m1_array=(x a r d)
m2_array=(q 3 fg d)
m3_array=(c e p)
Based on some logic...select which array prefix you need.
x=m1
for each element in ${x}_array
do
some-task
done
I'm doing some testing with #eduardo's solution to see if I can adapt the way he references the variables to get to my endgame.
** Addendum #2 December 14, 2018 **
Solution
I found it! Working with #eduardo's example I came up with the following:
#!/bin/bash
declare -a large_furry_mammals
#declare -a array_reference
large_furry_mammals=(horse zebra gorilla)
size=large
category=mammals
tmp="${size}_furry_${category}[#]"
for element in "${!tmp}"
do
echo $element
done
Here is what execution looks like. We successfully iterate over the elements of the array string that was built dynamically.
./example3b.sh
horse
zebra
gorilla
Thank you everyone.
If you have Bash 4.3 or newer, you can use namerefs:
large_furry_mammals=(horse zebra gorilla)
size=large
category=mammals
declare -n array_reference=${size}_furry_$category
printf '%s\n' "${array_reference[#]}"
with output
horse
zebra
gorilla
This is a reference, so changes are reflected in both large_furry_mammals and array_reference:
$ array_reference[0]='donkey'
$ large_furry_mammals[3]='llama'
$ printf '%s\n' "${array_reference[#]}"
donkey
zebra
gorilla
llama
$ printf '%s\n' "${large_furry_mammals[#]}"
donkey
zebra
gorilla
llama
declare -a large_furry_mammals
declare -a array_reference
large_furry_mammals=(horse zebra gorilla)
size=large
category=mammals
echo ${large_furry_mammals[#]}
tmp="${size}_furry_${category}"
array_reference=${tmp}"[1]"
eval ${array_reference}='bear'
echo tmp=$tmp
echo ${large_furry_mammals[#]}

Indirect referencing in Korn Shell

How can i refer an indirect variable in the korn shell
Suppose i've an Variable
FLAG_1=0
and i want this variable value to changed from the array that i've set
set -A Flags_array
Flags_array[0]=? #how to achieve this here
For example if i try
Flags_array[0]=$FLAG_1 # this won't work as this assign the value
I tried doing something like this
Flags_array[0]=FLAG_1
echo ${Flags_array[0]}
But this gives output FLAG_1
I tried using nameref but it's giving me error
$ nameref Flags_array[0]=FLAG_1
ksh: nameref: not found
Please tell me what is proper way to achieving this ?
because what i want to do is later
if i write Flags_array[0]=10 the value of FLAG_1 becomes 10
i just tried it on my console, and it seems that you are using an invalid option for set.... try using small a instead of capital:
set -a Flags_array
just did it like this in my console and worked:
set -a array
number=5
array[0]= $number
echo $array[0] --------> from which i got result : 5[0]
also just to know for future reference, you dont need the curly brackets {}, these you use only if you are calling a variable in a place were actual data would be expected (sorry for the expression), for example here you would need them:
were e.g. random=5
sed "${random}s/^.*$/1/" test.txt
Hope this clears things for you.. :)

Using a variable name to create an array bash, unix

First I should perhaps explain what I want to do...
I have 'n' amounts of files with 'n' amount of lines. All I know is
that the line count will be even.
The user selects the files that they want. This is saved into an
array called ${selected_sets[#]}.
The program will print to screen a randomly selected 'odd numbered'
line from a randomly selected file.
Once the line has been printed, I don't want it printed again...
Most of it is fine, but I am having trouble creating arrays based on the contents of ${selected_sets[#]}... I think I have got my syntax all wrong :)
for i in ${selected_sets[#]}
do
x=1
linecount=$(cat $desired_path/$i | wc -l) #get line count of every set
while [ $x -le $linecount ]
do ${i}[${#${i}[#]}]=$x
x=$(($x+2)) # only insert odd numbers up to max limit of linecount
done
done
The problem is ${i}[${#${i}[#]}]=$x
I know that I can use array[${#array[#]}]=$x but I don't know how to use a variable name.
Any ideas would be most welcome (I am really stumped)!!!
In general, this type is question is solved with eval. If you want a a variable named "foo" and have a variable bar="foo", you simply do:
eval $bar=5
Bash (or any sh) treats that as if you had typed
foo=5
So you may just need to write:
eval ${i}[\${#${i}[#]}]=$x
with suitable escapes. (A useful technique is to replace 'eval' with 'echo', run the script and examine the output and make sure it looks like what you want to be evaluated.)
You can create named variables using the declare command
declare -a name=${#${i}[#]}
I'm just not sure how you would then reference those variables, I don't have time to investigate that now.
Using an array:
declare -a myArray
for i in ${selected_sets[#]}
do
x=1
linecount=$(cat $desired_path/$i | wc -l) #get line count of every set
while [ $x -le $linecount ]
do
$myArray[${#${i}[#]}]=$x
let x=x+1 #This is a bit simpler!
done
done
Beware! I didn't test any of the above. HTH

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