Related
Not critical - but I'm trying to get a deeper understanding of bash scripting and this is driving me crazy!
My goal - in a bash script:
Define a function that accepts arguments
Set a bash variable (CMD)
with the function name & arguments
Execute it by $CMD
No problem if there is no whitespace in $args - but here's a minimal script to illustrate:
#!/bin/bash
function tstArgs () {
echo "In tstArgs: ArgCnt:$#; Arg1:[$1]; Arg2:[$2]"
}
ARG1=today
ARG2=tomorrow
CMD1="tstArgs $ARG1 $ARG2"
$CMD1 #Output - As Desired: In tstArgs: ArgCnt:2; Arg1:[today]; Arg2:[tomorrow]
ARGWS1="'today with spaces'"
ARGWS2="'tomorrow with spaces'"
CMD2="tstArgs $ARGWS1 $ARGWS2"
$CMD2 #Output: In tstArgs: ArgCnt:6; Arg1:[today]; Arg2:[with]
#The dream:
ARGARR=($ARGWS1 $ARGWS2)
CMD3="tstArgs ${ARGARR[#]}"
$CMD3 #Output: In tstArgs: ArgCnt:6; Arg1:[today]; Arg2:[with]
#ETC, ETC, ETC...
This doesn't show the COUNTLESS variations I tried - single quotes, double quotes, escaping quotes, changing IFS, using parameter escape operators ${ARG1#Q}, setting args w. echo XXX - and so much more - way too many to include here, but to be clear, I didn't just jump on stackoverflow without first spending HOURS.
Weirdly, I can use params w. whitespace if I call the function directly:
tstArgs $ARG1 $ARG2
#But no variation of anything like:
CMD="tstArgs $ARG1 $ARG2"
$CMD
I'm sure it must be possible, and probably simple - but it's some permutation I just haven't been able to crack.
Of course I can work around it - but I'm stubborn & persistent & hate to give up. If anyone has any insight, I'd be very grateful, and maybe even finally get some sleep...
Don't put arguments in a string. Put them in an array. Array elements handle spaces much more gracefully:
declare -a ARGS
ARGS+=( "today with spaces" )
ARGS+=( "tomorrow with spaces" )
CMD="tstArgs"
${CMD} "${ARGS[#]}"
Alternatively:
declare -a ARGS
ARGS[0]="today with spaces"
ARGS[1]="tomorrow with spaces"
CMD="tstArgs"
${CMD} "${ARGS[#]}"
Putting quotation marks around ${ARGS[#]} on the last line makes sure that each element of the array is quoted, thus preserving the spaces.
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
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=${prefix}_suffix
x11-common package installs a /etc/X11/Xsession.d/20x11-common_process-args script which is sourced in by /etc/X11/Xsession. This 20x11-common_process-args script contains following if-statement:
has_option() {
if [ "${OPTIONS#*
$1}" != "$OPTIONS" ]; then
return 0
else
return 1
fi
}
OPTIONS variable is a list of configuration options from a file separated by line-feeds(0a in ASCII). How to understand this if-statement? Literally, this parameter expansion part should modify the OPTIONS variable in a way that everything before the argument($1) is removed? This argument needs to match one of the configuration options. However, what is the general meaning of this if-statement?
This fragment of code:
xyz=gobbledegook
echo ${xyz#*de}
echoes "gook", so the ${OPTIONS#*$1} notation deletes everything from the start of $OPTIONS up to and including the $1.
In the script, it is checking whether $1 (the first argument to the function) is present in the list of options in $OPTIONS. If the value is different, then the option is matched; otherwise, it isn't. It's a fairly compact way of dealing with a lot of options all at once.
The newline in the test (well, strictly, it is in the string that is present in the test) is unorthodox but legitimate.
You can easily experiment for yourself, of course:
$ OPTIONS="-abc
> -def
> -ghi"
$ echo "${OPTIONS#*-abc}"
-def
-ghi
$ echo "${OPTIONS#*-def}"
-ghi
$ echo "${OPTIONS#*-ghi}"
$ echo "${OPTIONS#*-xyz}"
-abc
-def
-ghi
$
You are correct that the parameter expansion removes everything up to and including $1 from the expansion. If $1 is not present, then the expansion removes nothing, and the expansion is identical to simply expanding $OPTIONS without modification. So, the if statement simply returns 0 if OPTIONS contains $1 (that is, the two expansions are different), and returns 1 if it does not contain $1 (that is, the two expansions are the same).
In broader terms, has_option foo succeeds if foo is present in $OPTIONS, and fails if foo is not.
I'm relatively new to shell scripting and I've been stuck on this error for a couple days now. I'm trying to read in the contents of a file containing a list of strings and numbers, format it, and output the number of numbers below 50.
All the commands work when typed into the shell, however; in the script when I try and pass the filename in as an argument I keep getting a "No such file or directory" error.
Here is the function in question:
belowFifty(){
count=0
numbers=`cut -d : -f 3 < "$2"` #here is where the error occurs
for num in $numbers
do
if ((num<50));
then
count=$((count+1))
fi
done
echo $count
}
edit: sorry I forgot to mention the script does a couple things. $1 is the option, $2 is the file. I'm calling it like so:
./script.sh m filename
Try:
${2? 2 arguments are required to function belowFifty}
numbers=$( cut -d : -f 3 < $2 )
I suspect the problem is that you are calling the function
and not specifying the 2nd argument. Within the function,
$2 is the argument passed to the function, and not the argument
passed to the main script.
You specify "$2"; what's in the "$1" that's passed to the function and ignored? My strong suspicion is that you are trying to open the file with an empty string as the name, and there is no such file - hence the error message. The corollary is that you probably intended to reference "$1".
If so, you should probably write:
numbers=$(cut -d : -f 3 < "$1")
The back-tick notation should usually be avoided in favour of $(...).