This question already has answers here:
Create variable from string/nameonly parameter to extract data in bash?
(3 answers)
Closed 8 years ago.
Coding:
intVarNos=$#
strInp="Name1;Name2"
arrInp=(${strInp//;/ })
for ((i=0; i<=intVarNos-1; i++))
do
j=$((i+1))
echo "Value of " ${arrCTN[$i]} " is " $j
done
My requirement is to print the parameters or arguments in the following format
$./college.sh John Peter
Value of Name1 is John
Value of Name2 is Peter
But I am getting the result as
Value of Name1 is 1
Value of Name2 is 2
The usual way to print the parameter is $1,$2.... How could I print the value of my parameter in this case. $$j doesnt works
In Bash, you can perform an indirect lookup variable name lookup with:
${!j}
You could also assign the arguments to an array and then index the array.
args=("$#")
echo "${args[$j]}"
Related
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed 4 years ago.
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
# something
eval echo \"\$$bar\" would do it.
The accepted answer is great. However, #Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[#]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$#"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[#]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[#]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[#] format, you get the array and all its values expanded.
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[#]
echo "${!var}"
This question already has answers here:
How to get a variable value if variable name is stored as string?
(10 answers)
Closed 5 years ago.
#!/bin/bash
list="one two three"
one=1
two=2
three=3
for k in $list
do
echo $k
done
For the code above, output is:
one
two
three
But I always think it should output:
1
2
3
It's confusing. How to understand this?
The expansion $k just gives you the name of the variable as a string. If you want the value of the variable, you must use the parameter expansion syntax ${!k}. This is documented here.
#!/bin/bash
list="one two three"
one=1
two=2
three=3
for k in $list
do
echo "${!k}"
done
Output
1
2
3
This question already has answers here:
Usage of :- (colon dash) in bash
(2 answers)
Closed 5 years ago.
What means these declarations in bash script:
PREFIX=${1:-daily}
PROFILE=${2:-backup}
The key here is to understand what Parameter-Expansion ${PARAMETER:-WORD} syntax means here,
${PARAMETER:-WORD}
If the parameter PARAMETER is unset (never was defined) or null (empty), this one
expands to WORD, otherwise it expands to the value of PARAMETER, as if it
just was ${PARAMETER}
In your case the ${PARAMETER} being the positional arguments passed to a function or script,
Use the below script for a better understanding,
myTestFunction() {
PREFIX=${1:-daily}
PROFILE=${2:-backup}
printf "PREFIX value %s PROFILE value %s\n" "$PREFIX" "$PROFILE"
}
myTestFunction "some" "junk"
myTestFunction
which produces a result as
$ bash script.sh
PREFIX value some PROFILE value junk
PREFIX value daily PROFILE value backup
Also see the expanded debugger version of the script as
$ bash -x script.sh
+ myTestFunction some junk
+ PREFIX=some
+ PROFILE=junk
+ printf 'PREFIX value %s PROFILE value %s\n' some junk
PREFIX value some PROFILE value junk
+ myTestFunction
+ PREFIX=daily
+ PROFILE=backup
+ printf 'PREFIX value %s PROFILE value %s\n' daily backup
PREFIX value daily PROFILE value backup
how shell substitutes the values to the variables when $1 or $2 is not passed.
The syntax is generally used when by default you want to configure a variable with a certain value at the same time make it dynamically configurable also.
Assigns value of first argument to PREFIX variable if this first argument exist, "daily" if it does not.
It means, assign the first argument (if present), else daily to variable PREFIX and assign the second argument (if present), else backup to variable PROFILE
eg:
$ cat file.sh
#!/bin/bash
PREFIX=${1:-daily}
PROFILE=${2:-backup}
echo $PREFIX
echo $PROFILE
For the following command line args given, output would be like:
$ ./file.sh
daily
backup
$ ./file.sh abc
abc
backup
$ ./file.sh abc xyz
abc
xyz
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed 4 years ago.
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
# something
eval echo \"\$$bar\" would do it.
The accepted answer is great. However, #Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[#]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$#"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[#]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[#]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[#] format, you get the array and all its values expanded.
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[#]
echo "${!var}"
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed 4 years ago.
I have the variable $foo="something" and would like to use:
bar="foo"; echo $($bar)
to get "something" echoed.
In bash, you can use ${!variable} to use variable variables.
foo="something"
bar="foo"
echo "${!bar}"
# something
eval echo \"\$$bar\" would do it.
The accepted answer is great. However, #Edison asked how to do the same for arrays. The trick is that you want your variable holding the "[#]", so that the array is expanded with the "!". Check out this function to dump variables:
$ function dump_variables() {
for var in "$#"; do
echo "$var=${!var}"
done
}
$ STRING="Hello World"
$ ARRAY=("ab" "cd")
$ dump_variables STRING ARRAY ARRAY[#]
This outputs:
STRING=Hello World
ARRAY=ab
ARRAY[#]=ab cd
When given as just ARRAY, the first element is shown as that's what's expanded by the !. By giving the ARRAY[#] format, you get the array and all its values expanded.
To make it more clear how to do it with arrays:
arr=( 'a' 'b' 'c' )
# construct a var assigning the string representation
# of the variable (array) as its value:
var=arr[#]
echo "${!var}"