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
Related
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Search+replace strings in filenames
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How do I set a variable to the output of a command in Bash?
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Command not found error in Bash variable assignment
(5 answers)
Closed 1 year ago.
i have just tried to one of my first bash scripts, i need to find a substring(after the ? part) in a url and replaced with the replace_string,
#!/bin/bash
url="https://example.com/tfzzr?uhg"
# 123456 ...
first= echo `expr index "$url" ?`
last= expr length $url
replace_string="abc"
part_to_be_replace = echo ${url:($first+1):$last}//dont know how to use variable here
substring(url,part_to_be_replace,replace_string)
It does not work, i was able to find only the first accurance of ?, and the length of the string
Does this help?
url="https://example.com/tfzzr?uhg"
replace_string="abc"
echo "${url}"
https://example.com/tfzzr?uhg
echo "${url//\?*/${replace_string}}"
https://example.com/tfzzrabc
# If you still want the "?"
echo "${url//\?*/\?${replace_string}}"
https://example.com/tfzzr?abc
See https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html for further details.
Use parameter expansion:
#! /bin/bash
url='https://example.com/tfzzr?uhg'
replace_string=abc
new=${url%\?*}?$replace_string
echo "$new"
${url%\?*} removes the pattern (i.e. ? and anything following it) from $url. ? needs to be quoted, otherwise it would match a single character in the pattern. Double the percent sign to remove the longest possible substring, i.e. starting from the first ?.
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 "${!var}"
This question already has answers here:
Create variable from string/nameonly parameter to extract data in bash?
(3 answers)
Closed 8 years ago.
Coding:
intVarNos=$#
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]}"
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}"