I have to print a variable value which is art variable.
eg. variables are
A = X
Z_X = Test
Set in shell using
setenv A X
setenv Z_X Test
I want to print $Z_X value using $A
I am trying but without any success.
echo ${Z_${A}}
echo ${Z_[$A]}
could anyone tell me where I am wrong.
regards
A = X
Z_X = Test
This seems wrong; in csh, you need to use the set keyword to assign variables; in addition, the convention is also to use lower case for "normal" variables, and UPPER CASE for environment variables:
set a = x
set z_x = Test
You can then use eval to get what you want:
% eval echo \$z_$a
Test
% set x = `eval echo \$z_$a`
% echo $s
Test
This may be dangerous if you don't trust the source of $a, since it may also do a rm -rf / or something similarly dangerous (but if you trust the source of $a, it's perfectly fine).
You can get a list of all variables with set:
% set | grep ^z_$a
z_x Test
% set | grep ^z_$a | awk '{print $1}'
z_x
Which is the only safe way I can figure out to do what you want.
Generally this is a bad idea, and you should rather rethink your approach.
You can use indirect expansion:
name=Z_$A
echo ${!name}
From the manual:
If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and
${!name[#]} described below. The exclamation point must immediately follow the left brace
in order to introduce indirection.
Or indirect references:
name=Z_$A
eval echo \$$name
You can find more about indirect references in the guide.
Related
I would like to achieve this in Bash: echo $(a=1)and print the value of variable a
I test eval, $$a,{}, $() but none of them work as most of the times either I got literally a=1 or in one case (I don't remember which) it tried to execute the value.
I known that I can do: a=1;echo $a but because I'm little fun one command per line (even if sometimes is getting little complicated) I was wondering if is possible to do this either with echo or with printf
If you know that $a is previously unset, you can do this using the following syntax:
echo ${a:=1}
This, and other types of parameter expansion, are defined in the POSIX shell command language specification.
If you want to assign a numeric value, another option, which doesn't depend on the value previously being unset, would be to use an arithmetic expansion:
echo $(( a = 1 ))
This assigns the value and echoes the number that has been assigned.
It's worth mentioning that what you're trying to do cannot be done in a subshell by design because a child process cannot modify the environment of its parent.
I am trying to get the value of a variable to be selected by name at runtime, using eval, but I don't get its value if - (hyphen) is in the name.
ENV=dev
REGION=us-east-1
DBUSERNAME=DB_USER_${ENV}_$REGION
DBPASSWORD=DB_PASS_${ENV}_$REGION
eval "USERNAME=\${${DBUSERNAME}}"
eval "PASSWORD=\${${DBPASSWORD}}"
echo USERNAME=$USERNAME
echo PASSWORD=$PASSWORD
RESULT
echo USERNAME=east-1
echo PASSWORD=east-1
EXPECTED RESULT
echo USERNAME=DB_USER_dev_us-east-1
echo PASSWORD=DB_USER_dev_us-east-1
It's working fine if there is no hyphen present in the name.
Investigation
We can see what's happening by running this in shell with -x option to trace execution:
$ sh -x ./36332134.sh
+ ENV=dev
+ REGION=us-east-1
+ DBUSERNAME=DB_USER_dev_us-east-1
+ DBPASSWORD=DB_PASS_dev_us-east-1
+ eval USERNAME=${DB_USER_dev_us-east-1}
+ USERNAME=east-1
+ eval PASSWORD=${DB_PASS_dev_us-east-1}
+ PASSWORD=east-1
+ echo USERNAME=east-1
USERNAME=east-1
+ echo PASSWORD=east-1
PASSWORD=east-1
Notice that eval USERNAME=${DB_USER_dev_us-east-1} gives us USERNAME=east-1. That's parameter expansion in effect, as described in the Bash manual:
When not performing substring expansion, using the forms documented below (e.g., :-), bash tests for a parameter that is unset
or null. Omitting the colon results in a test only for a parameter
that is unset.
${parameter:-word}
Use Default Values. If parameter is unset or null, the expansion of word is substituted. Otherwise, the value of parameter
is substituted.
Since $DB_USER_dev_us is unset, then the expansion of ${DB_USER_dev_us-east-1} is east-1.
Workarounds
Shell doesn't allow - in variable names (including environment variables). I guess DB_USER_dev_us-east-1 was set by some non-shell program? In which case, you'll need a similar non-shell program to retrieve it, I think. I tested quoting the -, but to no avail.
If you can use Bash as your shell, you might want to use an associative array instead of composing variable names.
If you are able to change the environment variables, you might consider changing the - to (say) _, then using (Bash) ${REGION//-/_} or (otherwise) tr to transform the name:
REGION="${REGION//-/_}" # Bash
REGION="$(echo "$REGION"|tr - _)" # POSIX
You are evaluating/expanding a few too many times.
Toby's answer is exactly correct (and shows the proper debugging technique for this sort of issue) but the solution to the problem is to unwrap one level of expansion.
You wrote
eval "USERNAME=\${${DBUSERNAME}}"
which becomes
eval "USERNAME=\${DB_USER_dev_us-east-1}"
which then gets run through eval as
USERNAME=${DB_USER_dev_us-east-1}
which becomes
USERNAME=east-1
but you wanted to stop after the first expansion. That is
eval "USERNAME=\${DB_USER_dev_us-east-1}"
has already performed the expansion you wanted and gotten you the result you needed. So you don't want the \${...} bit or eval. Just
USERNAME=DB_USER_dev_us-east-1
which you get from
USERNAME=${DBUSERNAME}
Unless I've missed something or your example isn't accurate.
As they have explained the problem, here is what you can do
$> more a.sh
ENV=dev
REGION=us-east-1
DBUSERNAME=DB_USER_${ENV}_$REGION
DBPASSWORD=DB_PASS_${ENV}_$REGION
eval "USERNAME=${DBUSERNAME}"
eval "PASSWORD=${DBPASSWORD}"
echo USERNAME=$USERNAME
echo PASSWORD=$PASSWORD
Results
$> ./a.sh
USERNAME=DB_USER_dev_us-east-1
PASSWORD=DB_PASS_dev_us-east-1
Hi I would like to set a variable by concatenating two other variables.
Example
A=1
B=2
12=C
echo $A$B
desired result being C
however the answer I get is always 12
Is it possible?
UPDATED
Example
A=X
B=Y
D=$A$B
xy=test
echo $D
desired result being "test"
It looks like you want indirect variable references.
BASH allows you to expand a parameter indirectly -- that is, one variable may contain the name of another variable:
# Bash
realvariable=contents
ref=realvariable
echo "${!ref}" # prints the contents of the real variable
But as Pieter21 indicates in his comment 12 is not a valid variable name.
Since 12 is not a valid variable name, here's an example with string variables:
> a='hello'
> b='world'
> declare my_$a_$b='my string'
> echo $my_hello_world
my string
What you are trying to do is (almost) called indirection: http://wiki.bash-hackers.org/syntax/pe#indirection
...I did some quick tests, but it does not seem logical to do this without a third variable - you cannot do concatenated indirection directly as the variables/parts being concatenated do not evaluate to the result on their own - you would have to do another evaluation. I think concatenating them first might be the easiest. That said, there is a chance you could rethink what you're doing.
Oh, and you cannot use numbers (alone or as the starting character) for variable names.
Here we go:
cake="cheese"
var1="ca"
var2="ke"
# this does not work as the indirection sees "ca" and "ke", not "cake". No output.
echo ${!var1}${!var2}
# there might be some other ways of tricking it to do this, but they don't look to sensible as indirection probably needs to work on a real variable.
# ...this works, though:
var3=${var1}${var2}
echo ${!var3}
I was reading the bash advanced scripting guide (if memory serves me right), and it said something to the extent that eval can be used to force variable updates.
So I tried this:
randomPath="/path/$var/here/" # var is not defined at this point
echo $randomPath
/path//here/
var="is" # initially defining var
eval $randomPath
zsh: no such file or directory: /path//here/
I don't understand the error message, and I'm wondering if I'm using eval properly.
The output I was expecting is:
eval $randomPath
echo $randomPath
/path/is/here
The problem is that $var is already being substituted in randomPath="/path/$var/here/", and because it is blank, randomPath is set to /path//here. You want to use single quotes to prevent the early substitution:
randomPath='/path/$var/here/'
The second problem is that eval x runs x as a command. What you want to do is return the newly evaluated variable as a string:
eval echo $randomPath
You can store it in a variable in the usual way:
randomPath=`eval echo $randomPath`
Let's say, I have 3 variables, A=1, B=2, C=3 and finally a last variable containing the name of either of them (VAR=A). Is it possible to call A, B, C based on VAR's content without 'if's?
Like echo "${$VAR}"?
Yes.
eval echo \$$VAR
There is also a bash-only way of doing this, using indirect reference:
echo ${!VAR}
A=1
B=2
C=3
VAR=B
echo ${!VAR}
Output:
2
Documentation (man bash):
If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed
from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the
value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[#]} described below.
The exclamation point must immediately follow the left brace in order to introduce indirection.