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0

This question already has an answer here:

• 
  Indirectly access environment variable [duplicate]
  
  3 answers
  
  

I have a variable (var) which return value as VAR1 ($var has the value VAR1).
There is an input file which has VAR1 defined (VAR1=ABCDEF)

How can I use echo to get the value ABCDEF using $var??

I tried echo $(echo $var) and many other options, but I always get output as VAR1 or echo VAR1 but never ABCDEF. I used source in file with VAR1 declared, tried in command prompt etc.

variable echo

share|improve this question

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Kusalananda♦, DopeGhoti, jimmij, nwildner, muru 2 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

0

This question already has an answer here:

• 
  Indirectly access environment variable [duplicate]
  
  3 answers
  
  

I have a variable (var) which return value as VAR1 ($var has the value VAR1).
There is an input file which has VAR1 defined (VAR1=ABCDEF)

How can I use echo to get the value ABCDEF using $var??

I tried echo $(echo $var) and many other options, but I always get output as VAR1 or echo VAR1 but never ABCDEF. I used source in file with VAR1 declared, tried in command prompt etc.

variable echo

share|improve this question

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Kusalananda♦, DopeGhoti, jimmij, nwildner, muru 2 hours ago

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

This question already has an answer here:

• 
  Indirectly access environment variable [duplicate]
  
  3 answers
  
  

I have a variable (var) which return value as VAR1 ($var has the value VAR1).
There is an input file which has VAR1 defined (VAR1=ABCDEF)

How can I use echo to get the value ABCDEF using $var??

I tried echo $(echo $var) and many other options, but I always get output as VAR1 or echo VAR1 but never ABCDEF. I used source in file with VAR1 declared, tried in command prompt etc.

variable echo

share|improve this question

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

Kusalananda♦

138k17258426

edited yesterday

Kusalananda♦

138k17258426

asked yesterday

shashank barkishashank barki

41

New contributor

shashank barki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

shashank barkishashank barki

41

2 Answers
2

active

oldest

votes

6

Assuming that you are using the bash shell:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ echo "$!var"
ABCDEF

By using $!var you use variable indirection in bash. The value of the variable var is used to get the name of the variable to expand.

---

In bash you could also use a name reference variable:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ declare -n var="VAR1"
$ echo "$var"
ABCDEF

Here, the var variable refers to the VAR1 variable, so $var will expand to whatever $VAR1 expands to.

Name references are originally a ksh feature, and in that shell they are declare with typeset -n.

Name references are extremely useful fo passing references to arrays in calls to shell functions.

---

In any sh shell:

$ . ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ eval "echo "$$var""
ABCDEF

The eval utility takes a string which it will re-evaluate. Here, we give it the string echo "$VAR1" (after expansion of $var).

The issue with eval is that it's easy to introduce errors or vulnerabilities with it, by carelessly creating its argument string.

share|improve this answer

edited yesterday

answered yesterday

Kusalananda♦Kusalananda

138k17258426

add a comment | 

3

Assuming you are using bash, you can accomplish this easily with an indirect reference:

$ foo=bar
$ bar=magicword
$ printf "%sn" "$!foo"
magicword

The syntax $var (or $var) yields the contents of the variable var.

The syntax $!var yields the contents of the variable named in var.

share|improve this answer

edited yesterday

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

add a comment | 

6

Assuming that you are using the bash shell:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ echo "$!var"
ABCDEF

By using $!var you use variable indirection in bash. The value of the variable var is used to get the name of the variable to expand.

---

In bash you could also use a name reference variable:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ declare -n var="VAR1"
$ echo "$var"
ABCDEF

Here, the var variable refers to the VAR1 variable, so $var will expand to whatever $VAR1 expands to.

Name references are originally a ksh feature, and in that shell they are declare with typeset -n.

Name references are extremely useful fo passing references to arrays in calls to shell functions.

---

In any sh shell:

$ . ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ eval "echo "$$var""
ABCDEF

The eval utility takes a string which it will re-evaluate. Here, we give it the string echo "$VAR1" (after expansion of $var).

The issue with eval is that it's easy to introduce errors or vulnerabilities with it, by carelessly creating its argument string.

share|improve this answer

edited yesterday

answered yesterday

Kusalananda♦Kusalananda

138k17258426

add a comment | 

6

Assuming that you are using the bash shell:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ echo "$!var"
ABCDEF

By using $!var you use variable indirection in bash. The value of the variable var is used to get the name of the variable to expand.

---

In bash you could also use a name reference variable:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ declare -n var="VAR1"
$ echo "$var"
ABCDEF

Here, the var variable refers to the VAR1 variable, so $var will expand to whatever $VAR1 expands to.

Name references are originally a ksh feature, and in that shell they are declare with typeset -n.

Name references are extremely useful fo passing references to arrays in calls to shell functions.

---

In any sh shell:

$ . ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ eval "echo "$$var""
ABCDEF

The eval utility takes a string which it will re-evaluate. Here, we give it the string echo "$VAR1" (after expansion of $var).

The issue with eval is that it's easy to introduce errors or vulnerabilities with it, by carelessly creating its argument string.

share|improve this answer

edited yesterday

answered yesterday

Kusalananda♦Kusalananda

138k17258426

add a comment | 

Assuming that you are using the bash shell:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ echo "$!var"
ABCDEF

By using $!var you use variable indirection in bash. The value of the variable var is used to get the name of the variable to expand.

---

In bash you could also use a name reference variable:

$ source ./file
$ echo "$VAR1"
ABCDEF
$ declare -n var="VAR1"
$ echo "$var"
ABCDEF

Here, the var variable refers to the VAR1 variable, so $var will expand to whatever $VAR1 expands to.

Name references are originally a ksh feature, and in that shell they are declare with typeset -n.

Name references are extremely useful fo passing references to arrays in calls to shell functions.

---

In any sh shell:

$ . ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ eval "echo "$$var""
ABCDEF

The eval utility takes a string which it will re-evaluate. Here, we give it the string echo "$VAR1" (after expansion of $var).

The issue with eval is that it's easy to introduce errors or vulnerabilities with it, by carelessly creating its argument string.

share|improve this answer

edited yesterday

answered yesterday

Kusalananda♦Kusalananda

138k17258426

$ source ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ echo "$!var"
ABCDEF

$ source ./file
$ echo "$VAR1"
ABCDEF
$ declare -n var="VAR1"
$ echo "$var"
ABCDEF

$ . ./file
$ echo "$VAR1"
ABCDEF
$ var=VAR1
$ eval "echo "$$var""
ABCDEF

share|improve this answer

edited yesterday

answered yesterday

Kusalananda♦Kusalananda

138k17258426

answered yesterday

Kusalananda♦Kusalananda

138k17258426

answered yesterday

Kusalananda♦Kusalananda

138k17258426

3

Assuming you are using bash, you can accomplish this easily with an indirect reference:

$ foo=bar
$ bar=magicword
$ printf "%sn" "$!foo"
magicword

The syntax $var (or $var) yields the contents of the variable var.

The syntax $!var yields the contents of the variable named in var.

share|improve this answer

edited yesterday

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

add a comment | 

3

Assuming you are using bash, you can accomplish this easily with an indirect reference:

$ foo=bar
$ bar=magicword
$ printf "%sn" "$!foo"
magicword

The syntax $var (or $var) yields the contents of the variable var.

The syntax $!var yields the contents of the variable named in var.

share|improve this answer

edited yesterday

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

add a comment | 

Assuming you are using bash, you can accomplish this easily with an indirect reference:

$ foo=bar
$ bar=magicword
$ printf "%sn" "$!foo"
magicword

The syntax $var (or $var) yields the contents of the variable var.

The syntax $!var yields the contents of the variable named in var.

share|improve this answer

edited yesterday

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

$ foo=bar
$ bar=magicword
$ printf "%sn" "$!foo"
magicword

share|improve this answer

edited yesterday

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

answered yesterday

DopeGhotiDopeGhoti

46.6k56190

answered yesterday

DopeGhotiDopeGhoti

46.6k56190