Indirectly access environment variable [duplicate]
This question already has an answer here:
Variable substitution with an exclamation mark in bash
1 answer
Given I have in a bash script
ev=USER
How can I get the environment variable value for $USER using ev?
Tried naively doing:
echo ${"$"$ev}
which results in bad substitution.
I'd expect to get back whatever the value of $USER is.
bash variable
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
marked as duplicate by terdon♦ bash
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Mar 1 at 18:22
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:
Variable substitution with an exclamation mark in bash
1 answer
Given I have in a bash script
ev=USER
How can I get the environment variable value for $USER using ev?
Tried naively doing:
echo ${"$"$ev}
which results in bad substitution.
I'd expect to get back whatever the value of $USER is.
bash variable
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
marked as duplicate by terdon♦ bash
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Mar 1 at 18:22
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:
Variable substitution with an exclamation mark in bash
1 answer
Given I have in a bash script
ev=USER
How can I get the environment variable value for $USER using ev?
Tried naively doing:
echo ${"$"$ev}
which results in bad substitution.
I'd expect to get back whatever the value of $USER is.
bash variable
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
This question already has an answer here:
Variable substitution with an exclamation mark in bash
1 answer
Given I have in a bash script
ev=USER
How can I get the environment variable value for $USER using ev?
Tried naively doing:
echo ${"$"$ev}
which results in bad substitution.
I'd expect to get back whatever the value of $USER is.
This question already has an answer here:
Variable substitution with an exclamation mark in bash
1 answer
bash variable
bash variable
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
edited Mar 1 at 17:28
Rui F Ribeiro
41.9k1483142
41.9k1483142
asked Mar 1 at 10:21
PaulBPaulB
1234
asked Mar 1 at 10:21
PaulBPaulB
1234
asked Mar 1 at 10:21
PaulBPaulB
1234
1234
marked as duplicate by terdon♦ bash
Users with the bash badge can single-handedly close bash questions as duplicates and reopen them as needed.
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Mar 1 at 18:22
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Mar 1 at 18:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
By using an indirect expansion (also sometimes called "variable indirection"),
ev=USER
printf '%sn' "${!ev}"
This is described in the bash (5.0) manual, in the section titled "Parameter Expansion".
Or, by making ev a name reference (requires bash 4.3+),
declare -n ev=USER
printf '%sn' "$ev"
This is described in the bash (5.0) manual, just before the section called "Positional Parameters".
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
add a comment |
If it's only about environment variables, as opposed to shell variables, then on most systems, you can use:
printenv -- "$ev"
For shell variables, with any Bourne-like shell, you can do:
eval 'printf "%sn" "${'"$ev"}'}"'
Or with zsh:
printf '%sn' "${(P)ev}"
Or with bash:
printf '%sn' "${!ev}"
All 3 are arbitrary command injection vulnerabilities if the content of $ev is not under your control.
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
add a comment |
You can also evaluate the command after the vale for $ev has been substituted:
eval echo "$"$ev
The part "$"$ev resolves to $USER so eval executes echo $USER.
answered Mar 1 at 10:46
katoshkatosh
1619
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
By using an indirect expansion (also sometimes called "variable indirection"),
ev=USER
printf '%sn' "${!ev}"
This is described in the bash (5.0) manual, in the section titled "Parameter Expansion".
Or, by making ev a name reference (requires bash 4.3+),
declare -n ev=USER
printf '%sn' "$ev"
This is described in the bash (5.0) manual, just before the section called "Positional Parameters".
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
add a comment |
By using an indirect expansion (also sometimes called "variable indirection"),
ev=USER
printf '%sn' "${!ev}"
This is described in the bash (5.0) manual, in the section titled "Parameter Expansion".
Or, by making ev a name reference (requires bash 4.3+),
declare -n ev=USER
printf '%sn' "$ev"
This is described in the bash (5.0) manual, just before the section called "Positional Parameters".
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
add a comment |
By using an indirect expansion (also sometimes called "variable indirection"),
ev=USER
printf '%sn' "${!ev}"
This is described in the bash (5.0) manual, in the section titled "Parameter Expansion".
Or, by making ev a name reference (requires bash 4.3+),
declare -n ev=USER
printf '%sn' "$ev"
This is described in the bash (5.0) manual, just before the section called "Positional Parameters".
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
By using an indirect expansion (also sometimes called "variable indirection"),
ev=USER
printf '%sn' "${!ev}"
This is described in the bash (5.0) manual, in the section titled "Parameter Expansion".
Or, by making ev a name reference (requires bash 4.3+),
declare -n ev=USER
printf '%sn' "$ev"
This is described in the bash (5.0) manual, just before the section called "Positional Parameters".
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
edited Mar 1 at 10:37
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
answered Mar 1 at 10:26
Kusalananda♦Kusalananda
139k17260432
139k17260432
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
add a comment |
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
Perfect thanks ... env_val="${!ev}"
– PaulB
Mar 1 at 10:36
add a comment |
If it's only about environment variables, as opposed to shell variables, then on most systems, you can use:
printenv -- "$ev"
For shell variables, with any Bourne-like shell, you can do:
eval 'printf "%sn" "${'"$ev"}'}"'
Or with zsh:
printf '%sn' "${(P)ev}"
Or with bash:
printf '%sn' "${!ev}"
All 3 are arbitrary command injection vulnerabilities if the content of $ev is not under your control.
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
add a comment |
If it's only about environment variables, as opposed to shell variables, then on most systems, you can use:
printenv -- "$ev"
For shell variables, with any Bourne-like shell, you can do:
eval 'printf "%sn" "${'"$ev"}'}"'
Or with zsh:
printf '%sn' "${(P)ev}"
Or with bash:
printf '%sn' "${!ev}"
All 3 are arbitrary command injection vulnerabilities if the content of $ev is not under your control.
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
add a comment |
If it's only about environment variables, as opposed to shell variables, then on most systems, you can use:
printenv -- "$ev"
For shell variables, with any Bourne-like shell, you can do:
eval 'printf "%sn" "${'"$ev"}'}"'
Or with zsh:
printf '%sn' "${(P)ev}"
Or with bash:
printf '%sn' "${!ev}"
All 3 are arbitrary command injection vulnerabilities if the content of $ev is not under your control.
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
If it's only about environment variables, as opposed to shell variables, then on most systems, you can use:
printenv -- "$ev"
For shell variables, with any Bourne-like shell, you can do:
eval 'printf "%sn" "${'"$ev"}'}"'
Or with zsh:
printf '%sn' "${(P)ev}"
Or with bash:
printf '%sn' "${!ev}"
All 3 are arbitrary command injection vulnerabilities if the content of $ev is not under your control.
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
answered Mar 1 at 10:45
Stéphane ChazelasStéphane Chazelas
313k57592948
313k57592948
add a comment |
add a comment |
You can also evaluate the command after the vale for $ev has been substituted:
eval echo "$"$ev
The part "$"$ev resolves to $USER so eval executes echo $USER.
answered Mar 1 at 10:46
katoshkatosh
1619
add a comment |
You can also evaluate the command after the vale for $ev has been substituted:
eval echo "$"$ev
The part "$"$ev resolves to $USER so eval executes echo $USER.
answered Mar 1 at 10:46
katoshkatosh
1619
add a comment |
You can also evaluate the command after the vale for $ev has been substituted:
eval echo "$"$ev
The part "$"$ev resolves to $USER so eval executes echo $USER.
answered Mar 1 at 10:46
katoshkatosh
1619
You can also evaluate the command after the vale for $ev has been substituted:
eval echo "$"$ev
The part "$"$ev resolves to $USER so eval executes echo $USER.
answered Mar 1 at 10:46
katoshkatosh
1619
answered Mar 1 at 10:46
katoshkatosh
1619
answered Mar 1 at 10:46
katoshkatosh
1619
answered Mar 1 at 10:46
katoshkatosh
1619
1619
add a comment |
add a comment |
