How to get a bash script argument given its position from the end? [duplicate]
This question already has an answer here:
How to get the last argument to a /bin/sh function
11 answers
Get last element of $@ / argv / arguments array without eval [duplicate]
1 answer
I've got a shell script and I want to use the last argument:
#!/bin/bash
echo My last param is #somehow put it here
Here is what I've tried:
echo $$#
echo ${$#}
echo ${(($#))}
Unfortunately it did not work.
I'm specifically want to understand why my options did not work, I want to do something like double expansion.
In a broader sense, I would like to know how to access the Nth argument from the end. How do I achieve that?
shell-script shell
marked as duplicate by muru, Archemar, G-Man, telcoM, Anthony Geoghegan Dec 8 at 13:33
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How to get the last argument to a /bin/sh function
11 answers
Get last element of $@ / argv / arguments array without eval [duplicate]
1 answer
I've got a shell script and I want to use the last argument:
#!/bin/bash
echo My last param is #somehow put it here
Here is what I've tried:
echo $$#
echo ${$#}
echo ${(($#))}
Unfortunately it did not work.
I'm specifically want to understand why my options did not work, I want to do something like double expansion.
In a broader sense, I would like to know how to access the Nth argument from the end. How do I achieve that?
shell-script shell
marked as duplicate by muru, Archemar, G-Man, telcoM, Anthony Geoghegan Dec 8 at 13:33
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.
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42
add a comment |
This question already has an answer here:
How to get the last argument to a /bin/sh function
11 answers
Get last element of $@ / argv / arguments array without eval [duplicate]
1 answer
I've got a shell script and I want to use the last argument:
#!/bin/bash
echo My last param is #somehow put it here
Here is what I've tried:
echo $$#
echo ${$#}
echo ${(($#))}
Unfortunately it did not work.
I'm specifically want to understand why my options did not work, I want to do something like double expansion.
In a broader sense, I would like to know how to access the Nth argument from the end. How do I achieve that?
shell-script shell
This question already has an answer here:
How to get the last argument to a /bin/sh function
11 answers
Get last element of $@ / argv / arguments array without eval [duplicate]
1 answer
I've got a shell script and I want to use the last argument:
#!/bin/bash
echo My last param is #somehow put it here
Here is what I've tried:
echo $$#
echo ${$#}
echo ${(($#))}
Unfortunately it did not work.
I'm specifically want to understand why my options did not work, I want to do something like double expansion.
In a broader sense, I would like to know how to access the Nth argument from the end. How do I achieve that?
This question already has an answer here:
How to get the last argument to a /bin/sh function
11 answers
Get last element of $@ / argv / arguments array without eval [duplicate]
1 answer
shell-script shell
shell-script shell
edited Dec 8 at 8:17
asked Dec 7 at 19:51
YardenST
1606
1606
marked as duplicate by muru, Archemar, G-Man, telcoM, Anthony Geoghegan Dec 8 at 13:33
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.
marked as duplicate by muru, Archemar, G-Man, telcoM, Anthony Geoghegan Dec 8 at 13:33
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.
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42
add a comment |
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42
add a comment |
6 Answers
6
active
oldest
votes
In addition (to Kusalananda's answer):
#! /bin/bash
echo "(bash/ksh): ${@: -1}"
echo "(bash 3.x+): ${!#}"
echo "(bash 3.x+): $BASH_ARGV"
echo "(bash 3.x+/ksh): ${@:$#}"
echo "(bash 3.x+): ${BASH_ARGV[0]}"
and if you worry about portability:
#!/bin/bash
penultimate=''
ultimate=''
for p in "$@" ; do
penultimate="$ultimate"
ultimate="$p"
done
echo "penultimate=$penultimate"
echo "ultimate=$ultimate"
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion."
– JoL
Dec 7 at 22:29
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.
– JoL
Dec 9 at 5:20
|
show 4 more comments
To get any argument from bash
script, you can use slicing:
#!/bin/bash
# Get 3rd element from the end
from_end1=3
# Get last element
from_end2=1
# Get slice of array [end - from_end1 : end ] of length 1
echo "${@: -$from_end1: 1}"
echo "${@: -$from_end2: 1}"
You can also use this to get Nth element:
# Get 2nd element
from_beginning=2
echo "${@: $from_beginning: 1}"
Remember to check for length; this might return your program's name or an empty string.
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the#!
(and in particular didn't use bash tag).
– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
add a comment |
In bash
(release 4.3+), you may assign the passed parameters to an array and access the last one by the index -1
:
#!/bin/bash
params=( "$@" )
printf 'The last parameter given to the script is %sn' "${params[-1]}"
foo () {
params=( "$@" )
printf 'The last parameter given to the function is %sn' "${params[-1]}"
}
In general, negative indexes into arrays accesses the elements from the end of the array.
A negative index might work on bash 4.2+ depending on the patch level. But an slice${@: -1:1}
works with negative indexes since bash 3.0.
– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
add a comment |
Portable sh
, not bash-specific, and without O(n) loops:
eval x=$$(($#-1))
The -1
yields the penultimate argument; replace it with the position you want relative to the end, or drop it entirely if you want the very last one.
1
One of the very few uses ofeval
I've ever seen that actually seems merited and safe.
– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
add a comment |
In zsh, the most natural thing, I think, would be:
echo "ultimate: ${@[$#]}"
echo "ultimate: ${@[-1]}"
echo "penultimate: ${@[-2]}"
echo "penultimate: ${@[$(($# - 1))]}"
add a comment |
The clasic solutions for POSIX shells (which also work on ksh, zsh or bash) are:
for last do :;done; echo "last=$last"
eval "last=$$#"; echo "last=$last"
For newer shells (ksh93,zsh,bash):
echo "last=${@: -1}"
echo "last=${@:(-1)}"
echo "last=${@:~0}"
echo "last=${@:$#}"
Only for:
bash echo "last=${!#}"
bash echo "last=$BASH_ARGV"
zsh echo "last=${@[-1]}"
zsh echo "last=${@[#]}"
For the penultimate argument:
for arg do penultimate=$ultimate; ultimate=$arg; done; echo "$penultimate"
eval penultimate=$$((#-1))
echo "${@:$((#-1)):1}"
echo "${@: -2:1}"
echo "${@:~1:1}"
add a comment |
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
In addition (to Kusalananda's answer):
#! /bin/bash
echo "(bash/ksh): ${@: -1}"
echo "(bash 3.x+): ${!#}"
echo "(bash 3.x+): $BASH_ARGV"
echo "(bash 3.x+/ksh): ${@:$#}"
echo "(bash 3.x+): ${BASH_ARGV[0]}"
and if you worry about portability:
#!/bin/bash
penultimate=''
ultimate=''
for p in "$@" ; do
penultimate="$ultimate"
ultimate="$p"
done
echo "penultimate=$penultimate"
echo "ultimate=$ultimate"
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.
– JoL
Dec 9 at 5:20
|
show 4 more comments
In addition (to Kusalananda's answer):
#! /bin/bash
echo "(bash/ksh): ${@: -1}"
echo "(bash 3.x+): ${!#}"
echo "(bash 3.x+): $BASH_ARGV"
echo "(bash 3.x+/ksh): ${@:$#}"
echo "(bash 3.x+): ${BASH_ARGV[0]}"
and if you worry about portability:
#!/bin/bash
penultimate=''
ultimate=''
for p in "$@" ; do
penultimate="$ultimate"
ultimate="$p"
done
echo "penultimate=$penultimate"
echo "ultimate=$ultimate"
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.
– JoL
Dec 9 at 5:20
|
show 4 more comments
In addition (to Kusalananda's answer):
#! /bin/bash
echo "(bash/ksh): ${@: -1}"
echo "(bash 3.x+): ${!#}"
echo "(bash 3.x+): $BASH_ARGV"
echo "(bash 3.x+/ksh): ${@:$#}"
echo "(bash 3.x+): ${BASH_ARGV[0]}"
and if you worry about portability:
#!/bin/bash
penultimate=''
ultimate=''
for p in "$@" ; do
penultimate="$ultimate"
ultimate="$p"
done
echo "penultimate=$penultimate"
echo "ultimate=$ultimate"
In addition (to Kusalananda's answer):
#! /bin/bash
echo "(bash/ksh): ${@: -1}"
echo "(bash 3.x+): ${!#}"
echo "(bash 3.x+): $BASH_ARGV"
echo "(bash 3.x+/ksh): ${@:$#}"
echo "(bash 3.x+): ${BASH_ARGV[0]}"
and if you worry about portability:
#!/bin/bash
penultimate=''
ultimate=''
for p in "$@" ; do
penultimate="$ultimate"
ultimate="$p"
done
echo "penultimate=$penultimate"
echo "ultimate=$ultimate"
edited Dec 7 at 20:36
Kusalananda
121k16229372
121k16229372
answered Dec 7 at 20:18
Ljm Dullaart
59017
59017
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.
– JoL
Dec 9 at 5:20
|
show 4 more comments
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.
– JoL
Dec 9 at 5:20
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
to Kusalananda's answer.
– Ljm Dullaart
Dec 7 at 20:24
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
Yes, you're right; I added it in the answer.
– Ljm Dullaart
Dec 7 at 20:26
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
can you please explain about this notation? ${!#}
– YardenST
Dec 7 at 21:22
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
@YardenST From the bash manpage: "If the first character of parameter is an exclamation point (!), and parameter is not a nameref, it introduces a level of variable indirection. 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."
– JoL
Dec 7 at 22:29
1
1
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine
${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.– JoL
Dec 9 at 5:20
@YardenST Yes. Honestly, I think your attempts show what would have been the most obvious, intuitive syntax. I can only imagine
${!x}
was chosen to intentionally limit the syntax to only allow 1 level of indirection instead of multiple by nesting, maybe to ease the implementation. There's probably other subtle reasons, though, like arguments about consistency and how it implies that other thing could also be nested.– JoL
Dec 9 at 5:20
|
show 4 more comments
To get any argument from bash
script, you can use slicing:
#!/bin/bash
# Get 3rd element from the end
from_end1=3
# Get last element
from_end2=1
# Get slice of array [end - from_end1 : end ] of length 1
echo "${@: -$from_end1: 1}"
echo "${@: -$from_end2: 1}"
You can also use this to get Nth element:
# Get 2nd element
from_beginning=2
echo "${@: $from_beginning: 1}"
Remember to check for length; this might return your program's name or an empty string.
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the#!
(and in particular didn't use bash tag).
– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
add a comment |
To get any argument from bash
script, you can use slicing:
#!/bin/bash
# Get 3rd element from the end
from_end1=3
# Get last element
from_end2=1
# Get slice of array [end - from_end1 : end ] of length 1
echo "${@: -$from_end1: 1}"
echo "${@: -$from_end2: 1}"
You can also use this to get Nth element:
# Get 2nd element
from_beginning=2
echo "${@: $from_beginning: 1}"
Remember to check for length; this might return your program's name or an empty string.
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the#!
(and in particular didn't use bash tag).
– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
add a comment |
To get any argument from bash
script, you can use slicing:
#!/bin/bash
# Get 3rd element from the end
from_end1=3
# Get last element
from_end2=1
# Get slice of array [end - from_end1 : end ] of length 1
echo "${@: -$from_end1: 1}"
echo "${@: -$from_end2: 1}"
You can also use this to get Nth element:
# Get 2nd element
from_beginning=2
echo "${@: $from_beginning: 1}"
Remember to check for length; this might return your program's name or an empty string.
To get any argument from bash
script, you can use slicing:
#!/bin/bash
# Get 3rd element from the end
from_end1=3
# Get last element
from_end2=1
# Get slice of array [end - from_end1 : end ] of length 1
echo "${@: -$from_end1: 1}"
echo "${@: -$from_end2: 1}"
You can also use this to get Nth element:
# Get 2nd element
from_beginning=2
echo "${@: $from_beginning: 1}"
Remember to check for length; this might return your program's name or an empty string.
edited Dec 8 at 2:32
answered Dec 7 at 20:06
MatthewRock
3,88821847
3,88821847
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the#!
(and in particular didn't use bash tag).
– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
add a comment |
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the#!
(and in particular didn't use bash tag).
– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
1
1
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the
#!
(and in particular didn't use bash tag).– R..
Dec 8 at 0:10
This is bash-specific, right? You should probably note that since the OP's question did not mention bash except in the
#!
(and in particular didn't use bash tag).– R..
Dec 8 at 0:10
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
@R.. Fair point, information added.
– MatthewRock
Dec 8 at 2:33
add a comment |
In bash
(release 4.3+), you may assign the passed parameters to an array and access the last one by the index -1
:
#!/bin/bash
params=( "$@" )
printf 'The last parameter given to the script is %sn' "${params[-1]}"
foo () {
params=( "$@" )
printf 'The last parameter given to the function is %sn' "${params[-1]}"
}
In general, negative indexes into arrays accesses the elements from the end of the array.
A negative index might work on bash 4.2+ depending on the patch level. But an slice${@: -1:1}
works with negative indexes since bash 3.0.
– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
add a comment |
In bash
(release 4.3+), you may assign the passed parameters to an array and access the last one by the index -1
:
#!/bin/bash
params=( "$@" )
printf 'The last parameter given to the script is %sn' "${params[-1]}"
foo () {
params=( "$@" )
printf 'The last parameter given to the function is %sn' "${params[-1]}"
}
In general, negative indexes into arrays accesses the elements from the end of the array.
A negative index might work on bash 4.2+ depending on the patch level. But an slice${@: -1:1}
works with negative indexes since bash 3.0.
– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
add a comment |
In bash
(release 4.3+), you may assign the passed parameters to an array and access the last one by the index -1
:
#!/bin/bash
params=( "$@" )
printf 'The last parameter given to the script is %sn' "${params[-1]}"
foo () {
params=( "$@" )
printf 'The last parameter given to the function is %sn' "${params[-1]}"
}
In general, negative indexes into arrays accesses the elements from the end of the array.
In bash
(release 4.3+), you may assign the passed parameters to an array and access the last one by the index -1
:
#!/bin/bash
params=( "$@" )
printf 'The last parameter given to the script is %sn' "${params[-1]}"
foo () {
params=( "$@" )
printf 'The last parameter given to the function is %sn' "${params[-1]}"
}
In general, negative indexes into arrays accesses the elements from the end of the array.
edited Dec 7 at 21:05
answered Dec 7 at 20:03
Kusalananda
121k16229372
121k16229372
A negative index might work on bash 4.2+ depending on the patch level. But an slice${@: -1:1}
works with negative indexes since bash 3.0.
– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
add a comment |
A negative index might work on bash 4.2+ depending on the patch level. But an slice${@: -1:1}
works with negative indexes since bash 3.0.
– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
A negative index might work on bash 4.2+ depending on the patch level. But an slice
${@: -1:1}
works with negative indexes since bash 3.0.– Isaac
Dec 8 at 3:12
A negative index might work on bash 4.2+ depending on the patch level. But an slice
${@: -1:1}
works with negative indexes since bash 3.0.– Isaac
Dec 8 at 3:12
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
@Isaac That's good if you are code golfing, but not if you're writing code that others need to understand and maintain.
– Kusalananda
Dec 8 at 8:19
add a comment |
Portable sh
, not bash-specific, and without O(n) loops:
eval x=$$(($#-1))
The -1
yields the penultimate argument; replace it with the position you want relative to the end, or drop it entirely if you want the very last one.
1
One of the very few uses ofeval
I've ever seen that actually seems merited and safe.
– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
add a comment |
Portable sh
, not bash-specific, and without O(n) loops:
eval x=$$(($#-1))
The -1
yields the penultimate argument; replace it with the position you want relative to the end, or drop it entirely if you want the very last one.
1
One of the very few uses ofeval
I've ever seen that actually seems merited and safe.
– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
add a comment |
Portable sh
, not bash-specific, and without O(n) loops:
eval x=$$(($#-1))
The -1
yields the penultimate argument; replace it with the position you want relative to the end, or drop it entirely if you want the very last one.
Portable sh
, not bash-specific, and without O(n) loops:
eval x=$$(($#-1))
The -1
yields the penultimate argument; replace it with the position you want relative to the end, or drop it entirely if you want the very last one.
edited Dec 8 at 1:57
answered Dec 8 at 0:14
R..
1,85611014
1,85611014
1
One of the very few uses ofeval
I've ever seen that actually seems merited and safe.
– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
add a comment |
1
One of the very few uses ofeval
I've ever seen that actually seems merited and safe.
– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
1
1
One of the very few uses of
eval
I've ever seen that actually seems merited and safe.– Wildcard
Dec 8 at 0:35
One of the very few uses of
eval
I've ever seen that actually seems merited and safe.– Wildcard
Dec 8 at 0:35
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
@Isaac: Updated to note that. Thanks.
– R..
Dec 8 at 1:57
add a comment |
In zsh, the most natural thing, I think, would be:
echo "ultimate: ${@[$#]}"
echo "ultimate: ${@[-1]}"
echo "penultimate: ${@[-2]}"
echo "penultimate: ${@[$(($# - 1))]}"
add a comment |
In zsh, the most natural thing, I think, would be:
echo "ultimate: ${@[$#]}"
echo "ultimate: ${@[-1]}"
echo "penultimate: ${@[-2]}"
echo "penultimate: ${@[$(($# - 1))]}"
add a comment |
In zsh, the most natural thing, I think, would be:
echo "ultimate: ${@[$#]}"
echo "ultimate: ${@[-1]}"
echo "penultimate: ${@[-2]}"
echo "penultimate: ${@[$(($# - 1))]}"
In zsh, the most natural thing, I think, would be:
echo "ultimate: ${@[$#]}"
echo "ultimate: ${@[-1]}"
echo "penultimate: ${@[-2]}"
echo "penultimate: ${@[$(($# - 1))]}"
answered Dec 7 at 22:21
JoL
997310
997310
add a comment |
add a comment |
The clasic solutions for POSIX shells (which also work on ksh, zsh or bash) are:
for last do :;done; echo "last=$last"
eval "last=$$#"; echo "last=$last"
For newer shells (ksh93,zsh,bash):
echo "last=${@: -1}"
echo "last=${@:(-1)}"
echo "last=${@:~0}"
echo "last=${@:$#}"
Only for:
bash echo "last=${!#}"
bash echo "last=$BASH_ARGV"
zsh echo "last=${@[-1]}"
zsh echo "last=${@[#]}"
For the penultimate argument:
for arg do penultimate=$ultimate; ultimate=$arg; done; echo "$penultimate"
eval penultimate=$$((#-1))
echo "${@:$((#-1)):1}"
echo "${@: -2:1}"
echo "${@:~1:1}"
add a comment |
The clasic solutions for POSIX shells (which also work on ksh, zsh or bash) are:
for last do :;done; echo "last=$last"
eval "last=$$#"; echo "last=$last"
For newer shells (ksh93,zsh,bash):
echo "last=${@: -1}"
echo "last=${@:(-1)}"
echo "last=${@:~0}"
echo "last=${@:$#}"
Only for:
bash echo "last=${!#}"
bash echo "last=$BASH_ARGV"
zsh echo "last=${@[-1]}"
zsh echo "last=${@[#]}"
For the penultimate argument:
for arg do penultimate=$ultimate; ultimate=$arg; done; echo "$penultimate"
eval penultimate=$$((#-1))
echo "${@:$((#-1)):1}"
echo "${@: -2:1}"
echo "${@:~1:1}"
add a comment |
The clasic solutions for POSIX shells (which also work on ksh, zsh or bash) are:
for last do :;done; echo "last=$last"
eval "last=$$#"; echo "last=$last"
For newer shells (ksh93,zsh,bash):
echo "last=${@: -1}"
echo "last=${@:(-1)}"
echo "last=${@:~0}"
echo "last=${@:$#}"
Only for:
bash echo "last=${!#}"
bash echo "last=$BASH_ARGV"
zsh echo "last=${@[-1]}"
zsh echo "last=${@[#]}"
For the penultimate argument:
for arg do penultimate=$ultimate; ultimate=$arg; done; echo "$penultimate"
eval penultimate=$$((#-1))
echo "${@:$((#-1)):1}"
echo "${@: -2:1}"
echo "${@:~1:1}"
The clasic solutions for POSIX shells (which also work on ksh, zsh or bash) are:
for last do :;done; echo "last=$last"
eval "last=$$#"; echo "last=$last"
For newer shells (ksh93,zsh,bash):
echo "last=${@: -1}"
echo "last=${@:(-1)}"
echo "last=${@:~0}"
echo "last=${@:$#}"
Only for:
bash echo "last=${!#}"
bash echo "last=$BASH_ARGV"
zsh echo "last=${@[-1]}"
zsh echo "last=${@[#]}"
For the penultimate argument:
for arg do penultimate=$ultimate; ultimate=$arg; done; echo "$penultimate"
eval penultimate=$$((#-1))
echo "${@:$((#-1)):1}"
echo "${@: -2:1}"
echo "${@:~1:1}"
answered Dec 8 at 2:14
Isaac
11.1k11648
11.1k11648
add a comment |
add a comment |
Is your question for a solution specific to bash?.
– Isaac
Dec 8 at 1:42